strength of materials and structures lecturer guide

STRENGTH OF MATERIALS AND STRUCTURES

N5Strength of Materials

and StructuresLecturer Guide

Henty Wickens

0843 - Futuremanagers - N5 Strength of materials LG.indd 1 2019/03/05 12:32 PM

© Future Managers 2018

All rights reserved. No part of this book may be reproduced in any form, electronic, mechanical, photocopying or otherwise, without prior permission of the copyright owner.

ISBN 978-0-6391-0078-4

First edition 2018

To copy any part of this publication, you may contact DALRO for information and copyright clearance. Any unauthorised copying could lead to civil liability and/or criminal sanctions.

Telephone: 086 12 DALRO (from within South Africa); +27 (0)11 712-8000Telefax: +27 (0)11 403-9094Postal address: P O Box 31627, Braamfontein, 2017, South Africawww.dalro.co.za

Every effort has been made to trace the copyright holders. In the event of unintentional omissions or errors, any information that would enable the publisher to make the proper arrangements would be appreciated.

Published byFuture Managers (Pty) LtdPO Box 13194, Mowbray, 7705Tel (021) 462 3572Fax (021) 462 3681E-mail: [email protected]: www.futuremanagers.com


Contents

Module 1 Stress and strain and testing of materials 1

Module 2 Strain energy 9

Module 3 Compound bars and temperature-induced stresses 24

Module 4 Thin cylinders and riveted joints 44

Module 5 Loading of beams 59

Module 6 Simple bending of beams 80

Module 7 Columns and struts 105

Module 8 Shafts 113

Module 9 Forces 127

Module 10 Structural frameworks 135

Module 11 Concrete and foundations 148


MODULE

Pre-knowledge for this module

Knowledge assumed to be in place before starting with this module:• Good knowledge of Engineering Science N4• Basic knowledge of calculating stress• Drawing of straight-line graphs• Know the conditions for equilibrium• Newton’s third rule

Learning outcomes

When students have completed this learning module, they should be able to:• Identify the diff erent stresses• Calculate direct and shear stress• Know what strain is and calculate it• Know what modulus of elasticity (Young’s modulus) for materials is and to

calculate it• Draw a stress-strain graph and use it to obtain information about a material• Describe diff erent tests on materials for diff erent strengths

Guidelines for students

• Make sure they understand the basic concept of the theory about stress and strain• Always work from basic principles and avoid formulas not directly connected to

fi rst principles• Always make sketches of the member or whatever is described in a problem• Read through the module and underline important facts• Make sure they know the answers to the self-check, before doing the exercises

Stress and strain and testing of materials1


2 N5 Strength of materials and structures Lecturer Guide

Exercise 1.1 SB page 19

1. It is very important to sketch the problem.

1.1 Working stress = maximum stress __________ FOS

= 120 ___ 3

= 40 MPa

1.2 Safe load = σ.A

= 40 × π _ 4 (0,0094562)

= 2,809 kN

2. Effective area = (50 – 12) × 15

= 570 mm2

∴ σ = F __ A = 60k _______ 570 × 1 0 −6

= 105,26 MPa

3.

ø60 ø20

50 100

20 kN 20 kN

(1)

(2)

Basic equation: E = σ __ ε = F __ A __ x _ L = FL __ Ax = σ L __ x

3.1 Maximum stress smallest area = σ = F __ A = 20k ____ π _ 4 90 2 2

= 63,662 MPa

3.2 x T = x 1 + x 2

= F L 1 ___ A 1 E + F L 2 ___ A 2 E

= 20k ____ 200G [ 0,05 ____ π _ 4 0,0 6 2

+ 0,1 ____ π _ 4 0,0 2 2

] = 0,0336 mm

3.3 ε T = ε 1 + ε 2 ∴ ε 1 = σ 1 __ E = F ___ A 1 E = 20k ____

π _ 4 0,0 6 2 × 200G = 3,537 × 10–5

ε2 = σ 2 __ E = F ___ A 2 E = 20k ____

π _ 4 0,0 2 2 × 200G = 3,183 × 10–4

∴ ε T = ε 1 + ε 2 = 3,537 × 10–5 + 3,183 × 10–4 = 3,537 × 10−4

15

5060 kN

151250


3Module 1 • Stress and strain and testing of materials

4. E = 200 GPa

ø60 ø80ø20F F

(1)

(2)

(3)

σmax = 100 MPa

4.1 Maximum stress smallest area ∴ F = σ max A 2

= 100 M × π _ 4 0,0 2 2 = 31,416 kN This is the max load which can be applied due to the weakest section.

4.2 σ1 = F __ A 1 = 31,416k _____

π _ 4 0,0 6 2 = 11,11 MPa

σ3 = F __ A 3 = 31,416k _____

π _ 4 0,0 8 2 = 6,25 MPa

5. σAL = 155 MPa

Eal = 69 GPa

σs = 465 MPa

Es = 207 GPa

σc = 247 MPa

EC = 110 GPa

5.1 Aluminium maximum allowable on working stress = σ max ___ FOS = 155 ___ 3 = 51,67 MPa

Steel maximum allowable on working stress = σ max ___ FOS = 465 ___ 3 = 155 MPa

Copper maximum allowable on working stress = σ max ___ FOS = 247 ___ 3 = 82,33 MPa

5.2 FAL = σAAA = 51,67 M × π _ 4 (0,052 – 0,042) = 36,523 kN

FS = σSAS = 155 M × π _ 4 (0,032 – 0,022) = 60,868 kN

FC = σcAc = 82,33 M × π _ 4 (0,092) = 523,76 kN

∴ Max force = 36,523 kN

The smallest load is the max because the other loads will damage the aluminium. The strain in aluminium will be higher as the max allowable for aluminium is 51,67 MPa.

5.3 σa = 51,67 MPa acting force

σs = F __ A = 36,523k __________ π _ 4 (0,0 3 2 − 0,0 2 2 )

= 93,01 MPa

σc = F __ A = 36,523k _____ π _ 4 0,0 9 2

= 5,74 MPa

This will be the stresses in the other materials when the max load is applied.

ø90ø40

80 100 60

OD = 50

L

OD = 30

IO = 20F F

AL

SteelCopperAL



5.4 Total extension: x T = x a + x s + x c

x s = σ s L s ___ E s = 93,01M × 0,1 _________ 207G = 0,0449 mm

x c = σ c L c ___ E c = 5,74M × 0,06 _________ 110G = 3,131 × 10–3 mm

From : x a = x T – ( x s + x c )

= 0,0981 – (0,0449 + 3,131 × 10–3)

= 0,050069

But x a = x hollow + x solid due to the two different areas.

= F L H ____ A H E a + F L s ___ A s E a

∴ 0,050069 × 10–3 = 36 523 _____ 69G [ L H __________

π _ 4 (0,0 5 2 − 0,0 4 2 ) + (0,08 − L H )

_______ π _ 4 0,0 5 2

] (∴ LS = (80 – LH))

∴ 94,5914 = 1 414,711 LH + 509,296(0,08 – LH)

53,848 = 905,415 LH

LH = 59,473 mm

6.

Force

Extension

6.1

6.2yield point

max load point

point of fracturelimit of proportionality

6.3

6.4

7. 70

ø15

7.1 σ = F __ A = 7k _____

π _ 4 0,01 5 2 = 39,612 MPa (see 3.1 on page 14 in the Student Book)

7.2 E = σL __ x = 39,612 M × 0,07 __________ 35 × 1 0 −6

= 79,224 GPa (see 3.2 on page 14 in the Student Book)

8. 8.1 σ = F __ A = 28k _____ π _ 4 0,01 2 2

= 247,57 MPa

8.2 E = σL __ x = 247,57 M × 0,06 __________ 0,19 × 1 0 −3

= 78,18 GPa

8.3 σmax = F max ____ Area = 52k _____ π _ 4 0,01 2 2

= 459,78 MPa

8.4 Actual stress = 40k _____ π _ 4 0,00 9 2

= 628,76 MPa

8.5 % elongation = x _ L × 100 ___ 1 = 8 __ 60 × 100 ___ 1 = 13,33%

8.6 % reduction in area = ( 1 2 2 − 9 2 ______ 1 2 2

) × 100 ___ 1 = 43,75%

Basic equation:E = σ _ ε = F/A ___ x/L = FL __ Ax = σL __ x (see pages 6 and 9 in the Student Book)

Basic equation:E = σ _ ε = F/A ___ x/L = FL __ Ax = σL __ x

x = 35 × 10–6 m F = 7 kN



9.

Change units of extension to metres.

F(kN) 3,5 8,5 13,5 18,8 23,5 28,5 31 33,5 35 36

x × 10–3 9 22 35 48 61 74 87 93 104 157

9.1 E = FL __ Ax = 28,5k × 0,060 ____________ π _ 4 0,01 2 2 × 74 × 1 0 −6

= 204,32 GPa

9.2 σ = F __ A = 27k _____ π _ 4 0,01 2 2

= 238,73 MPa

9.3 σ = F __ A = 46k _____ π _ 4 0,01 2 2

= 406,73 MPa

9.4 % x = x _ L × 100 ___ 1 = 9 __ 60 × 100 ___ 1 = 15%

9.5 % A = 1 2 2 − 8 2 ______ 1 2 2

× 100 ___ 1 = 55,56%

10. 10.1

Forc

e kN

x × 10–3

0

5

10

20 40 60 80 100 120 140 160

15

20

25

30

35

40

–A27

70

σ M

Pa

Strain × 10–4

0

40

20

10 20 30 40 50 60

60

80

100

120

140

160

180



10.2 E = FL __ Ax = 42 × 1 0 3 ______ 6 × 1 0 −4

= 70 GPa

Study the graph and select any value on the straight line of the graph. Take values from the table given which is on the straight line (see page 8 in the Student Book).

10.3 σ = F __ A = 21k _______ 100 × 1 0 −6

= 210 MPa (read pages 7–8 in the Student Book)

11.

F 25 55 80 95 104 109 114 117 118 120x mm 0,08 0,176 0,255 0,303 0,332 0,35 0,41 0,52 0,88 1,75MPa σ 62,87 138,33 201,2 238,9 261,56 274,14 286,7 294,3 296,8 301,8

ε = x _ L 3,2 × 10–4

7,04 × 10–4

1,02 × 10–3

1,212 × 10–3

1,328 × 10–3

1,4 × 10–3

1,64 × 10–3

2,08 × 10–3

3,52 × 10–3

7 × 10–3

× 10–4 3,2 7,04 10,2 12,12 13,28 14 16,4 20,8 35,2 70

11.1 E = σ __ ε = 201,2 M _______ 1,02 × 1 0 −3

= ± 197,3 GPa 80 kN force on straight line ∴ use 201,2 MPa stress. If the stress for the

55 kN or 95 kN was used, the answer would be close to 197,3 GPa. 11.2 σ = F __ A = ± 261,56 MPa Read from graph where straight line stops. 11.3 ± 286,7 MPa Read from graph just after straight line stops.

12. Safe stress = 465 ___ 5 = 93 MPa See page 9 in Student Book.

OD = 100

ID = 220

Force in wall = σ.A = 93M × π _ 4 ( 0,2 2 2 − 0, 1 2 ) = 2 804,814 kN

σ M

Pa

Strain × 10–4

0

50

10 20 30 40 50 60 70

250

200

100

150

300



∴ Internal pressure = F __ A = 2 804,814k _______ π _ 4 01 1 2

= 357,12 MPa

• Calculate the area and store in memory of calculator. ∴ σ = F __ A Divide each force by the area to obtain stress for each force. • To calculate the strain, use calculator and divide the extension by the gauge

length for each extention (both can be in mm not to change to m – keep the units the same, m or mm).


1. σ = 80 MPa

ε = 1,054 × 10–3

L = 120 mm

1.1 E = σ __ ε = 80 M ________ 1,054 × 1 0 −3

= 75,9 GPa 1.2 ε = x _ L ∴ x = Lε = 120 × 1,054 × 10–3 = 0,1265 mm 1.3 σ = F __ A ∴ A = F _ σ = 24k ___ 80M = 3 × 10–4 m2

A = wt ∴ 3 × 10–4 = 0,05 × t ∴ t = 6 mm

2.

σmax = 100 MPa

xT = 0,0718 mm

E = 200 GPa 2.1 σ = F __ A ∴ A = F _ σ = 60k ____ 100M = 6 × 10–4 = π _ 4 D2

D = 27,64 mm

2.2 x T = x 1 + x 2 x 2 = x T – x 1 = 0,0718 – ( σL __ E ) = 0,0718 – ( 100M × 0,08 ________ 200G ) = 0,0718 – 0,04 = 0,0318 mm x 2 = FL ___ AE = 60k × 3L _________

π _ 4 0,0 6 2 × 200G = 0,0318 × 10–3

∴ L = 100 mm ∴ 2L = 200 mm

t

120

W

ø60 ø6060 kN 60 kN

(2)

L 8 2L

(1)

(2)



3. 3.1 Allowable stress = 84 __ 4 = 21 MPa See page 8 in the Student Book. Point C is the point that the particles in

the material separate before setting and the safe stress must be less. ∴ Use F.O.S.

3.2 F = σ.A = 21M × π _ 4 0,022 = 6,597 kN 3.3 x = σL __ E = 21M × 0,25 _______ 200G = 0,026 mm


1. F = σ.A = 300 M × πD × t (use the shear area πDt) = 300 M × π × 0,01 × 0,02 = 188,496 kN

2. 2.1 OD = 110

ID = 90 Figure 1

F = σ.A = 60 M × π × 0,09 × 0,012 (shear area in collar = πdt) = 203,575 kN

2.2 σ = F __ A = 203,575 _____ π _ 4 0,0 9 2

= 32 MPa (shaft diameter)

2.3 σ = F __ A = 203,575k __________ π _ 4 (0,1 1 2 − 0,0 9 2 )

(contact area of collar – see Figure 1)

= 64,8 MPa

3. 3.1 Area: σ = F __ A ∴ F _ σ = 100k ____ 100M = 1 × 10–3 = π _ 4 D2

D = 35,68 mm

3.2 τ = F __ A = 100k _______ 2 × π _ 4 0,0 2 2

= 159,155 MPa (pin shear in two areas ∴ 2 × area of pin [cross-sectional area])

3.3

Figure 2

Two legs of fork ∴ Area = 2(80 – 20)15 = 1 800 mm2

σt = F __ A = 100k ________ 1 800 × 1 0 −6

= 55,56 MPa

15

80

8020

Remember answers for graph questions will always be ± values (questions 9–11).

In practice more tests are done to get the correct answers. We only do one calculation.


MODULE

2 Strain energy


Knowledge assumed to be in place before starting with this learning module:• Knowledge of the formulae used in Module 1 of this book• How to calculate work done• What potential energy is

Learning outcomes

When students have completed this learning module, they should know:• What a gradually applied load is• What a shock load is• What a suddenly applied load is• What strain energy is• What resilience is• How to calculate stresses in single and compound members due to the three ways

in which loads can be applied


• Make use of sketches when you do calculations and always work from fi rst principles

• Working from fi rst principles will give students more knowledge of the subject and they will also understand the theory of the subject better

• Make a good study of the work content and underline all important facts• Before starting with the exercise, make sure they understand the theory of the

module and know the answers of the self-check exercises




1. U = 1 _ 2 Fx = 1 _ 2 F ( FL ___ AG ) = F 2 L ___ 2AG

= (60k ) 2 × 0,5 ___________ 2 × π _ 4 0,04 × 206G

= 3,477 J

2. 2.1

UT = U1 + U2 = 1 _ 2 F x 1 + 1 _ 2 F x 2

= 1 _ 2 F( x 1 + x 2 )

= 1 _ 2 F [ F L 1 ___ A 1 E + F L 2 ___ A 2 E

] = FL __ 2E [ L 1 __ A 1

+ L 2 __ A 2 ]

= ( 100k ) 2 ______ 2 × 206G [ 0,6 ____ 0,0 5 2

+ 1 ____ 0,0 8 2

] = 9,618 J

2.2 xT = x1 + x2

= F L 1 ___ A 1 E + F L 2 ___ A 2 E

= 100k ____ 206G [ 0,6 ____ 0,0 5 2

+ 1 ____ 0,0 8 2

] = 0,192 mm

2.3 εT = ε1 + ε2

= x 1 __ L 1 + x 2 __ L 2

= F L 1 ____ A 1 E L 1 + F L 2 ____ A 2 E L 2

= 100k ____ 206G [ 1 ____ 0,0 5 2

+ 1 ____ 0,0 8 2

] = 2,7 × 10–4

2.4 σmax = F __ A 1 = 100k ____

0,0 5 2 = 40 MPa (max stress in smallest area)

3. ID = 50

OD = 80

E = 200 GPa

U = 30 J

801 000

600 80

50

50

2

1


11Module 2 • Strain energy

3.1 U = 1 _ 2 Fx

= 1 _ 2 F ( FL ___ AE ) ∴ 30 = F 2 L ___ 2AE = F 2 × 0,7 ________________

2 × π _ 4 (0,0 8 2 − 0,0 5 2 )200G

∴ F2 = 3,676 × 1 0 10 ________ 0,7

F = √ __________

5,251 × 1 0 10 = 229,149 kN

3.2 σ = F __ A = 229,149k __________ π _ 4 (0,0 8 2 − 0,0 5 2 )

= 74,811 MPa

3.3 x = σL __ E = 74,811 M × 0,7 _________ 200G = 0,262 mm

3.4 ε = σ _ E = 74,811M ______ 200G = 3,741 × 10–4

4.

E = 210 GPa

4.1 εT = ε1 + ε2 + ε3

= σ 1 __ E + σ 2 __ E + σ 3 __ E

= 1 _ E [ F __ A 1 + F __ A 2

+ F __ A 3 ]

= F _ E [ 1 __ A 1 + 1 __ A 2

+ 1 __ A 3 ]

∴ 7,548 ×10–4 = 200k ____ 210G [ 1 __ A 1 + 1 ____

π _ 4 0, 1 2 + 1 ____

0,0 5 2 ]

∴ 1 __ A 1 = 265,216

A1 = 3,771 × 10–3

∴ π _ 4 (0,0 8 2 − d 2 ) = 3,771 × 10–3

d = 40 mm 4.2 xT = x1 + x2 + x3

= F L 1 ___ A 1 E + F L 2 ___ A 2 E

+ F L 3 ___ A 3 E

= F _ E [ L 1 __ A 1 + L 2 __ A 2

+ L 3 __ A 3 ]

= 200k ____ 210G [ 0,6 __________ π _ 4 (0,0 8 2 − 0,0 4 2 )

+ 0,2 ____ π _ 4 0, 1 2

+ 0,5 ____ 0,0 5 2

] = 0,366 mm 4.3 Check for smallest area: A1 = π _ 4 (0,0 8 2 − 0,0 4 2 ) = 3,77 × 10–3 m2

A3 = 0,052 = 2,5 ×10–3 m2

600200

50 × 50OD = 80 ø0 = 200

ID = ?

500

200 kN 200 kN1

2 3



σmax = F ___ A min

= 200k _______ 2,5 × 1 0 −3

= 80 MPa

We must check for the smallest area to calculate because the hollow area can be smaller, depending on diameters.

4.4 UT = 1 _ 2 F x T = 1 _ 2 × 200k × 0,366 × 10–3

= 36,6 J

5.

ECi = 80 GPa EC = 100 GPa

5.1 UT = U1 + U2

= 1 _ 2 F x 1 + 1 _ 2 F x 2 = 1 _ 2 F [ x 1 + x 2 ]

∴ = F _ 2 [ F L 1 ___ A 1 E 1 + F L 2 ___ A 2 E 2

] = F 2 __ 2 [ L 1 ___ A 1 E 1

+ L 2 ___ A 2 E 2 ]

= (250k ) 2 _____ 2 [ 2,5 ________ π _ 4 0, 1 2 × 80G

+ 2,5 ________ π _ 4 0, 1 2 × 100G

] = 223,812 J

5.2 UT = 1 _ 2 F x T

= 1 _ 2 F [ FL ___ AE ] ∴ 223,812 = F 2 L ___ 2AE = (250k ) 2 × 5 _______ 2A × 210G

∴ A = 3,324 × 10–3

∴ π _ 4 d 2 = 3,324 × 10–3

d = 65,06 mm


E = σ __ ε = F/A ___ x/L = FL __ Ax = σL __ x (basic equation still applies)

d = 80

E = 206 GPa

W = 4 kN

h = 200 mm

ø100 Ci ø100 C

2,5 m

250 250

2,5 m

d = ?

Students must show more calculation steps.

L = 4 m



1. 1.1 PE = U

W(h + xT) = 1 _ 2 F x T ( x T = FL ___ AE ) ∴ 4k ( 0,2 + F × 4 _________

π _ 4 0,0 8 2 × 200G ) = F 2 × 4 ___________

2 × π _ 4 0,0 8 2 × 206G

∴ 4k(0,2 + 3,863 × 10–9 F) = 1,931 × 10–9 F2

∴ 800 + 1,545 × 10–5 F = 1,931 × 10–9 F2

÷ 1,931 × 10–9: ∴ F2 – 8 002,072 F – 4,143 × 1011 = 0

F = −(−8 002,072 ± √ ______________________________

(−8 002,072 ) 2 − 4(1)(−4,143 × 1 0 11 ) _________________________________ 2 × 1

F = 8 002,072 ± 1 287 337 ______________ 2

= 647,67 kN

∴ σ = F __ A = 647,67k _____ π _ 4 0,0 8 2

= 128,85 MPa

1.2 PE = U W(h + x) = 1 _ 2 Fx ∴ 40k(0,02 + 3,863 × 10–9 F) = 1,931 × 10–9 F2

∴ 800 + 1,545 × 10–4 F = 1,931 × 10–9 F2

÷ 1,931 × 10–9: ∴ F2 – 80 010,357 F – 4,143 × 1011

F = 80 010,357 ± √ _______

b 2 − 4ac ______________ 2(1)

F = 80 010,357 ± 1 289 796,1 _______________ 2 = 684,9 kN

σ = F __ A = 684,9 ____ π _ 4 0,0 8 2

= 136,2 MPa

2. E = 210 GPa

W = 200 × 9,81 = 1 962 N

2.1 PE = U

W(h + xT) = 1 _ 2 F x T

xT = x1 + x2

= F L 1 ___ A 1 E + F L 2 ___ A 2 E

= F ____ 210G [ 1 ____ 0,0 4 2

+ O 1 8 ____ π _ 4 0,0 2 2

] xT = 1,51 × 10–8 F

1 40 × 46

2L = 0,8 m d = 20 h = 300

1 m



Substitute in :

∴ 1 962(0,3 + 1,51 × 10–8 F) = 1 _ 2 F × 1,51 × 1 0 −8 F

∴ 588,6 + 2,963 × 10–5 F = 7,55 × 10–9 F2

÷ 7,55 × 10–9: ∴ F2 – 3 924,503 F – 7,796 × 10–10 = 0

F = − (−3 924,503) ± √ _______

b 2 − 4ac _________________ 2(1)

F = 3 924,503 ± 558 441,1 ______________ 2

= 281,183 kN

A1 = 0,062 = 1,6 × 10–3 m

A2 = π _ 4 0,0 2 2 = 3,162 × 10–4

∴ σmax = 281,183k ________ 3,142 × 1 0 −4

= 894,92 MPa

2.2 From xT = 1,51 × 10–8 F

= 1,51 × 10–8 × 281,183 k

= 4,246 mm

2.3 UT = 1 _ 2 F x T = 1 _ 2 × 281,183 × 4,246 × 10–3

= 596,932 J

2.4 σ1 = W __ A 1 = 1 962 ____

0,0 4 2 = 1,226 MPa

σ2 = W __ A 2 = 1 962 ________

3,1 4 2 × 1 0 −4 = 6,244 MPa

2.5 U1 = 1 _ 2 F x 1 = F 2 L ___ 2AE = 1 96 2 2 × 1 ___________ 2 × 0,0 4 2 × 210G

= 5,728 × 10–3 J

U2 = F 2 L ___ 2AE = 1 96 2 2 × 0,8 ___________ 2 × π _ 4 0,0 2 2 × 210G

= 0,0233 J

UT = U1 + U2 = 0,0291 J Final strain is the strain under load gradually applied.

3. A = 700 mm2

xT = 2,5 mm

E = 206 GPa

PE = U

3.1 E = σL __ x

∴ 206G = σ3,5 _______ 2,5 × 1 0 −3

σ = 147,43 MPa

3.2 PE = U ∴ W(h + x T ) = 1 _ 2 F x T

L = 3,5 mh = 15 mm



∴ W(0,015 + 2,5 × 10–3) = 1 _ 2 σ.A × 2,5 × 10–3

0,0175W = 1 _ 2 × 147,143M × 700 × 10–6 × 2,5 × 10–3

W = 7,357 kN

3.3 Mass = weight _____ g = 7,357k _____ 9,81

= 749,96 kg

4. 4.1 F = 2W

∴ F = 2 × 50 = 100k

∴ σ = F __ A = 100k _____ π _ 4 0,02 6 2

= 188,349 MPa

4.2 xT = σL __ E = 188,349M × 3 _________ 200G = 2,825 mm

5.

5.1 σA = 50k ____ π _ 4 0,0 5 2

= 25,46 MPa

σB = 50k ____ π _ 4 0,0 3 2

= 70,74 MPa

σC = F __ A = 50k ____ π _ 4 0,0 7 2

= 12,99 MPa

5.2 xA = σ A L A ____ E = 25,46M × 0,4 ________ 200G = 0,051 mm

xB = σ B L B ___ E = 71,74M × 0,03 _________ 200G = 0,106 mm

xC = σ C L C ___ E = 12,99M × 1 0 6 × 0,9 ____________ 200G = 0,058 mm

5.3 UA = 1 _ 2 F x A = 1 _ 2 × 50k × 0,051 × 1 0 −3 = 1,275 J

UB = 1 _ 2 F x B = 1 _ 2 × 50k × 0,106 × 1 0 −3 = 2,65 J

UC = 1 _ 2 F x C = 1 _ 2 × 50k × 0,058 × 1 0 −3 = 1,45 J

5.4 Strain

εA = σ A __ E = 25,46M _____ 200G = 1,273 × 10–4

εB = σ B __ E = x B

__ L B = 0,106 × 1 0 −3 ________ 0,3 = 3,53 × 10–4

εC = x C __ L C = 0,058 × 1 0 −3 ________ 0,9 = 6,44 × 10–5

5.5 UT = U1 + U2 + U3 = 5,375 J (add the values in 5.3)

OR

UT = 1 _ 2 F x T

= 1 _ 2 × 50k (0,051 + 0,106 + 0,058)10–3

= 5,375 J

ø50 ø30 ø70

400 300

50 kN 50 kN

900

A B C



6. 6.1 U = 1 _ 2 Fx = 1 _ 2 F FL ___ AE

∴ 3 = F 2 × 0,6 ___________ 2 × π _ 4 0,0 5 2 × 200G

F = 62,666 kN

6.2 σ = F __ A = 62,666k _____ π _ 4 0,0 5 2

= 31,92 MPa

7.

30 x 30 10 x 10

600 400

F F

7.1 xT = x1 + x2

= FL1 ___ A 1 E + F L 2 ___ A 2 E

∴ 0,15 × 10–3 = F ____ 209G [ 0,6 ____ 0,0 3 2

+ 0,4 ____ 0,0 1 2

] 0,15 × 10–3 = 2,2329 × 10–8 F F = 6,718 kN

7.2 σmax = F ___ A min = 6,718k _____ 0,0 1 2

(max stress in smallest area)

= 67,18 MPa

8.

L = 1,5 mø = 30

W = 4 kgx = 0,08 mm

E = FL __ Ax

= 4 × 9,81 × 1,5 _____________ π _ 4 0,0 3 2 × 0,08 × 1 0 −3

= 1,041 GPa

∴ PE = U

∴ W(h + xT) = 1 _ 2 F x T

∴ 4 × 9,81 ( 0,003 + FL ___ AE ) = 1 _ 2 F ( FL ___ AE ) ∴ 4 × 9,81 ( 0,003 + F × 1,5 ___________

π _ 4 0,0 3 2 × 1,041G ) = F 2 × 1,5 _____ 2AE

∴ 4 × 9,81(0,003 + 2,0385 × 10–6 F) = 1,019 × 10–6 F2

∴ 0,11772 + 7,999 × 10–5 F = 1,019 × 10–6 F2

÷ 1,019 × 10–6: ∴ F2 – 78,5 F – 115 525,02 = 0

F = 78,5 ± √ _______

b 2 − 4ac ___________ 2(1)

= 78,5 ± 684,297 _________ 2

= 381,4 N

∴ σ = F __ A = 381,4 ____ π _ 4 0,0 3 2

= 539,569 kPa

9. PE = U

∴ W(h + x) = 1 _ 2 Fx

25k ( h + σL __ E ) = 1 _ 2 σ.A × σL __ E



25k ( h + 102M × 0,8 _______ 205G ) = (102M)2 ______ 2 × π _ 4 0,1 2 2 × 0,8 ____ 205G

∴ 25k (h + 3,98 × 10–4) = 229,593

h = 8,786 mm

10. 10.1 U = 1 _ 2 F x T = 1 _ 2 σ.A. σL __ E = σ 2 AL ____ 2E

8 = σ 2 × 0,0 3 2 × 2 _________ 2 × 198G

σ2 = 1,76 × 1015

σ = 41,95 MPa

10.2 10.2.1 U = 1 _ 2 Fx ∴ 8 = F 2 L ___ 2AE = F 2 × 2 ___________ 2 × 0,0 3 2 × 198G

∴ F = 37,757 kW

10.2.2 U = Wx = W(2W)L ______ EA = 2WL ____ EA (F = 2W for suddenly applied)

8 = 2 × W 2 × 2 ________ 0,0 3 2 × 198G

W = 18,879 kN

10.2.3 W(h + x) = U

W(0,02 + xT) = 8

But xT ∴ 8 = 1 _ 2 F x T

∴ 8 = 1 _ 2 × 37,757 k x T

∴ xT = 4,238 × 10–4

Substitute in ∴ W(0,02 × 4,238 × 10–4) = 8

W = 391,7 N11. 11.1 U = 1 _ 2 Fx

= 1 _ 2 × 2W × 2WL ____ AE (F = 2W)

∴ 8 = 2 W 2 L ____ AE

= 2 W 2 × 1,5 ____________ 320 × 1 0 −6 × 200G

∴ W = 13,064 kN

∴ Force in rod = F = 2W = 26,128 kN

11.2 U = 1 _ 2 Fx (F = W)

8 = F 2 L ___ 2AE

= F 2 × 1,5 __________ 2 × 320 × 1 0 −6 E

(E = 200G)

∴ F = 26,128 kN

IMPORTANT For U to be constant the force F must also be constant.∴ From 10.2.1 F = 37,757 kxT



11.3 PE = U

∴ W(h + x) = 1 _ 2 Fx

W(h + x) = F 2 L ___ 2AE = 8

∴ F 2 × 1,5 _____ 2AE = 8

F = 26,128 kN

11.4 11.4.1 F = 2W x = FL ___ AE

∴ = 26,128k × 1,5 _________ AE = 0,612 mm

11.4.2 x = FL ___ AE = 26,128k × 1,5 _________ AE = 0,612 mm

11.4.3 x = W(h + x) = 1 _ 2 Fx

W(h + x) = 8 = 1 _ 2 Fx

= 1 _ 2 × 26,128k

∴ x = 0,612 mm

12. xT = 1 mm

W = 300 × 9,81

E = 210 GPa

PE = U W(h + xT) = 1 _ 2 F x T

xT = x1 + x2

= F _ E [ L 1 __ A 1 + L 2 __ A 2

] 1 × 10–3 = F ____ 210G [ 0,8 _______

350 × 1 0 −6 + 0,6 _______

400 × 1 0 −6 ]

∴ F = 55,472 kN

Substitute in : 300 × 9,87(h + 1 × 10–3) = 1 _ 2 55,472k × 1 × 10–3

h = 8,424 mm

13. 13.1 U = 1 _ 2 Fx 4 = 1 _ 2 F × FL ___ AE = F 2 × 2 _______________

2 × 2 500 × 1 0 −6 × 200G

∴ F = 44,721 kN

13.2.1 F = 2W = 2 × 44,721 = 89,442 kN

IMPORTANT Students must show all values.

IMPORTANT Students must show all values. This is only showing the metod of calculation.

L = 600 h = ?400 mm2

L = 800 350 mm2



13.2 13.2.2 PE = ∪ W(h + xT) = 1 _ 2 Fx

44,721 ( 5 × 1 0 −3 + F × 2 _________ 2 500 × 1 0 −6 E

) = F L 2 ___ 2AE ∴ 223,605 + 1,789 × 10–4 F = 2 × 10–9 F2 ÷ 2 × 10–9: ∴ F2 – 89 450F – 1,118 × 1011 = 0

∴ F = 89 450 ± √ _______

b 2 − 4ac ____________ 2(1)

= 89 450 ± 674 693,49 _____________ 2 = 382,072 kN

14. 14.1 F Fø50 ø30

60100

U = 10 J

UT = U1 + U2

∴ 10 = 1 _ 2 F( x 1 + x 2 ) x = FL ___ AE

∴ 10 = F 2 __ 2E [ 0,1 ____ π _ 4 0,0 5 2

+ 0,06 ____ π _ 4 0,0 3 2

] ∴ 10 × 2 × 200G = F2 [50,93 + 84,883]

F = 171,617 kN

14.2 σ1 = 171,617k ______ π _ 4 0,0 5 2

= 87,4 MPa

σ2 = 171,617k ______ π _ 4 0,0 3 2

= 242,79 MPa

14.3 x1 = σ 1 L 1 ___ E = 87,4M × 0,1 ________ 200G = 0,044 mm

x2 = σ 2 L 2 ___ E = 242,79M × 0,06 __________ 200G = 0,0728 mm

15.

ø25

ø12,5

600 = L

600 = L

If stress = 220 MPa when load falls h metres, stress is maximum in smallest area.

∴ Force in rod will be:

F = σ.A = 220 M × π _ 4 0,01252

= 26,998 kN

∴ xT = x1 + x2 = F L 1 ___ A 1 E + F L 2 ___ A 2 E

∴ xT = 26 998 × 0,6 __________ π _ 4 0,02 5 2 × 207G

+ 26 998 × 0,6 ________ π _ 4 0,012 5 2 × E

(E = 207G)

= 7,971 × 10–4 m



But: PE = U

W(h + xT) = 1 _ 2 F x T

∴ 12 × 9,81(h + 7,971 × 10–4) = 1 _ 2 × 26 998 × 7,971 × 10–4

h = 90,6 mm

16. 16.1

ø40 ø30 ø20

10060 80

E = 210 GPa

xT = x1 + x2 + x3 (x = FL ___ AE )

∴ 0,177 × 10–3 = F _ E [ 0,1 ____ π _ 4 0,0 4 2

+ 0,06 ____ π _ 4 0,0 3 2

+ 0,08 ____ π _ 4 0,0 2 2

] ∴ F = 88,68 kN

16.2 D = 1,2d

xT = FL ___ AE

∴ A = 70k × 924 _____________ 0,177 × 1 0 −3 × 210G

= 451,9774 mm2

A = π _ 4 ( D 2 − d 2 ) = 451,9774

D2 – d 2 = 575,4755

Substitute in : (1,2d)2 – d 2 = 575,4755

∴ 1,44d 2 – d

2 = 575,4755

d = 36,165 mm

D = 43,4 mm

17. 80 kN 80 kN

2 000 mm2

1 m 2 m

1 000 mm2

E = 200 GPa

UT = U1 + U2

= F L 1 ___ A 1 E + F L 2 ___ A 2 E

= 80k ____ 200G [ 1 ________ 2 000 × 1 0 −6

+ 2 ________ 1 000 × 1 0 −6

] = 40 J

Uniform bar

Volume of step bar = V1 + V2 = A1L1 + A2L2

= (2 000 × 1 000) + (1 000 × 2 000)

= 4 × 106 mm3

Volume of step bar = Volume single bar

IMPORTANT Students must show all steps.



∴ 4 × 106 = AS × 3 000

AS = 1 333,33 m2

∴ U single = 1 _ 2 F x T

= F 2 L ___ 2AE

= (80k ) 2 × 3 _________________ 2 × 1 333,33 × 1 0 −6 × 200G

= 36 J

∴ 2 sections more energy.

18.

18.1 Max stress ⇒ small area ∴ F = σ.A = 200 m × 0,012

= 20 kN 18.2 xT = x1 + x2

But U = 1 _ 2 F x T

∴ xT = 2U __ F

= 2 × 1,3 _____ 20k

= 0,13 mm 18.3 Stress in round bar = F __ A 2

= 20k _____ π _ 4 0,02 5 2

= 40,744 MPa

xT = x1 + x2

= σ 1 L 1 ___ E + σ 2 L 2 ___ E

0,13 × 10–3 = 200M L 1 ______ 200G + 40,744M (0,25 − L 1 ) _____________ 200G

∴ × 200G: ∴ 26 × 106 = 200M L1 + 40,744M(0,25 – L1)

÷ 106: ∴ 26 = 200L1 + 10,186 – 40,744L1

15,814 = 159,256L1

L1 = 99,3 mm

10 × 10

ø25

250 – L 1

L 1

250



19. 19.1

S C S

d = 20

OD = 40ID = 35 Steel

W = ?

L = 400

h = 80

xT = 0,16 mm

xcT = xsT

∴ Use values of copper: ∴ σc ÷ EC = σ C L ___ x c

∴ σc = E C x c ___ L

∴ σc = 100G × 0,16 × 1 0 −3 ____________ 0,4

= 40 MPa

∴ σ Steel = E S x S ___ L

= 198G × 0,16 × 1 0 −3 ____________ 0,4

= 79,2 MPa

19.2 UT = UC + Usteel

∴ UC = 1 _ 2 Fx

= 1 _ 2 σ.Ax

= 1 _ 2 × 40M × π _ 4 0,0 2 2 × 0,16 × 1 0 −3

= 1,005 J

US = 1 _ 2 Fx

= 1 _ 2 σ.Ax

= 1 _ 2 × 79,2M × π _ 4 (0,0 4 2 − 0,03 5 2 ) × 0,16 × 1 0 −3

= 1,866 J

19.3 PE = U

W(h + x) = U1 + U2

W(0,08 + 0,16 × 10–3) = 1,005 + 1,866

0,08016W = 2,871

W = 35,82 N

(Can also use values of steel: LS = LC)



20.

σ = 80 mPa

d = 25

E = 200G

PE = U

W(h + x) = 1 _ 2 Fx

∴ x = σL __ E

= 80M × 1,5 _______ 200G

= 6 × 10–4 m

Sub. in : W(h + x) = 1 _ 2 σ.Ax

∴ W(0,15 + 6 × 10–4) = 1 _ 2 × 80M × π _ 4 0,2 5 2 × 6 × 1 0 −4

W = 78,227 N

Mass = 78,227 _____ 9,81

= 7,97 kg

h = 150 mm

L = 1,5 m


MODULE

Compound bars and temperature-induced stresses3


Knowledge assumed to be in place before starting with this module:• Good knowledge of Modules 1 and 2• Ability to calculate for stress, strain and the change in length• Linear expansion of material subjected to change in temperature• Knowledge of screw thread terms• What a compound bar is• Nature of stresses due to cooling and an increase in temperature• Th e knowledge of Module 1• Must be able to use simultaneous equations• Ability to calculate moments of force

Learning outcomes

When students have completed this module they should be able to:• Calculate stresses in compound bars due to applied loads• Calculate the stresses in compound bars due to change in temperature• Calculate the resultant stresses in compound bars subjected to change in

temperature and an external force

Guidelines for the students

• To do the calculations in this module students must have a good knowledge of Module 1

• Th e same calculation principles are going to be used as they learnt in Module 1 and 2

• Th e knowledge will only be applied to a new theory for compound bars• Work from fi rst principles and do not rely on their memory and make sure they

understand the work from fi rst principles• Make use of sketches when they do calculations and always work from fi rst

principles• Working from fi rst principles will give students more knowledge of the subject

and they will also understand the theory of the subject better

IMPORTANT• Make a good study of the work content and underline all important facts• Before starting with the exercises make sure they understand the theory of the module and know the

answers of the self-check exercises


25Module 3 • Compound bars and temperature-induced stresses


1. C ø50S ø50200 kN 200 kN

1.1 FT = FC + FS

∴ 20k = σcAc + σsAs

= π _ 4 0, 0 2 ( σ c − σ s )

∴ 101,859 × 106 = σc + σs

xc = xs

∴ σ c L c ___ E c = σ s L s ___ E s

∴ σs = σ c E s ___ E c = σ c 207

____ 100 = 2,07σc

Substitute in ∴ 101,859 × 106 = 2,07σc

∴ σc = 33,179 MPa

σc = 68,68 MPa

1.2 Final length: LF = LO + x

= 100 + xc

∴ 0,100 + σ c L c ___ E c = 0,1 + 33,179 × 1 0 6 × 0,1 ___________ 100G

= 0,1 + 3,318 × 10–5

= 100,033 mm

2. S 200 mm2A 300 mm2 ε = 0,0005

ES = 209 GPa FT = FA + FS EA = 104,5 GPa

2.1 xs = xc

But x = σL __ E

∴σ = Ex __ L

= Eε ( x _ L = ε)

∴ σs = Esε and σA = Ecε

∴ σs = 209G × 0,0005 = 103,5 MPa

∴ σA = 104,5G × 0,0005 = 52,25 MPa

2.2 Fs = σsAs

= 103,5 × 106 × 200 × 10–6 = 20,7 kN

FA = σAAA

= 52,25M × 300 × 10–6 = 15,675 kN

Es = 207 GPaEc = 100 GPa



3. Copper

ø12 Steel

1505 5Copper

OD = 30

Id = 25

p = 1,5 mm pitch

ES = 200 GPa

EC = 100 GPa

3.1 FS = FC

∴ σcAs = σcAc

∴ σ s π _ 4 0,01 2 2 = σ c

π _ 4 (0,0 3 2 − 0,02 5 2 )

∴ σs = 1,91 σc

xT = xs + xc

xT = 2 _ 8 × 1,5 = 0,375 mm

∴ 0,375 × 10–3 = σ s L s ___ E s = σ c L c ___ E c

= σ s (150 + 10) ________ 200G + σ c (0,15)

_____ 100G

× 200G: ∴ 75 × 106 = 0,16 σs + 0,3 σc

Substitute in ∴ 75 × 106 = 0,16 (1,91σc) + 0,3 σc

= 0,6056 σc

∴ σc = 123,84 MPa

∴ σs = 236,54 MPa

3.2 xs = σ s L s ___ E s = 236,54M × 0,16 __________ 200G = 0,189 mm

xc = σ c L c ___ E c = 123,84M × 0,15 __________ 100G = 0,186 mm

4. OD = 500 200 kg x 9,81 N ES = 140G

Id = 420 EC = 14G

L = 2 m

4.1 FT = FI + FC

200 × 9,81 = σIAI + σcAc

= σ 1 π _ 4 ( 0, 5 2 − 0,4 2 2 ) + σ c ( π _ 4 0,4 2 2 )

1 962 = 0,058 σc1 + 0,139 σc

xS = xc

∴ σ s L s ___ E s = σ c L c ___ E c



∴ σS = σ c 140 ____ 14

= 10 σc

Substitute in ∴ 1 962 = 0,058(10 σc) + 0,139 σc

= 0,719 σc

∴ σc = 2,729 kPa

From σS = 27,29 kPa

4.2 xS = xc = σ c L c ___ E c = 2 729 × 2 ______ 14G = 3,9 × 10–4 mm

4.3 U = 1 _ 2 F x T = 1 _ 2 (200 × 9,81) × 3,9 × 10–7

= 3,826 × 10–4 J

5.

20ø

D = 100d = 80

D = 300

AT = π _ 4 0,32 = 0,07069 m2

AS = AP + AR

= π _ 4 ( 0, 1 2 − 0, 3 2 ) + ( π _ 4 × 0,0 2 2 × 6 ) = 2,827 × 10–3 + 1,885 × 10–3

AS = 4,712 × 10–3 m3

∴ AC = 0,07069 – 4,712 × 10–3 – π _ 4 0,082

AC = 0,061 m2

FT = FS + FC

300k = σsAs + σcAc

= σc4,712 × 10–3 + σc0,061

xs = xc

σ s __ E s = σ c __ G c

∴ σs = 15 σc

Substitute in ∴ 300k = 15 σc × 4,712 × 10–3 + σc 0,061

= 0,13168 σc

σc = 2,278 MPa

From σs = 34,17 MPa



6.

10 kN

Steelø4 ø6 ø4

1 2 3

ES = 200G

EC = 100G

x1 = x2 = x3

L1 = L2 = L3 = 3 m

No3 = d = 4

F = 30 kN

x = 60

L = 3

6.1 E3 = FL __ Ax

= 30k × 3 _________ π _ 4 0,00 4 2 × 0,06

∴ FT = F1 + F2 + F3

E3 = 119,37 GPa 10k = F1 + F2 + F3

x1 = x2 = x3

∴ x1 = x2

F 1 L 1 ___ A 1 E 1 = F 2 L 2 ___ A 2 E 2

∴ F 1 __________ π _ 4 0,00 4 2 × 200G

= F 2 __________ π _ 4 0,00 6 2 × 100G

∴ F1 = F 2 π _ 4 0,00 6 2 × 200G

___________ π _ 4 0,00 6 2 × 100G

F1 = 0,889 F2

x3 = x2

∴ F 3 L 3 ___ A 3 E 3 = F 2 L 2 ___ A 2 E 2

∴ F 3 ____________ π _ 4 0,00 4 2 × 119,37G

= F 2 __________ π _ 4 0,00 6 2 × 100G

∴ F3 = F 2 0,00 6 2 × 119,37G ____________

0,00 6 2 × 100G

= 0,531 F2

Substitute and in ∴ 10k = 0,889 F2 + F2 + 0,531 F2

∴ F2 = 4,132 kN

From ∴ F1 = 3,674 kN

From ∴ F3 = 2,194 kN

6.2 xT = x1 = σ 1 L 1 ___ E 1

= F 1 L 1 ___ A 1 E = 3,674k × 3 _________

π _ 4 0,00 4 2 200G

= 4,386 mm



7.

12

6 m8 m

8 kN

A = 900 m2

x1 = x2 E = 1 GPa

7.1 Load in each rope = 8 _ 2 = 4 kN

∴ σ = F __ A = 4k _______ 900 × 1 0 −6

= 4,44 MPa

7.2 Because it is a continuous rope.

∴ Average strength of rope = 6 + 8 ____ 2 = 7 m

∴ x = σ L AV ___ E

= 4,44m × 7 _______ 1G

= 31,08 mm

8.

10 N

9 m5 m

8.1 ∴ FT = F1 + F2

x1 = x2

∴ F 1 L 1 ____ A 1 G 1 = F 2 L 2 ____ A 2 G 2

[ A 1 = A 2 G 1 = G 2 ]

∴ × AG: ∴ F1L1 = F2L2

∴ F2 = F 1 9 ___ 5 = 1,8 F1

Substitute in ∴ 10k = F1 + 1,8 F1

∴ F1 = 3,571 kN ∴ F2 = 6,429 kN

σ1 = F 1 __ A = 3,571k _____ π _ 4 0,0 6 2

= 2,84 MPa

σ2 = F 2 __ A = 6,429k _____ π _ 4 0,0 4 2

= 5,12 MPa

8.2 ε1 = σ 1 __ E = 2,84M ____ 1,1G = 2,582 × 10–3

ε2 = σ 2 __ E = 5,12M ____ 1,1G = 4,655 × 10–3

8.3 x1 = x2 and ε2 = x 2 __ L 2

∴ x1 = ε1L1 ∴ x2 = ε2L2



x1 = 2,582 × 10–3 × 9 x2 = 4,655 × 10–3 × 5

= 23,24 mm = 23,28 mm

∴ U1 = 1 _ 2 F 1 x 1 = 1 _ 2 3,571k × 23,28 × 10–3

= 41,57 J

U2 = 1 _ 2 F 1 x 1 = 1 _ 2 × 6,429k × 23,28 ×10–3

= 74,83 J

8.4 3,571

6,419

RHx

10

(1 – x)

1

Moments about right ∴ 10x = 3,571 × 1

x = 0,3571 m

∴ 10 kN load is 357,1 mm from right

9. p = 1,2 mm

σB = 20 MPa

GS = 210 GPa

EB = 110 GPa

LB = 1,2 m

LT = 1,188 m

9.1 FB = FT xT = xB + xT

σBAB = σTAT

∴ 20 × π _ 4 0,01 5 2 = σT π _ 4 (0,02 4 2 − 0,01 6 2 )

σT = 14,063 MPa

9.2 FB = FT

σBAB = σTAT

σB = σT ( 0,024 2 − 0,016 2 __________ 0,015 2

) σB = 1,42 σT

xT = xB + xTU (xT = length × pitch)

(1,2 × 1,5)10–3 = σ B L B ___ E B + σ T L T

___ E T

1,8 × 10–3 = σ B 1,2 ____ 210G + σ T 1,188

_____ 110G

B

OD = 24 T

ø15

Id = 16



× 210G: ∴ 378 × 106 = 1,2σB + 2,268σT

Substitute in ∴ 378 × 106 = 1,2(1,42σT) + 2,268σT

∴ σbrass = 378M ____ 3072 = 95,17 MPa

σbolt = 135,14 MPa

Resultant stress = Stress tight nut + initial stress

∴ σbrass = 95,17 + 14,06 = 109,23 MPa

σsteel = 135,14 + 20 = 155,14 MPa


IMPORTANT A sketch of the question must be made for the temperature question that will solve 90% of the problem.

1. L = 500L = 800

S Cø30 ø50 Δt = 90° – 20° = 70°

Free expansion: ∴ ∆LC = αctL = 18 × 10–6 × 70 × 0,8 = 1,008 mm

∴ ∆LS = αstL = 12 × 10–6 × 70 × 0,5 = 1,42 mm and LT = 1 300 mm ∴ LF = LT + ∆LC + ∆LS = 1 300 + 1,008 + 1,42 = 1 301,428 mm

2.

t = 50°

ε = 175 GPa

α = 15 × 10–6

2.1 With no free expansion: Stress = strain × E σ = εE

= αtL ___ L × E (∴ ε = change in lenghth ___________ original length = αtL ___ L = αt)

= αtE

= 15 × 10–6 × 50 × 175G

= 131,25 MPa



2.2 ε = αtL ___ L = αt = 15 × 10–6 × 50 = 7,5 × 1 0 −4

3.

C S

30

30 30

50

L = 400 t = 90°

Steel 30 × 50

Copper 30 × 30

3.1 FC = FS

δxc = δxs

∴ αtL – xc = αtL + xs

αctLc – αstLs = xc + xs

= F C L C ____ A C E C + F S L S ___ A S E S

÷ L: ∴ 90(17,5 – 13,5)10–6 = F ________ 0,0 3 2 × 100G

+ F ____________ 0,03 × 0,05 × 209G

× 209G: ∴ 75,24 × 106 = 2 322,22F + 666,66F

∴ 75,24 × 106 = 2 988,89F

∴ FS = 25,173 kN = FC

∴ σs = F C __ A C = 25,173k _______ 0,03 × 0,05 = 16,78 (T)

FC = 25,173k _____ 0,0 3 2

= 27,97 MPa (C)

xS

xC

C

400

S

αtLC

αtLS

δxC

δxS



3.2 t = T1 – T2 = 20 – (–5) = 25°C

C

S

L = 400αtLS

αtLC

xS

xC

δxC = δxS

FS = FC

∴ δxs = dxc

αtLs + xs = αtLc + xc

∴ αtLc – αtLs = xs + xc

÷L: ∴ LS = LC ∴ 25(17,5 – 13,5)10–6 = F C ___ A S E S + F C

____ A C E C

= F _________ (30 × 50)209G + F _________ (30 × 30)100G

× 209G: 20,9 × 106 = 666,67F + 2 322,22F

F = 6,993 kN ∴ σs = F __ A S

= 6 993 _____ 30 × 50 = 4,66 MPa (C)

∴ σc = F __ A C = 6 993 ____ ( 3 0 2 ) = 7,77 MPa (T)

3.3 εc = σ c __ E = 27,97M _____ 100G = 2,797 × 10–4

εs = σ s __ E = 16,78M _____ 209G = 8,029 × 10–5

4.

xC xS

αtLS

ø80 ø60

LS = 400LC = 300

S

δx

CαtLC

4.1 FC = FS

δxc = δxs

∴ αtLc – xc = xs – αtLs αctL + αstL = xs + xc

(18 × 10–6 × 80 × 0,3) + (12 × 10–6 × 80 × 0,4) (t = 80°) = F × 0,4 _________

π _ 4 0,0 8 2 × 100G + F × 0,3 _________

π _ 4 0,0 6 2 × 100G

× 209G: 171,36M = 79,58F + 222,82F = 302,39F

IMPORTANT Students must show all calculations.



∴ F = 566,68 kN

∴ σs = F __ A S = 566,68 × 1 0 3 ________

π _ 4 0,0 8 2 = 112,74 MPa

∴ σc = F __ A C = 566,68 × 1 0 3 ________ π _ 4 0,0 6 2

= 200,42 MPa

4.2 Use info for steel or use info for copper

∴ δxs = x – αtL

= σsL ___ E – αtL

= 112,74M × 0,4 _________ 210G – (12 × 10–5 × 80 × 0,4)

= 0,169 mm (shorter)

4.3 εs = σ s __ E s = 112,74M ______ 210G = 5,369 × 10–4

εc = σ c __ E c = 200,42M ______ 100G = 2,004 × 10–3

4.4 U = 1 _ 2 Fx = 1 _ 2 × 566,68k × 0,169 × 10–3

= 47,88 J

5.

xSxC

C

S

αtLC

αtLS

δx

5.1 δxc = δxs

αtLc – xc = αtLs + xs

αctL – αstL = xs + xc

(18 × 10–6 × 30 – 12 × 10–6 × 30) = F ____________ 400 × 1 0 −6 × 200G

+ F ____________ 400 × 1 0 −6 × 100G

× 200G: ∴ 36M = 2 500F + 5 000F ∴ F = 4,8 kN

σS = F __ A = 4,8k _______ 400 × 1 0 −6

= 12 MPa (C)

σc = F __ A = 4,8k _______ 400 × 1 0 −6

= 12 MPa (T)

Compound bar FT = FC + FS

100k = FC + FS

xs = xc

FL ___ AE = FL ___ AE

F S ____ 200G = F C ____ 100G



FS = 2FC

Substitute in ∴ 100k = FC + 2FC

∴ FC = 33,333 kN

FS = 66,667 kN

∴ σc = 33,333k _____ 400 = 83,33 MPa (T)

σs = 66,667k _____ 400 = 166,67 MPa (T)

∴ Rs σc = –12 – 83,33 = 95,33 MPa (T)

Rs σs = +12 – 166,67 = 154,67 MPa (T)

5.2 Use info of copper for cooling: δxc = αtL – xc

= 18 × 10–6 × 30 × 0,6 – FL ___ AE

= 3,24 × 10–4 – 4,8k × 0,6 ____________ 400 × 1 0 −6 × 100G

= 3,24 × 10–4 – 7,2 × 10–5

= 2,52 × 10–4 m = 0,252 mm ∴ LF = LO – δx = 600 – 0,252 = 599,748 mm

For tension of 100 kN, use steel:

xS = xC

∴ xS = σSLF ___ ES

= 166,67M × 0,599748 _____________ 200G

= 0,5 mm

∴ Final total length after coding and load:

LFT = FF + xload

= 599,748 + 0,5

= 600,248 mm

6.

xS

xC

S 300 mm2

C 400 mm2

400

αtLC

αtLS

δx

For Question 5.2 info of steel can also be used



6.1 δxc = δxs

αtL – x = αtL + x

αctL – αstL = xs + xc

∴ 40(18 –12)10–6 = F ________ 300 × 200G + F ________ 400 × 100G

× 200G: ∴ 48m = 3 333,31F + 5 000F

∴ F = 5 760 N

∴ σs = F S __ A S = 5 760 ______ 300 − m = 19,2 MPa (T)

σc = F S __ A S = 5 760 ______ 400 − m = 14,4 MPa (C)

6.2 FT = FS + FC

xs = xc

F S L ___ A S E S = F C L

____ A C E C

∴ FS = F C 300 × 200 ________ 400 × 100

= 1,5 FC

Substitute in ∴ 80 = 1,5 FC + FC

∴ FC = 32 kN ∴ FS = 48 kN

∴ σS = F S __ A S = 48k _______ 300 × 10–6 = 160 MPa (C)

σC = F C __ A C = 32k _______ 400 × 10–6 = 80 MPa (C)

6.3 Res: σc = +80 + 14,4 = 94,6 MPa (C)

Res: σs = +160 – 19,2 = 140,8 MPa (C)

6.4 LF = LO + δxc – xc (use info of one material)

Copper info: 400 + [αtL – x] – σ c L ___ E C

= 400 + [ 18 × 1 0 −6 × 40 × 94 − 14,4M × 0,4 ________ 100G ] – ( 80M × 0,4 _______ 100G ) = 400 + 0,23 – 0,32 = 399,91 mm

7. D = 40

d = 30

FC = FS

xT = xS + xC

Subjected to load means it is compressed.

1,5 m

ø12

20 20



7.1 Stress due to nuts tightening Total change in length:

xT = 80 ___ 360 × 1,75 = 0,389 mm

∴ xT = xs + xc

∴ 0,389 × 10–3 = F S L S ___ A S E S + F C L C

____ A C E C

= F 1,54 ________ π _ 4 0,01 2 2 200G

+ F 1,5 _____________ π _ 4 ( 0,0 4 2 − 0,0 3 2 ) 100G

× 200G: ∴ 77,8M = 13 616,59F + 5 456,74F ∴ F = 4,079 kN

∴ σs = F __ A S = 4,079k _____

π _ 4 0,01 2 2 = –36,07 MPa (T)

σc = F __ A C = 4,079k __________ π _ 4 (0,0 4 2 − 0,0 3 2 )

= +7,42 MPa (C)

Stress due to temperature cooling

C

C

S

20 20

xS

xC

αtLC

αtLS

δx

LC = 1,5

LS = 1,5 + 0,04

= 1,54

Cooling: FS = FC ∴ δxc = δxs

αtLc – xc = αtLs + xss

αctL – αstL = xs + xc [ ∵ x − − FL ___ AE ] ∴ (18 × 10–6 × 20 × 1,5) – (12 × 10–6 × 20 × 1,54)

= F1,54 ________ π _ 4 1 2 2 × 200G

+ F1,5 _____________ π _ 4 (4 0 2 − 3 0 2 ) × 100G

× 200G: ∴ 34,08M = 13 616,59F + 5 456,74F ∴ F = 1,788 kN

∴ σs = F __ A S = 1,788k _____

π _ 4 1 2 2 = 15,81 MPa (C)

σs = F __ A C = 1,788k ________ π _ 4 (4 0 2 − 3 0 2 )

= 3,25 MPa (T)

∴ Res: σc = +7,42 – 3,25 = +4,17 MPa (C)

Res: σs = –36,07 + 15,81 = –20,26 MPa (T)

7.2 Steel info



xF = xnut – δxs

δxs = δtL + x = (12 × 10–6 × 20 × 1,54) + 20,26M × 1,54 _________ 200G = 0,526 mm (cooled)

7.3 After nut was turned 80°:

xs = F S L S ___ A S E S = 4,079k × 1,54 _________ A E S

= 0,278 mm ∴ xc = xT – xs

= 0,389 – 0,278 = 0,111 mm Us = 1 _ 2 F x s

= 1 _ 2 × 4,079k × 0,278 × 10–3

= 0,567 J

Uc = 1 _ 2 F x c

= 1 _ 2 × 4,079k × 0,111 × 10–3

= 0,226 J

8.

3 000 kN

Sø3

1

Cø6

2

Sø3

3

8.1 FT = F1 + F2 + F3

FT = F1 + F2 + F3

∴ 3 000k

x1 = x2 = x3

∴ x1 = x2

∴ F 1 L 1 ___ A 1 E 1 = F 2 L 2 ___ A 2 E 2

(÷ L as L1 = L2)

∴ F1 = F 2 A 1 E 1 _____ A 2 E 2

= 3 2 200G _____ 6 2 100G

F2



F1 = 0,5 F2

x1 = x3

∴ F 1 ___ A 1 E 1 = F 3 ___ A 1 E 1

∴ F1 = F3 = 0,5 F2

Substitute and in : ∴ 3k = 0,5 F2 + F2 + 0,5 F2

= 2F2

∴ F2 = 1 500 N ∴ FC = 1,5 kN FS = 0,5 × 1,5 = 750 kN/wire

8.2 ∴ σc = 1 500 _____ π _ 4 0,00 6 2

= –53,05 MPa (T)

σs = 750 _____ π _ 4 0,00 3 2

= –106,1 MPa

FC = FS

Temp. δxc = δxs

αctL – αstL = xs + xc ( x = FL ___ AE ) 100(18 – 12)10–6 = F ___ A S E S

+ F ____ A C E C

∴ 6 × 10–4 = F ________ 2 × π _ 4 3 2 200G

+ F ______ π _ 4 6 2 100G

× 200G: ∴ 120M = 141 471,06F

∴ F = 848,23 N

∴ σc = 848,23 _____ 2 × π _ 4 3 2

= –60 MPa

σs = 848,3 ____ π _ 4 6 2

= –30 MPa

σres σc = –53,05 + 30 = –23,05 MPa (T)

Res σs = –106 – 60 = –166 MPa (T)

9. d = 20

D = 28

Ltube = 600 Lbolt = 620

σB = 30 MPa σT = 127,3 MPa

E = 200G 12 × 10–6

600

ø18

10 10

LB = 620



E = 100G 12 × 10–6

t = (30° – t2)

Stress in bolt and tube will be zero when their final lengths are equal.

Also

Res σB = 0 = –30 nut + 30 temp

σT = 0 = + 27,3 nut – 27,3 temp

But δxB = δxT

∴ αTtLT – αBtLB = xB + xT

∴ t[18 × 10–6 × 0,6 – 12 × 10–6 × 0,620] = σ B L B ___ E B + σ T L T

___ E T

∴ 3,36 × 10–6t = 30M × 0,62 _______ 200G + 27,3 × 0,6 ______ 100G

= 9,3 × 10–5 + 1,638 × 10–4

= 2,568 × 10–4

t = 76,43°

Tube compressive and bolt tensile when nut was tightened:

t = t1 – t2

76,43 = 30 – t2

∴ t2 = 30 – 76,43

= –46,43 °C

10.

800

800, 288

C

S

t1 = 20 °C

10.1 Bar must be cooled for both to have the same length.

αtL

αtL

0,288

C

S

∴ αctL = αstL + 0,288 × 10–3

∴ t(αcLc = αsLs) = 0,288 × 10–3

∴ t(18 × 10–6 × 0,800288 – 12 × 10–6 × 0,8) = 0,288 × 10–3

Tube length = 600Bolt length = 600 + 10 + 10 = 620



∴ t = 0,288 × 1 0 −3 __________ 4,805184 × 1 0 −6

= 59,94 °C t = t1 – t2

∴ t2 = t1 – t

= 20 – 59,94

= –39,94°C

10.2 αtLs = 12 × 10–6 × 0,8 × 59,96

= 0,575424 mm

∴ LN = 800 – 0,575424

= 799,425 mm

11.

Sø50

LS = 250 LC = 200t = 80 – 20 = 60°

Final

Cø50

αtLC

αtLS

xS

xC

δx

11.1 δtLc = 20 × 10–6 × 60 × 0,2 = 0,24 mm

δtLs = 12 × 10–6 × 60 × 0,25 = 0,18 mm

11.2 δxc = δxs

αtLc – xc = xs – αtLs

αtLc + αtLs = xs + xc

(From 11.1)

(0,24 + 0,18)10–3 = σ S L S ___ E S + σ C L C

___ E C

= σ S 0,25 ____ 200G + σ C 0,2

____ 95G

× 200G: ∴ 84M = 0,25 σs + 0,421 σc

But FC = FS

∴ σsAs = σcAc (AS = AC)

∴ σs = σc

Substitute in : ∴ 84M = 0,25 σc + 0,421 σc

∴ σc = σs = 125,19 MPa



12.

C

800

ø30

C

S

xC

xS

αtLC

αtLS

δx

OD 45ID 32

Final 12.1 FS = FC

δxc = δxs

αtLc + xc = αtLs + xs

αtLc – αtLs = xs + xc ( x = FL ___ AE ) 80(18 – 12)10–6 = F _______

π _ 4 3 0 2 200G + F ___________

π _ 4 (4 5 2 − 3 2 2 )100G

× 200G: ∴ 96M = 1 414,711F + 2 543,935F ∴ F = 24,251 kN

∴ σs = F __ AS = 24,251k _____

π _ 4 3 0 2 = 34,31 MPa (T)

σs = F __ A C = 24,251k ________ π _ 4 (4 5 2 − 3 2 2 )

= 30,85 MPa (C)

12.2 FF = LO + δxs

= 800 + αtL + x x = σL __ E

= 800 + (80 × 12 × 10–6 × 0,8) + 34,31M × 0,8 _________ 200G = 800 + 0,905 = 800,905 mm 12.3

C

C

SFT FT

FT = FC + FS

xs = xc = 0,5 mm

F S L ___ A S E S = 0,5 × 10–3

∴ FS = A S E S × 0,5 × 1 0 −3 ___________ L

= ( π _ 4 3 0 2 ) ( 200G ) ( 0,5 × 1 0 −3 )

_______________ 0,800905

= 88,257 kN

∴ x = F C L ____ A C E C = 0,5 × 10–3

∴ FC = π _ 4 ( 4 5 2 − 3 2 2 ) ( 100G ) 0,5 × 1 0 −3

__________________ 0,800905

= 49,081 kN



∴ F = FC + FS

= 49,081 + 88,257 = 137,338 kN

12.4 σs = F __ A S = 88,257k _____

π _ 4 3 0 2 = +124,86 MPa

σc = F __ A C = 49,081 k ________ π _ 4 (4 5 2 + 3 2 2 )

= +62,43 MPa

∴ Res σc = +62,43 + 30,85 = 93,28 MPa (C) Res σs = +124,86 + 34,31 = 90,55 MPa (C)13.

xT = 0,3

∆t = t °C

FC = 0

AS = 110 mm2

AC = 210 mm2

13.1 No load in copper means there is NO stress in the copper. After a change in temperature, the resultant stress in the copper is zero, which is equal to free expansion (free expansion = no stress).

∴ ∆x = αtL

∴ ∆x = xT = 0,3 = αtL

∴ ∆t = 0,3 × 1 0 −3 _______ α c L

= 0,3 × 1 0 −3 __________ 18 × 1 0 −6 × 0,12

= 138,89 °C

13.2 Change in length of steel due to temperature = αstLs

∴ ∆L = 12 × 10–6 × 138,89 × 0,12

= 0,2 mm

∴ Distance that F must compress steel

= 0,3 – 0,2

x = 0,1 mm

∴ Force required:

x = FL ___ AE = 0,1 × 10–3

∴ F = xAE ___ L

= 0,1 × 1 0 −3 × 110 × 1 0 −6 × 200G ___________________ 0,12

= 183,33 kN

CSL = 120

F

F


MODULE


Knowledge assumed to be in place before starting with this module:• Knowledge of diff erent rivets• Ability to calculate force and the diff erent areas, cross-sectional area and

shear area• Knowledge of Module 1

Learning outcomes

When students have completed this module, they should be able to:• Describe a thin cylinder• Calculate the tensile and longitudinal stresses in thin cylinders• Calculate the safe internal pressure for a thin cylinder• Calculate the three failing criteria for a riveted joint• Calculate the joint effi ciency for a riveted joint• Identify diff erent riveted joints


• Make sure they know the diff erent riveted joints• Understand the principles of the module• Work from fi rst principles• Always make sketches of the joint under consideration

Thin cylinders and riveted joints4


45Module 4 • Thin cylinders and riveted joints


1. Ød = 2 m

P = 1,2 MPa

t = 25 mm

1.1 Tensile stress σt = PiD ___ 2t

= 1,2M × 2 ______ 2 × 0,025

= 48 MPa

1.2 Longitudinal stress σL = PiD ___ 4t

= 1,2M × 2 ______ 4 × 0,025

= 24 MPa

2. Pi 4 MPa D = 2 m Yield stress = 500 MPa

FOS

∴ Safe stress = 500 ___ 5 = 100 MPa

∴ Tensile stress: σt = PiD ___ 2t

∴ t = 4M × 2 ______ 2 × 100M

= 40 mm

Longitudinal stress σL = PiD ___ 4t

∴ t = 4M × 2 ______ 4 × 100M

= 0,02 m

= 20 mm

Use 40 mm plate thickness.

3. 2 MPa σallowable 80 MPa

d = 1,2 m of 72%

Spherical shell; no tensile stress only longitudinal stress.

∴ σL = PiD ___ 4tη

∴ t = 4M × 1,2 __________ 4 × 80M × 0,72

= 10,42 mm

4 m

Two stresses∴ Calculate a plate thickness for each.



4. D = 1,6 m t = 12 mm

ηL = 72% ηC = 68%

70 MPa

σt = PiD ___ 2t η L

∴ Pi = 70M × 2 × 0,012 × 0,72 ______________ 1,6

= 756 kPa

σL = PiD ___ 4t η C

∴ Pi = 70M × 4 × 0,012 × 0,68 ______________ 1,6

= 1,428 MPa

Maximum allowable = 756 kPa 5. D = ? Yield stress = 550 MPa η = 75% t = 16 Pi = 1,8 MPa FOS = 6

∴ Allowable stress = 550 ___ 6 = 91,67 MPa

σt = PiD ___ 2tη

∴ D = 91,67M × 2 × 0,016 × 0,75 ________________ 1,8M

= 1,22 m

σC = PiD ___ 4tη

∴ D = 91,67M × 4 × 0,016 × 0,75 ________________ 1,8M

= 2,44 m

Use 1,22 m.

6. t = 12 mm

d = 115 m

6.1 σt = PiD ___ 2t = 2,1M × 1,5 _______ 2 × 0,012 = 131,25 MPa

6.2 σC = PiD ___ 4t = 2,1M × 1,5 _______ 4 × 0,012 = 65,63 MPa

6.3 F = Pi DL

= 2,1M × 1,5 × 3

= 9,45 MN

Smallest pressure is the max because the higher pressure will destroy the cylinder.

If a bigger diameter is used, the stress will be more than the allowable stress and destroy the cylinder.

L = 3 m

Pi = 2,1 MPa



6.4 F = Pi π _ 4 d 2

= 2,1M × π _ 4 1,52

= 3,71 MN

7. F = 14 MN

7.1 σ = F __ A = 14M ___ πDt = 14M __________ π × 2,5 × 0,025

= 71,3 MPa

7.2 σ = PiD ___ 4t

∴ Pi = 71,3M × 4 × 0,025 ___________ 2,5

= 2,85 MPa safe pressure

8. L = 2,5 m Pi = 1,2 MPa t =10

D = 1,2 m σL = 36 MPa σt = 72 MPa

8.1 Force resist longitudinal

Fresist = σt 2tL

= 72M × 2 × 0,01 × 2,5

= 3,6 MN

8.2 Force acting longitudinal joint

Facting = Pi × DL

= 1,2M × 2,5 × 1,2

= 3,6 MN

8.3 Circum. joint

Force resisting

Fresist = σc × πDt

= 36M × π × 1,2 × 0,01

= 1,357 MN

8.4 Force acting

Fact = Pi × π _ 4 d 2

= 1,2M × π _ 4 1,22

= 1,357 MN



9. d = 2 m Pi = 1,7 MPa σt = 80 MPa η = 75%

σt = PiD ___ 2t η c

∴ t = PiD ____ 2σcη

= 1,7M × 2 __________ 2 × 80M × 0,75

= 28,33 mm

10. σy = 450 MPa

t = 15 ηt = 50%

d = 2,1 Pi = 2 MPa

ηL = 75%

∴ σt = Pid ___ 2t η c = 2M × 2,1 __________ 2 × 0,015 × 0,75

= 186,67 MPa

∴ FOS = σ y __ σ t = 450 _____ 186,67 = 2,41

Take 3

σC = Pid ___ 4t η t = 2M × 2,1 _________ 4 × 0,015 × 0,5

= 140 MPa

∴ FOS = σ y __ σ L = 450 ___ 140 = 3,21

Use 4 as the FOS.

11. σ = 120 MPa

D = 2,5 t = 14

ηL = 80% ηC = 40%

11.1 σt = PiD ___ 2t η L

Pi = σ2t η L ____ D

= 120M × 2 × 0,014 × 0,8 ______________ 2,5

= 1,0752 MPa

σC = PiD ___ 4t η C

Pi = 120M × 4 × 0,014 × 0,4 ______________ 2,5

= 1,0752 MPa

11.2 1,0752 MPa




1. t = 16 mm σt = 80 MPa σC = 190 MPa τ = 35 MPa

1.1 d = 6,05 √ _ t

= 6,05 √ ___

16 = 24,2 Use 25 mm diameter

1.2 Ft = Fs

(p – d)t σt = τη π _ 4 d2

(p – 25)16 × 80 = 35 × 2 × π _ 4 252

p – 25 = 26,845 p = 51,845 mm

1.3 Ft = (p – d)tσt

= (51,845 – 25)105 × 80

= 21,476 kN

Fs = n π _ 4 d 2τ

= 2 × π _ 4 0,0252 × 35 m

= 34,36 kN

FC = ndtσC

= 2 × 0,025 × 0,16 × 190M

= 152 kN

Joint is safe against crash FC > Ft an Fs.

1.4 F = ptσt

= 0,051845 × 0,016 × 80M

= 66,36 kN

∴ ηt = F C __ F = 21,476 _____ 66,36 = 32,36%

ηS = F S __ F = 34,36 ____ 66,35 = 51,78%

ηC = F C __ F = 152 ____ 66,35 = 229,1%

∴ Joint eff. = 32,36%



2.

2.1 d = 6,05 √ _ t

= 6,05 √ ___

18

= 26 mm

2.2 Ft = Fs

(p – d)t σt = η π _ 4 d2 τ

(p – 26)18 × 80 = 3 × π _ 4 262 × 50

p – 26 = 55,305

p = 81,31 mm

2.3 F = σ.A

= σt × pt

= 80M × 0,08131 × 0,018

= 117,09 kN

2.4 Ft = (p – d)t σt

= (81,31 – 26)18 × 80

= 79,649 kN

2.5 Fs = n π _ 4 d2τ

= 3 × π _ 4 262 × 50

= 79,639 kN

2.6 FC = ndt σC

= 3 × 26 × 18 × 180

= 252,72 kN

2.7 Ft = 79,639 kN

t = 18 mm σt = 80 MPa τ = 50 MPa σC = 180 MPa

mm can be used if the 106 of the stress is left out.



2.8 η = Ft __ F × 100 ___ 1

= 79,639 _____ 117,09 × 100 ___ 1

= 68,02%

3. t = 25

d = 32

σt = 440

τ = 377

σC = 660 MPa FOS = 6

3.1 Ft = Fs

(p – d)t σt __ 6 = η π _ 4 d2 × 1,75 τ _ 6

(p – 32)25 × 440 ___ 6 = 2 × π _ 4 322 × 1,75 × 377 ___ 6

∴ p – 32 = 96,47

p = 128,47 mm

3.2 F = ptσt

= 128,47 × 25 × 440 ___ 6

= 235,53 kN

Ft = (p – d)t σt

= (128,47 – 32)25 × 440 ___ 6

= 176,86 kN

Fs = n π _ 4 d2 × 1,75τ

= 2 × π _ 4 322 × 1,75 × 377 ___ 6

= 176,86 kN

FC = ndtσC

= 2 × 32 × 25 × 660 ___ 6

= 176 kN

Safe load is 176 kN.

3.3 η = 176 _____ 235,73

= 74,73%



4. t = 18

d = 25

p = 90

4.1 F = ptσt = 100 kN

η = p − d ____ p × 100 ___ 1

= 90 − 25 _____ 90 × 100 ___ 1

= 72,22%

4.2 Ft = (p – d) t σt

∴ σt = Ft _____ (p − d)t

= 100k ____________ (0,09 − 0,025)0,018

= 85,47 MPa

Fs = n π _ 4 d2 × 1,75τ

100k = 2 × π _ 4 0,0252 × 1,75τ ∴ τ = 58,21 MPa

FC = ndt σC

100k = 2 × 0,025 × 0,018 × σC

∴ σC = 111,11 MPa

FOS (T) = 500 ____ 85,47 = 5,85 = 6

FOS (C) = 800 _____ 111,11 = 7,2 = 8

FOS (S) = 400 ____ 58,21 = 6,8 = 7

Use a FOS of 8.

4.3 F = 2(p – d)t1 σt

100k = 2(0,09 – 0,025)0,012 σt

∴ σt = 64,1 MPa

5. D = 1,8 t = 20 p = 100 Pi = 1,5 MPa d = 30 L = 3,5 m

5.1 η = p − d ____ p × 100 ___ 1

= 100 − 30 ______ 100

= 70%

100 kN

12 = t1

If a given load is acting on a joint then the efficiency will always be p – d

____ p × 100 ___ 1 .

If pitch and rivet diameter are given then the joint efficiency will always be p – d

____ p × 100 ___ 1 .This also means that the force Ft = (p – d)tσt is the acting force on the joint.



∴ σt = Pid ___ 2tη

= 1,5M × 1,8 _________ 2 × 0,02 × 0,7

= 96,43 MPa

5.2 Ft = Fs

∴ (p – d)t σt = η π _ 4 d2 τ

∴ (100 – 30)20 × 96,43 = 2 × π _ 4 302 × τ

∴ τ = 95,49 MPa

5.3 Ft = FC

(p – d)t σt = ndtσC

(100 – 30)20 × 96,43 = 2 × 30 × 20σC

∴ σC = 112,5 MPa

6.

300

t = 12

d = 22

σt = 59

τ = 49

σC = 79

6.1 Ft = Fs

(p – d)tσt = n π _ 4 d2 1,75τ

(p – 22)12,59 = 2 π _ 4 222 × 1,75 × 49

p – 22 = 92,08

p = 114,1 mm

6.2 Ft = (p – d)tσt

= (114,1 – 22)12 × 59

= 65,21 kN

6.3 Fs = n π _ 4 d2 1,57τ

= 2 × π _ 4 222 × 1,57 × 49

= 65,19 kN

6.4 FC = ndtσC

= 2 × 22 × 12 × 79

= 41,712 kN

Ft acting force on joint and Ft = Fs = FC.



6.5 F = ptσt

= 114,1 × 12 × 59

= 80,78 kN

η = 41,712 _____ 80,78 × 100 ___ 1

= 51,62%

7. D = 2,1 m

d = 25 mm

τ = 80 MPa

σt = 100 MPa

7.1 Ft = Fs

(p – d)tσt = η π _ 4 d2 × 1,75τ

(p – 25)18 × 100 = 2 × π _ 4 252 × 1,75 × 80

p – 25 = 76,36

p = 101,36 mm

7.2 η = p − d ____ p × 100 ___ 1

= 101,36 − 25 ________ 101,36 × 100 ___ 1

= 75,33%

7.3 σt = Pid ___ 2tη

100M = Pi 2,1 ____________ 2 × 0,018 × 0,7533

∴ Pi = 1,29 MPa

7.4 Ft = FC

(p – d)t σt = ndtσC

(101,36 – 25)18 × 100 = 2 × 25 × 18 × σC

∴ σC = 152,72 MPa

8. t = 12

d = 18

σt = 80

τ = 35

σC = 190

E = 18

Because Ft is acting force on joint.



8.1 Ft = Fs

(p – d)tσt = n π _ 4 d2 τ

(p – 18)12 × 80 = 2 × π _ 4 182 × 35

p – 18 = 18,56

p = 36,55 mm

8.2 Ft = (p – d)t × σt

= (36,55 – 18)12 × 80

= 17,808 kN

Fs = 2 π _ 4 182 × 35

= 17,81 kN

FC = ndtσC

= 2 × 18 × 12 × 190

= 82,05 kN

Yes, safe against crushing.

8.3 F = ptσt

= 36,55 × 12 × 80

= 35,088 kN

η = 17,808 _____ 35,088 × 100 ___ 1

= 50,75%

9. D = 2 m

Pi = 800 kPa

η = 70%

σt = 80

σC = 92

τ = 70

9.1 σt = PiD ___ 2tη

80M = 800k × 2 ______ 2t 0,7

∴ t = 800k × 2 _________ 2 × 80M × 0,7

= 14,29 mm

≈ 15 mm

Ft is the acting force and for

crushing a force of 82,05 kN is needed.



9.2 FC = FS

ndtσC = n π _ 4 d2 τ

2 × d × 15 × 92 = 2 × π _ 4 d2 × 70

÷ d: ∴ d = 25,1

Say 26 mm

9.3 Ft = Fs

(p – d)tσt = n π _ 4 d2 τ

(p – 26)15 × 80 = 2 × π _ 4 262 × 70

p – 26 = 61,94

p = 87,94 mm

9.4 η = p − d ____ p × 100 ___ 1

= 87,94 − 26 _______ 87,94 × 100 ___ 1

= 70,43%

10. D = 220

t = 5

Pi = 5 MPa

10.1 σt = PiD ___ 2t

= 5M × 0,220 _______ 2 × 0,005

= 110 MPa

10.2 σL = PiD ___ 4t

= 5M × 0,220 _______ 4 × 0,005

= 55 MPa

10.3 F = σ × A

= 55M × πDt

= 55M × π × 0,22 × 0,005

= 190,07 kN

11. 200 240 kN

t = 14

d = 20

11.1 FS = n π _ 4 d2 × 1,75 τ

240k = n π _ 4 0,022 × 1,75 × 75M

Students must show all steps for calculations.



∴ η = 5,82

Say 6 rivets

11.2 F = (W – 3d)t1σt

240k = 2[0,200 – (3 × 0,02)]t1 × 105M

∴ t1 = 8,16 mm

∴ 8 mm

11.3 240k = 6 × 1,75 × π _ 4 0,022 τ

∴ τ = 72,76 MPa

240k = (w – d)t σc

240k = (200 – 3 × 20)14 σt

σt = 85,74 MPa

240k = 64d × tσc

∴ σc = 142,86 MPa

12. d = 21

t = 12

τ = 55 MPa

σt = 70 MPa

σc = 100 MPa

Rivet pitch: Ft = FS

σt(p – d)t = n π d 2 ___ 4 τ (single shear)

70(p – 21)12 = 2 π2 1 2 ___ 4 × 55

p – 21 = 45,36

p = 66,36

Use 67 mm pitch

Because Ft = FS ∴ shearing and tearing efficiencies are the same.

∴ ηs = ηt = p − d ____ p × 100 ___ 1

= 67 − 21 _____ 67 × 100 ___ 1

= 68,66%

ηC = F C __ F = σCdtη

____ σt pt × 100 ___ 1

= σcdη ___ σt p × 100 ___ 1

Two cover strips together must be more than plate thickness.∴ Use 8 total 16 and t = 14.



= 100 × 21 × 2 ________ 70 × 67 × 100 ___ 1

= 89,55%

13.

pi = 210 kPa D = 2,5 m

PC

13.1 Shear stress rivets = 360 MPa max stress

∴ Safe shear stress = 360 ___ FOS = 360 ___ 6 = 60 MPa

Shear force in rivets is caused by internal pressure acting at end of drum.

∴ Force acting at end of drum = Shear force in rivets

∴ pi π _ 4 D2 = n π _ 4 d2τ

÷ π _ 4 : ∴ piD2 = nd2τ

210k × 2,52 = n × 0,0162 × 60M

∴ Number of rivets = n = 85,45 rivets

∴ Use 86 rivets

Two rows of rivets ∴ 86 __ 2 = 43 rivets per row

13.2 Pitch of rivets = πD ___ 43 = π(2 500) ______ 43 = 182,651 mm

∴ Rivets/m = 1 000 _____ 182,651 = 5,475 rivets/m

We cannot make it 6 because it is taken per metre length.

∴ There will be part of a rivet per metre.


MODULE


Knowledge assumed to be in place before starting with this module:• How to calculate a moment• Ability to calculate the reactions for a simply supported beam• Ability to draw a simple shear force diagram (N4 Engineering Science)• Ability to draw a simple bending moment diagram

Learning outcomes

When students have completed this module, they should be able to:• Calculate the shear force at any point on a beam• Draw a shear force diagram and determine the position of maximum bending

moment• Calculate the bending moment at any point on the beam• Draw a bending moment diagram and determine the point of zero bending

moment


• Make sure they understand the basic concept of the theory about shear force and bending moment diagrams

• Make sure they understand the rules and defi nitions for shear force and bending moment diagrams

• Always make decent sketches of the beams and cantilevers and their shear force and bending moment diagrams

• Make a good study of the module and underline important facts

5 Loading of beams



Exercise 5 SB page 151

1.

Determine C moments about A

‹ __

M = __

› M

6C = (20 × 2) + (10 × 9)

C = 21,67 N

Determine A moments about C

__

› M =

‹ __

M

6A + (3 × 10) = (20 × 4)

6A = 8,33 N

BM – D left

BMA – BMD = 0

∴ BMB = + 8,33 × 2 = +16,66 Nm

BM right MC = –10 × 3 = –30 Nm

2.

24

A B DC3

21,67

–11,67+16,66

SF – DN

BM – DNm

8,33

8,33

20

–30

10

10

2 2 2 2 2

A B D E FC

SF – DN

BM – D

15

–10–15

+10

10

–20–20

30 10

10

25 N25 N


61Module 5 • Loading of beams

Moments about B

‹ __

M = __

› M

4D + (10 × 2) = (30 × 2) + (10 × 6) D = 25 N

Moments about D

__

› M =

‹ __

M

4B + (10 × 2) = (30 × 2) + (10 × 6) D = 25 N

BM left

MA = MF = 0

MB = –10 × 2

= –20 Nm

MC = +(25 × 2) – (10 × 4)

= + 10 Nm

BM right

MD = 10 × 2

= –20 Nm

ME = 0

3. 3.1 and 3.2

0

0

0

5030

8080125

50 2 23 3A B C D E

50

SF – D0 N

BM – D

–50–30

10 N/m

Total UDL = 10 × 10 = 100 N

Reaction: ∴ A = E = 50 N each

SFC = + 50 – (5 × 10) = 0

SFB = + 50 – (2 × 10) = 30 (left)

SFD = + 50 – (2 × 10) = –30 (right)



BM left

MA = ME = 0

MB = +(50 × 2) – ( 10 × 2 × 2 _ 2 ) = 100 – 20

= 80 Nm

MC = +(50 × 5) – ( 10 × 5 × 5 _ 2 ) = 250 – 125

= +125 Nm

BM right MD = (50 × 2) – ( 10 × 2 × 2 _ 2 ) = +80 Nm

4.

2 2 21A B C D

HE F G

13

41,89

24,2232,33

37,81

16,11

16,11

168,11

–9,89

–25,89

29,6619,78

–16

0

BM – DNm

SF – DN

0

4 N/m 6 N/m 8 N/m

x

Moments about A

‹ __

M = __

› M

∴ 9F = ( 4 × 2 × 2 _ 2 ) + [ 6 × 3 ( 3 × 3 _ 2 ) ] + [8 × 4 × 9]

= 8 + 81 + 288

F = 41,89 N Moments about F ∴

__ › M =

‹ __

M ∴ 9A = 4 × 2(8) + (6 × 3 × 4,5] = 64 + 81 A = 16,11 N



BM left

MA = MG = 0

∴ MB = +(16,11 × 2) – ( 4 × 2 × 2 _ 2 ) = 34,22 – 8

= +24,22 Nm

∴ MC = +(16,11 × 3) – (4 × 2 × 2)

= +48,33 – 16

= +32,33 Nm

∴ MD = (16,11 × 6) – (4 × 2 × 5) – (6 × 3 × 1,5)

= 96,66 – 40 – 27

= +29,66 Nm

BM right

ME = (41,89 × 2) – (8 × 4 × 2)

= 83,78 – 64

= +19,78 Nm

∴ MF = –8 × 2 × 1

= –16 Nm

H position: x = 8,11 ___ 6 = 1,352 m from C (x = S.F. at C ________ UDL C to D )

∴ MH = (16,11 × 4,352) – (4 × 2 × 3,352) – ( 6 × 1,352 × 1,352 ____ 2 ) = 70,111 – 26,816 – 5,484

= 37,811 Nm



5.

Moments about A

‹ __

M = __

› M

∴ 8E = ( 4 × 3 × 3 _ 2 ) + (20 × 6) + 8 × 3 ( 7 + 3 _ 2 ) + 6 × 12

= 18 + 120 + 204 + 72

E = 414 ___ 8

= 51,75 N

Moments about E

__

› M =

‹ __

M

∴ 9A +(6 × 4) + (8 × 3 × 0,5) = (2 × 20) + (4 × 3 × 6,5)

8A + 24 + 12 = 40 + 78

A = 82 __ 8

= 10,25 N

Position of turning point H where S.F. = 0 (BM can be a max there):

Position of H = S.F. at A ______________ UDL between A and B

= 10,25 ____ 4

= 2,563 m from A

MA = MG = 0

‹ ___

MH = (10,25 × 2,563) – ( 4 × 2,563 × 2,563 ____ 2 ) = 26,27 – 13,14

= +13,13 Nm

BM – DNm

SF – DN

4 N/m 8 N/m

2,563

10,25

13,13 12,757,5

0

0

–1,75

–21,75–29,75

–14,5–40 –12

6

22

10,25 3 3 1 1 21,5

E = 51,75

20

2

6

A H B C D 12 F G

0

x



‹ __

MB = (10,25 × 3) – (4 × 3 × 1,5)

= 30,75 – 18

= +12,75 Nm

‹ __

MC = (10,25 × 6) – (4 × 3 × 4,5)

= 61,5 – 54

= +7,5 Nm

___

› MD = (51,75 × 1) – (8 × 3 × 1,5) – (6 × 5)

= 51,75 – 36 – 30

= –14,25 Nm

ME = –(6 × 4) – (8 × 2 × 1)

= –24 – 16

= –40 Nm

MF = –6 × 2

= –12 Nm

6.

Moments about B

‹ __

M = __

› M

4E + (2 × 2 × 1) = (4 × 1) + (8 × 3) + (4 × 2 × 5)

∴ E = 16 kN

4

12A B8

C D16E F2 21

4

8

–8

–8

–8–4

–4

0

0

0

0

4 kN/m

SF – DkN

BM – DkN/m

8 kN4 kN2 kN/m



Moments about E

__

› M =

‹ __

M

4B + (4 × 2 × 1) = (8 × 1) + (4 × 3) + (2 × 2 × 5)

∴ B = 8 kN

BM = at a point

MA = ME = 0

∴ MB = –2 × 2 × 1

= –4 kN/m

∴ MC = (8 × 1) – (2 × 2 × 2)

= 0 kN/m

∴ MD = (16 × 1) – (4 × 2 × 2)

= 16 – 16

= 0 kN/m

∴ ME = – 4 × 2 × 1

= –8 kN/m

7.

4A E8

B C297,5(y – 2)

122,5

122,5

–37,5

–157,5170

187,57

–220

x = 3,0625 m140

D4 2

0

0

0

30 kN/m80 kN

80SF – D

kN

BM – DkNm

40 kN/my

All lines in a BM diagram are curved lines.



Moments about A

‹ __

M = __

› M

∴ 8C = (40 × 4 × 2) + (30 × 6 × 7) + 80 × 10 C = 297,5 kN

Moments about C

__

› M =

‹ __

M ∴ 8A + (80 × 2) = (40 × 4 × 6) + (30 × 6 × 1) A = 122,5 kN

Position of E

∴ x = 122,5 ____ 40 (x = S.F. at A ________ UDL A to B )

= 3,0625 m from A

MA = MD = 0

∴ ME = (122,5 × 3,0625)

= − 40 × 3,062 5 2 ________ 2

= +187,57 kN/m

∴ MB = (122,5 × 4) – (40 × 4 × 2)

= +170

∴ MC = –(80 × 2) – (30 × 2 × 1)

= −220 kNm

Moments at inflection point (right) (BM = 0)

MIMF = 0 = 297,5(y – 2) – 80y – 30 × y × y _ 2

∴ –15y2 + 217,5y – 595 = 0

∴ y = −217,5 ± √ ___________________

217, 5 2 − 4(−15)(−595) _____________________ 2(−15)

= −217,5 ± 107,73 __________ −30

y = 3,659 m from D

80 kN

297,5 kN

30 kN/m

2y – 2

y

D

C

Inflection Point

Figure 1



8.

Moments about B

‹ __

M = __

› M

4D + (80 × 1) + (60 × 1 × 0,5) = (20 × 4 × 2) + (100 × 2) + (100 × 1 × 4,5)

∴ 4D + 110 = 160 + 200 + 450

D = 175 kN

Moments about D

__

› M =

‹ __

M

4B + (100 × 1 × 0,5) = (100 × 2) + (20 × 4 × 2) + (60 × 4,5) + 80 × 5

4B + 50 = 200 + 160 + 270 + 400

B = 245 kN

MA = ME = 0

MB = –(80 × 1) – ( 60 × 1 × 1 _ 2 ) = –110 kNm

MC = (245 × 2) – (20 × 2 × 1) – ( 60 × 1 × 2 1 _ 2 ) – (80 × 3)

= 490 – 40 – 150 – 240

= +60 kNm

100 kN

2452

175

100

0

0

0

0

–80

–140 max

SF – DkN

BM – DkNm

–35–75

–50

–110

–60

10565

80 kN

A B C D E1 2F 1

60 kN/my – 0,5

y – 1

100 kN/m20 kN/m

y



MD = –100 × 1 × 0,5

= –50 kNm

y80 kN

245 kN

1 mA

B

F20 kN/m60 kN/m

(y – 1 _ 2 )

y – 1

y – 1 ___ 2

Figure 2

BM at inflection point F = 0

∴ MF = 0 = +245(y – 1) – 60 × 1(y – 0,5) – 80y –20(y – 1) ( y − 1 ) ____ 2

= 245y – 245 – 60y + 30 – 80y – 10(y2 – 2y + 1)

÷ 10 = 24,5y – 24,5 – 6y + 3 – 8y – y2 + 2y – 1

= –y2 + 12,5y – 22,5

∴ y = −12,5 ± √ _________________

12, 5 2 − 4(−1)(−22,5) ___________________ 2(−1)

= −125 ± 8,14 ________ −2

= 10,32 m and 2,18 m

∴ Inflection point 2,18 m from left.

Beam length = 6 m

∴ 10,32 m not applicable.

9.

SF – DkNm

BM – DkNm

4 kN/m2 kN/m

A B C D E F22222 8

8 max

max

zeroBM

TPTP

TP

4

–4–8

–8–4

0

0

0

0

16



Moments about E

__

› M =

‹ __

M

∴ 6B = 2 × 4 × 6

B = 8 kN

Moments about B

‹ __

M = __

› M

∴ 6E = (4 × 4 × 6)

E = 16 kN

MA = M1 = 0

MB = –2 × 2 × 1

= –4 kNm

MC = 8 × 2 – 2 × 4 × 2

= 0 kNm

MD = (16 × 2) – (4 × 4 × 2)

= 0 kNm

ME = –4 × 2 × 1

= –8 kNm

10.

E

1 kN/m

30 kN20

CB50

22

–25

37,5

–42

–28 max

5

10

10

20

00

0 0

DA 5 23

SF – DkN

E



10.1 Moments about A

‹ __

M = __

› M

∴ 8B = (30 × 5) + (20 × 10) + (1 × 10 × 5)

8B = 150 + 200 + 50

B = 50 kN

Moments about B

__

› M =

‹ __

M

∴ 8A + (20 × 2) = (30 × 3) + (1 × 10 × 3)

8A + 40 = 90 + 30

A = 10 kN

10.2 MA = MC = 0

MD = +10 × 5 – 1 × 5 × 5 _ 2

= +37,5 kNm

MB = –(20 × 2) – ( 1 × 2 × 2 _ 2 ) = –42 kNm

10.3 ∴ Inflection point at E.

1 kN/m20 kN

250(y – 2)

y

E C

B

ME = 0 = 50(y – 2) – 20y – ( y − y _ 2 ) = 50y – 100 – 20y – 0,5y2

= –0,5y2 + 30y – 100

y = −30 ± √ _________________

3 0 2 − 4(−0,5)(−100) _________________ 2(−0,5)

= −300 ± 26,46 ________ −1

= 3,54 m from C



11.

11.1 MG = 0

Moments about C calculate value of P

∴ ‹ __

M = __

› M

10F + (10 × 4) + (4 × 3 × 2,5) = (10 × 2) + 5P + (4 × 5 × 7.5)

10F + 40 + 30 = 20 + 5P + 150

∴ 10F = 100 + 5P

∴ F = (10 + 0,5P)

MG = 0 = +6,875(10 + 0,5P) – (4 × 5 × 4,375) – P × 1,875

0 = 68,75 + 3,4375P – 87,5 – 1,875P

∴ 18,75 = 1,5625P

Substitute in ∴ P = 12 kN

11.2 From : reaction F = 10 + (0,5 × 12)

= 16 kN

Fup = Fdown

C + 16 = 10 + 12 + 10 + 12 + 20

= 64

10 kN

6,875

5 – x

10 kN4 kN/m 4 kN/m

SF kN

BM kNm

CA D

PH

GB3 1 2 3 5

1648

26

3230

16

40

0

0

0

–16–22

–18

–70

–48

–22

–10

E F

x

H



C = 48 kN

11.3 MA = MF = 0

MB = –10 × 3 – 4 × 3 × 1,5 = –48 kNm

MC = –10 × 4 – 4 × 3 × 2,5 = –70 kNm

MD = –(10 × 6) – (4 × 3 × 4,5) + (48 × 2)

= –60 – 54 + 96

= −18 kNm

ME = 16 × 5 – 4 × 5 × 2,5

= 80 – 50

= 30 kNm

Position of H

∴ x = CF at E ___________ UDL from E to F

= 4 _ 4

= 1 m

Right: MH = (16 × 4) – 4 × 4 × 2

= +32 kNm

12.

MA = 0

MB = –3 × 2 = –6

MC = –(6 × 4) – (3 × 6)

= –42

Shear forcesFA = –3 NFB = –3 – 6 = –9 NFC = –3 – 6 = –9 N

0

0 0

–9 –9

–6

–42

0

–3–3

36

A

SF – D

BM – D

BC 4 2



13.

MA = 0

MB = –2 × 3 × 1,5

= –9 Nm

BM – D

MC = –2 × 3 × 2,5

= –15 Nm

MD = –(6 × 2) – (2 × 3 × 4,5)

= –39 Nm

ME = –(2 × 3 × 7,5) – (6 × 5) – (4 × 3 × 1,5)

= –93 Nm

Shear forcesFA = 0FB = –(2 × 3) = –6 NFC = –(2 × 3) – 6 = –12 NFD = – 6 – (2 × 3) = –12 NFE = –(2 × 3) – 6 – (3 × 4) = –24 N

4 N/m 2 N/m

0 0SF – D

BM – D

6 N

D

–24

–93

–39

–15

–9

–12–12

–6–6

C B A3 32 1E



14.

MA = 0

MB = 0

MC = –9 × 3 = –27 Nm

MD = –9 × 8 – ( 4 × 5 × 5 _ 2 ) = –122 Nm

ME = –9 × 11 – (4 × 5 × 5,5) = 209 Nm

MF = –(9 × 14) – (4 × 5 × 8,5) – (9 × 3) = –323 Nm

15.

–38

F 3

9 9

3 5 3 3E D C B A

–38

–29–29

–9 –9

0

0

0

0

SF – D(N)

BM – D(Nm)

–323–209

–122

–27

4 N/m

Shear forcesFA = 0FB = –9 NFC = –9 NFD = –9 – (4 × 5) = –29 NFE = –(4 × 5) – 9 – 9 = –38 NFF = –9 – (4 × 5) – 9 = –38 N

8 kN/m

E

30 kN

35,14

11,1411,14

27,14

–8

–4

–18,86 34,28

–122,06

–18,86

SF – D (kN)

BM – D (kNm)

D C B A

00

3 12 2



Reaction at B – moments about E

∴ ME = 0 = –(30 × 3) – (8 × 3 × 6,5) + B × 7

= –90 – 156 + 7B

∴ B = 35,14 kN

MA = 0

MB = – ( 8 × 1 × 1 _ 2 ) = –4 kNm

MC = +(35,14 × 2) – ( 8 × 3 × 3 _ 2 ) = +34,28 kNm

MD = +(35,14 × 4) – (8 × 3 × 3,5) = –122,06 m

ME = (–30 × 3) – (8 × 3 × 6,5) + (35,14 × 7) = 0

16.

24 4

2 2

0

A B

2 N 4 N 2 N

C D E

0

0

0

2 2 2

–2 –2 –2 –2

–4 –4

SF – D(N)

SF from A

BM – D(Nm)

x x

Reactions at B and D.

Symmetrical length and load.

∴ Fup = Fdown

B + D = 2 + 4 + 2 = 8

∴ B = D = 4

MC = 0 = + (4 × x) – 2 × (2 + x)

0 = 4x – 4 – 2x

4 = 2x

∴ x = 2 ∴ B to D = 4 m

MA = ME = 0

MD = –(2 × 2) = –4 Nm = MB MC = +(4 × 2) – (2 × 4) = 0

Shear forcesFA = 0FB1 = –(8 × 1) = –8 kNFB2 = –(8 × 1) + 35,14 = +27,14 kNFC = –(8 × 3) + 35,14 = +11,14 kNFD = –(8 × 3) + 35,14 – 30 = –18,86 kNFE = –30 – (8 × 3) + 35,14 = –18,86 kN



17. 17.1 Reaction A

__

› M =

‹ __

M about D

A × 15 = ( 2 × 15 × 15 __ 2 ) + (5 × 4 × 9)

= 225 + 180 A = 27 N

Reaction D Moments about A

‹ __

M = __

› M

15D = ( 2 × 15 × 15 __ 2 ) + 5 × 4 × 6 D = 23 N

MB = +(27 × 2) – ( 2 × 2 × 2 _ 2 ) = 54 – 4 = 50 Nm

17.2 MC = +(23 × 7) – ( 2 × 7 × 7 _ 2 ) = 161 – 49 = +112 Nm

17.3 ME = +27 × 4 – ( 2 × 4 × 4 _ 2 ) = 108 – 16

= +92 Nm

x = 19 __ 7 = 2,714 m (x = S.F. at E ________ UDL E to C = S + 2)

∴ MF = (27 × 6,714) – (2 × 4 × 4,714) – ( 7 × 2,714 × 2,714 ____ 2 ) = 181,278 – 37,712 – 25,78

= 117,786 Nm



17.4

2 2A B E F C

9 mD

2,714

2 N/m 2 N/m

23

SF.D(N)

BM – D(Nm)

27

27

0 0

2319

–9

11292

50

117,786 –23

5 N/m

x

4 7

MA = MD = 0

Total UDL = 20 N

∴ Load/m = 20 __ 4 = 5 N/m

18.

F-DiagF

ED

8

4

A0 B C

D EF

0

–4

–8

SF. – D

CBA

2 kN/m 4 kN/m

4 kN

2 m 2 m 2 m1 m 1 m

8 kN

8 kN 16 kN



19.

A B

80 kN 100 kN

20 kN/m60 kN/m 100 kN/m

245 kN

2 m 2 m1 m

0

–80B

–140

–35

C

65

100105

D0E

–75SF-DkNGiven

1 m

175 kNC D E

20.

10 kN

3 m 3 m2 m 5 m1 m

4 kN/m 4 kN/m10 kN

A

A B

B

26

–10

–22–16

SF–DkN

C D4

16

E0 F

0

B C D E F48 16 kn

F-Diag

12 kN


MODULE


Knowledge assumed to be in place before starting with this module:• How to calculate a moment• How to calculate the reactions for a simple supported beam• How to draw a simple shear force diagram• How to draw a simple bending moment diagram• Know the diff erent stresses• Engineering Science N4• Good understanding of Module 5

Learning outcomes

When students have completed this module, they should be able to calculate:• Th e size of a beam needed• Centroid of a section• Moment of inertia of standard beams and built-up beams• Maximum safe load for a beam• Section modules for a section• And be able to use hot rolled section tables


• Make sure they understand the basic concept of the theory about simple bending• Make sure they understand how to use the shear force and bending moment

diagrams to calculate information for beams• Always make decent sketches of the beams and cantilevers and their shear force

and bending moment diagrams• Make a good study of the module and underline important facts

Simple bending of beams6


81Module 6 • Simple bending of beams


1. OD = 500 M __ I = σ _ y

ID = 460

I = π __ 64 (0,54 – 0,464) = 8,701 × 10–4 m6

y = 500 ___ 2 = 0,25

∴ M = σI __ y = 60M × 8,701 × 1 0 −4 ____________ 0,25

= 208,823 kNm

But Mmax = ML + MW

= w L 2 ___ 8 + w L 2 ___ 8

208 823 = L 2 __ 8 [1 304 + 2 308]

L = √ ______

462,51

= 21,51 m

2.

400

480

y

3

L = 3 mW

4 kN/m

M __ I L = σ _ y = 100 ___ 200

M = σI __ y = 100 × 1 0 6 × 0,08 × 0, 4 3 ______________ 0,2 × 12

= 213,33 kNm

M = w L 2 ___ 8 + WL ___ 4

213,333 = 4 × 1 0 3 × 3 2 ________ 8 + W3 ___ 4

W = 278,44 kN

+



3.

3 m

300 N/m

D = 50 M __ I = σ _ y : M = σI __ y = 120M × π0,0 5 4 _________ 0,025 × 64 = 1 472,63 Nm

1 472,63 = w L 2 ___ 8 + w L 2 ___ 8

= 300 × 3 2 ______ 8 + w 3 2 ___ 8

w = 1,009 kN/m

4.

- y AT = A1y1 + A2y2

4.1 - y (400 × 50 + 60 × 500) = (500 × 60 × 250) + (400 × 50 × 525)

50 000 - y = 18 × 106

- y = 360 mm

h1 = - y – g = 360 – 250 = 110 mm

h2 = y2 – - y = 525 – 360 = 165 mm

4.2 Ixx = I1xx + A1h12 + I2yy + A2h2

2

= [ 0,06 × 0, 5 3 _______ 12 + 0,06 × 0,5 × 0,1 1 2 ] + [ 0,4 × 0,0 5 3 _______ 12 + 0,4 × 0,05 × 0,1652 ] = 9,88 × 10–4 + 5,487 × 10–4

= 1 536,7 × 10–6 m4

4.3 Iyy = I1 + A1h12 + I2 + A2h2

2 h1 = h2 = 0 = I1xx + I2yy

= 0,05 × 0, 4 3 _______ 12 × 0,5 × 0,0 6 3 _______ 12 = 275,67 × 10–6 m4

x

x

xx

x

y

y

y

y

y

_ y

y50

y2 = 525

y1 = 250

400

500

60

1

2



4.4

∴ σ M __ I = σ _ y

M = WL ___ 4 = 200k × 5 ______ 4 = 250 kNm

∴ σmax = M y max ____ I xx = 250k × 0,36 _________

1 536,7 × 1 0 −6

= 58,57 MPa

σmin = M y min ____ I xx

= 250k × 0,19 _________ 1 536,7 × 1 0 −6

= 30,91 MPa

4.5

σmax = σmin ∴ xt = xc

∴ σ = 250k × 0,2 _________ 275,67 × 1 0 −6

σ = Mx ___ Iyy

σ = 181,38 MPa

4.6 Zmax = M ___ σ max = 250k _____ 58,57M

= 4,268 × 1 0 −3 m 3

Zmin = M ___ σ min = 250k _____ 30,91M

= 8,088 × 1 0 −3 m 3

5.

5.1 6R = (80 × 2) + ( 12 × 6 × 6 _ 2 ) R = 62,67 kN 6L = (80 × 4) + ( 12 × 6 × 6 _ 2 ) L = 89,33 kN

190

360 y max

y minNA

200yy

x

x 200

6

12 kN/m

80 kN

2

RL

4



5.2

Mmax = (89,33 × 2) – ( 12 × 2 × 2 _ 2 ) (at 80 kN point of inflection)

= 154,66 kNm

5.3

Ixx = 1 __ 12 [0,2 × 0,63 – 0,18 × 0,583]

= 6,7332 × 10–4 m4

M __ I = σ _ y σ = My ___ I

= 154,66 × 1 0 3 × 0,3 ___________ 6,7332 × 1 0 −4

= 68,91 MPa

5.4 σ at 150 mm ∴ y = 150

∴ σ = My ___ I = 154,66 × 1 0 3 × 0,15 ____________

6,7332 × 1 0 −4

= 34,45 MPa

6. 6.1

12 kN/m

–62,67

–14,67

89,3365,33

89,33 62,67

80 kN

kN.SF.D

2

00

42

y

y

200

180

580600

150

3

2

2080 1

20x3

x

x

x x

x

x

x2

x1

y

y

yy

y

y

y y 10

100

168,64

300

y1 = 5

y2 = 150 _ y

y3 = 290



- y AT = A1y1 + A2y2 + A3y3

No. Area y Ay

1 10 × 80 800 5 4 000

2 20 × 300 6 000 150 900k

3 20 × 100 2 000 290 580k

Total 8 800 ΣA-max 1 484k

∴ yAT = Σ Area-moment

- y 8,8k = 1 484k

- y = 168,64 mm

- x AT = Σ Area-moments

- x 8 800 = Σ Area-moments

No. Area x Ax

1 800 60 48k

2 6 000 10 60k

3 2 000 70 140k

ΣA-max 248k

∴ - x AT = Σ A-moment

- x 8 800 = 248 × 103

- x = 28,18 mm

6.2 Ixx = I1yy + A1h12 + I2xx + A2h2

2 + I3yy + A3h32

h1 = 168,64 – 5 = 163,64 mm

h2 = 168,64 – 150 = 18,64 mm

h3 = 290 – 168,64 = 121,64 m

I1T = I1yy + A1h12 = 0,08 × 0,0 1 3 ________ 12 + 0,08 × 0,01 × 0,163642

= 2,143 × 10–5 m4

I2T = I2yy + A2h22 = 0,02 × 0, 3 3 _______ 12 + 0,02 × 0,3 × 0,018642

= 4,708 × 10–5 m4

I3T = I3yy + A3h32 = 0,1 × 0,0 2 3 _______ 12 + 0,1 × 0,02 × 0,121642

= 2,966 × 10–5



Ixx = I1T + I2T + I3T

= (2,143 + 4,708 + 2,966)10–5

= 9,817 × 1 0 −5 m 4

6.3

100

70

y yx

x

Y Y

20

h3h1

h2

10

_

x = 28,18

13

2

80

2060

300

X

X X

X

X

X

y

yy

y

h1 = 60 – 28,18 = 31,82

h2 = 28,18 – 10 = 18,18

h3 = 70 – 28,18 = 41,82

Iyy = I1xx + A1h12 + I2yy + A2h2

2 + I3xx + A3h32

I1T = Ixx + A1h12

= 0,01 × 0,0 8 3 ________ 12 + [0,01 × 0,08 × 0,031822]

= 1,2367 × 10–6 m4

I2T = Iyy + A2h22

= 0,3 × 0,0 2 3 _______ 12 + [0,3 × 0,02 × 0,018182]

= 2,183 × 1 0 −6 m 4

I3T = Ixx + A3h32

= 0,02 × 0, 1 3 _______ 12 + [0,02 × 0,1 × 0,041822]

= 5,164 × 1 0 −6 m 4

Iyy = I1T + I2T + I3T

= (1,2367 + 2,183 + 5,164)10–6

= 8,584 × 10–6 m4



6.4

M = WL + w L 2 ___ 2 = (20 × 4) – ( 10 × 4 × 4 _ 2 ) = 160 kNm

20 kn

4

10 kN/m

M __ F = σ _ y

σt = My ___ I xx = 160 × 1 0 3 × 0,13136 _____________

98,17 × 1 0 −6

131,36

168,64 = ymax

Tensile

Compressive

x x

y

y

ymin

= 214,1 MPa

6.5 σmax = My ___ I = 160 × 103 kNm

Zmax = I xx ___ y max = 98,17 × 1 0 −6 ________ 0,16864

= 5,821 × 10–4 m3

6.6 kxx = √ __

I xx __ A T = √ ________

98,17 × 1 0 −6 ________ 8 800 × 1 0 −6

= 105,62 mm

kyy = √ __

I yg __ A T = √

________ 8,584 × 1 0 −6 ________

8 800 × 1 0 −6 = 31,23 mm

7.

WN2 3

5 m σ = 80 mPa

15

15

115

11510

1

1

2X X 230

x x

y

y

y

y

y

y

y100

y = 115 + 15 = 130 mm

Ixx = 2[I1yy + A1h12] + [I2xx + A2h2

2]

h1 = 122,5 mm h2 = 0

∴ Ixx = 2 [ 0,1 × 0,01 5 3 ________ 12 + 0,1 × 0,015 × 0,122 5 2 ] + 0,01 × 0,2 3 3 ________ 12

= 45,075 × 10–6 + 10,139 × 10–6

= 55,214 × 10–6 m4



M = σI __ y = 80M × 55,214 × 1 0 −6 _____________ 0,13

= 33,978 kNm

M = Wab ____ L

33 978 = W × 2 × 3 _______ 5 ∴ W = 28,315 kN

8.

1 m 1 m 1 m

3 kN/m 2 kn5 kN 30

Dx x

Ixx = 0,03 × D 3 ______ 12 = 2,5 × 1 0 −3 D 3

At fixed end: Mmax = (2 × 3) + (5 × 1) + (3 × 1 × 1,5)

= 15,5 kNm

8.1 M __ I = σ _ y Ixx = 2,5 × 10–3 D3

y = D __ 2

σ = 120 MPa

BM 1 m from free end: M1 = 2 × 1 = 2 kNm

∴ 2 × 1 0 3 ________ 2,5 × 1 0 −3 D 3

= 120 × 1 0 6 ______ 0,5D

∴ 2,5 × 10–3 D3 × 120 × 106 = 2 × 103 × 0,5 D (M × 0,5 D)

÷ D: ∴ 300k D2 = 1 000

D = √ __________

3,33 × 1 0 −3

= 57,74 mm

8.2 BM 2 m from free end: M2 = (2 × 2) + ( 3 × 1 × 1 _ 2 ) = 5,5 kNm

∴ M 2 __ I = σ _ y ∴ 5,5 × 1 0 3 ________ 2,5 × 1 0 −3 D 3

= 120 × 1 0 6 ______ 0,5D

2,5 × 10–3 D3 × 120 × 106 = 0,5D × 5,5 × 103 (M × 0,5 D)

÷ D: ∴ 300k D2 = 2 750

D = 95,74 mm

8.3 M3 = 15,5 kNm

M 3 __ I xx = σ _ y

∴ 300k D2 = 0,5 × 15,5 × 103 (M × 0,5 D)

D = 160,73 mm

In all three cases the formula stays the same and values only B.M. that change.

W

25 m

L

a b

Students must show all steps in calculations.



9. 9.1

1

1

160300

200

200

160

•

•

• 570x x

Ixx = IxxT – 2[Ixx1 + A1h12]

= 0,3 × 0,5 7 3 _______ 12 – 2 [ 0,16 × 0, 2 3 _______ 12 + 0,16 × 0,2 × 0,12 5 2 ] = 4,63 × 10–3 – 1,213 × 10–3

= 3,4165 × 1 0 −3 m 4

M __ I = σ _ y ∴ M = σI __ y

= 130 × 1 0 6 × 3,4165 × 1 0 −3 × 2 __________________ 0,57

= 1,558 MNm

9.2 M __ I = σ _ y

∴ Z = M __ σ

= 1,588 × 1 0 6 _______ 130 × 1 0 6

= 0,0119876 m3

Zshaf = I _ y = π D 4 × 2 _____ 64D = π D 3 ___ 32

∴ π D 3 ___ 32 = 0,0119876 D3 = 0,1221 D = 496,11 mm

10.

200

200

350

500

200

hT

100

200

100 700

200

100

200

400

532

•

•

••x x

• •

• •

_ y

Y

y

y1

y2

y3

y y

yx

X X

x

xx

Y

31

2



10.1

No. Area A y Ay

1 0,532 × 0,7 0,3724 0,35 0,13034

2 0,2 × 0,4 –0,08 0,2 –0,016

3 0,2 × 0,2 –0,04 0,5 –0,02

AT 0,2524 ΣAmom 0,09434

- y AT = ΣAmom

- y 0,2524 = 0,09434

- y = 373,772 mm

10.2 Ixx = [IxxT + AThT2] – [Iyy2 + A2h2

2] – [Ixx3 + A3h32]

hT = - y – yT = 373,772 – 350 = 23,772

∴ IxxT = 0,532 × 0,70 0 3 _________ 12 + 0,532 × 0,7 × 0,0237722 (for total cross-section including the holes) = 15,416 × 10–3 m4

I2T = (Iyy + A2h22) h2 = 373,772 – 200 = 173,772

= 0,4 × 0, 2 3 ______ 12 + 0,4 × 0,2 × 0,1737222

= 2,682 × 10–3 m4

I3T = I3xy + A3h32 h3 = 500 – 373,772 = 126,228

= 0, 2 4 ___ 12 + 0,22 × 0,126,2282

= 7,7067 × 10–4 m4

∴ True Ixx = IxxT – I2T – I3T

∴ True: Ixx = 15,416 × 10–3 – 2,682 × 10–3 – 7,7067 × 10–4

= 11,96 × 1 0 −3 m 4

10.3 Iyy = Iyysolid – Ixx2 – Iyy3 (h = 0)

= 0,7 × 0,53 2 3 ________ 12 – 0,2 × 0, 4 3 ______ 12 – 0, 2 4 ___ 12

= 8,783 × 10–3 – 1,2 × 10–3

= 7,583 × 10–3 m4



10.4 M = wL __ 4 + w L 2 ___ 8

= 10 × 5 ____ 4 + 12 × 5 2 _____ 8

= 50 kNm

σmax= M y max ____ Ixx

S

10 kW

12 kN/m

= 50k × 0,373772 __________ 11,96 × 1 0 −3

= 1,56 MPa

σmin = M y min ____ Ixx

= 50k × 0,326228 __________ 11,96 × 1 0 −3

= 1,36 MPa

10.5

11.

(a) Zxx = I _ y = π ( D 4 − d 4 ) × 2 _________ 64 D

= π __ 32 ( 0,0 8 4 − 0,01698 4 4 ___________ 0,08 ) = 50,163 × 10–6 m3

(b)

15

150

60 • • 60

• 20

15

1 1

2X

Y Y

X

42,560

_ y = 17,5

NA •

_ y

326,228326,228

700

ymax

ymin

373,772

Y

Y 1,36 mPa

1,56 mPa

x x x x

D = 80

d = 16,986x x



- y AT = ΣAmom

- y (2 × 15 × 60) + (150 × 20) = 2(15 × 60 × 30) + (150 × 20 × 10)

4 800 - y = 84 000

- y = 17,5 mm

Iyy = 2(I1T + A1h12) + I2 + A2h2

2

= 2 [ 0,015 × 0,0 6 3 ________ 12 + 0,015 × 0,06 × 0,012 5 2 ] + [ 0,15 × 0,0 2 3 ________ 12 + 0,15 × 0,02 × 0,007 5 2 ] = 8,2125 × 10–7 + 2,6875 × 10–7

= 1,09 × 10–6 m3

Zmax = Iyy ___ y max = 1,09 × 1 0 −6 _______ 0,0425

= 25,647 × 1 0 −6 m 3

(c) Zxx = I _ y

= B D 3 × 2 ______ 12D

= B D 2 ___ 6

= 0,06 × 0,0 8 2 ________ 6

= 64 × 1 0 −3 m 3

(d) Ixx = 1 __ 12 [BD3 – bd 3]

= 1 __ 12 [(0,06 × 0,183) – (0,04 × 0,153)]

= 1,791 × 10–5

Zxx = I xx __ y = 1,791 × 1 0 −5 ________ 0,09

= 199 × 1 0 −6 m 3

Channel = 23,647 × 10–6 m3

Pipe = 50,163 × 10–6 m3

Solid rectangular = 64 × 10–6 m3

Hollow rectangular = 199 × 10–6 m3

Y X

XX




1. - y AT = A1y1 + A2y2

TAPER

X

x

x x

x

Xh2

h1

101,

6

418,

03

734,

3

_ y =

367

,0768

3,5

341,

7

1

2

1.1 - y (17,84 × 10–3 + 1,23 × 10–3) = 17,84 × 10–3 × 0,6835 _____ 2 + 1,23 × 10–3 × 0,7343 = 7,000009 × 10–3

- y = 367,07 mm (418,03)

1.2 ∴ Ixx = Ixx + A1h12 + Ixx + A2h2

2

h1 = 734,3 – 367,07 = 367,23

h2 = 367,07 – 341,75 = 25,32

Ixx = (2,176 × 10–6 + 1,23 × 10–3 × 0,367232) + (1 363 × 10–6 + 17,34 × 10–3 × 0,025322)

= 1,681 × 10–4 + 1,374 × 10–3

= 1 542 × 10–3 m4 1 542 × 10–6 m4

Iyy =Iyy1T + Iyy2B

= 0,2528 × 10–6 + 51,83 × 10–6

= 52,083 × 1 0 −6 m 6

1.3 M __ I = σ _ y ∴ σmax = M y max ____ Ixx = 400k × 0,41803 __________ 1 542 × 1 0 −6

= 108,44 MPa



2.

y y

x x

x x

1

2

•

•h = 609,6y2 = 594,7

y1 = 304,8

25,3237,1

382,9

10,4

h2

h1 = 78,1

381 × 102 × 55,1

610 × 305 × 149

__

y

x

y

x

y

2.1 - y AT = A1y1 + A2y2

- y (19,03 × 10–3 + 7,019 × 10–3) = (19,03 × 10–3 × 0,3048) + (7,019 × 10–3 × 0,5947)

26,049 × 10–3 - y = 9,975 × 10–3

- y = 382,9 mm

2.2 Ixx = I1 + A1h12 + I2 + A2h2

2

h1 = 382,9 – 304,8 = 78,1 mm ( _ y – y1 = h1)

h2 = 594,7 – 382,9 = 211,8 mm (y2 – _ y = h2)

∴ Ixx = Ixx + A1h12 + Iyy + A2h2

2

= (1 247 × 10–6 + 19,03 × 10–3 × 0,07812) + (5,849 × 10–6 + 7,019 × 10–3 × 0,21182)

= 1,363 × 10–3 + 3,207 × 10–4

= 1 683,792 × 10–6 m4

2.3 IyyT = Ixx2 + Iyy1

= 149,1 × 10–6 + 93,08 × 10–6

= 242,18 × 1 0 −6 m 4

2.4 Zmax = I xx ___ y max = 1 683 × 1 0 −6 ________ 0,3829

= 4 397,47 × 10–6

Zmin = I xx ___ y min = 1 683 × 1 0 −6 ________ 0,2371

= 7 101,6 × 1 0 −6

2.5 M __ Iv = σv __ yv

M = σI ___ y max = 120M × 1 683,792 × 1 0 −6 ________________ 0,3829

= 527,7 kNm



Mmax = MPL + Mweight

∴ Mmax = WL ___ 4 + W L 2 ___ 8

527,7 × 103 = W6 ___ 4 + (55,1 + 149)9,81 × 6 2 _____________ 8

W = 518 690,01 × 4 _________ 6

= 345,79 kN

3.

h = 50 __ 2 + 14,5 = 39,5

x • • • x

50 × 100

80 × 45 × 8,64 kg/m 50 kN

6 m

10 kN/mh

1

2 2

3.1 IxxT = 2Ixx2 + Ixx1

= (2 × 1,059 × 10–6) × 0,05 × 0, 1 3 _______ 12

= 6,285 × 10–6 m4

IyyT = Iyy1 + 2[Iyy2 + A2h22]

= 0,1 × 0,0 5 3 _______ 12 + 2[0,1936 × 10–6 + 1,102 × 10–3 × 0,03952] = 1,0417 × 10–6 + 3,826 × 10–6

= 4,8677 × 10–6 m4

3.2 Zxxmax = I xx ___ y max = 6,285 × 1 0 −6 ________ 0,05

= 1,257 × 1 0 −4 m 3

3.3 M = WL ___ 4 + w L 2 ___ 8

= 50 × 1 0 3 × 6 ________ 4 + 10k × 6 2 ______ 8

= 75k + 45k = 120 kNm

3.4 M __ I = σ _ y : σmax = M y max ____ I xx

= 120 × 1 0 3 × 0,05 __________ 6,285 × 1 0 −6

= 954,65 MPa



3.5 Zxx = M __ σ = 120k ________ 954,65 × 1 0 6

= 125,7 × 10–6 m3

(T.flange) 178 × 102 × 21,5 kg/m

Zexx = 170,2 × 10–6 m3 (nearest bigger value)

4.

σ = 80 MPa

7 kN/m

4 m

3 m

6 m

2 m

6 kN 3 kN

4.1 M = (3× 6) + (6 × 3) + ( 7 × 4 × 4 _ 2 ) = 18 + 18 + 56

= 92 kNm

∴ Zxx = M __ σ

= 92k ___ 80M

= 1 150 × 10–6 m3

406 × 178 × 67,2 kg/m (1 189 × 10–6 m3)

4.2 Zxx = M __ σ

∴ σ = M __ 2

= 92 × 1 0 3 ________ 1 189 × 1 0 −6

= 77,38 MPa Actual stress must be smaller than max stress. ∴ Select a section with a bigger nearest Z-value.



5.

• •

•

•

•

y2

173,4

y1

50,8

y3

265,6

y4

70 + 206,8 + 8 + 21,7 = 306,5

y

g

x x

70

191,46

141,14

h = 206,8 _ y

306,5+

26,1= 332,6

8 3

4

2

1

5.1 - y AT = Σ A-moments

No. A × 10–3 y Ay

1 2,797 0,0508 1,421 × 10–4

2 3,8 0,1734 6,5892 × 10–4

3 2,797 0,2656 7,429 × 10–4

4 2 × 1,107 0,3065 6,7859× 10–4

AT 11,608 Σ A-mom 2,2224818 × 10–3

∴ - y AT = Σ A-mom

- y 11,608 × 10–3 = 2,2224818 × 10–3

- y = 191,46 mm

5.2 Ixx = IT1 + IT2 + IT3 + IT4

∴ IT1 = Iyy1 + A1h12 (h1 = 191,46 – 50,8 = 140,66)

= 1,135 × 10–6 + 2,797 × 10–3 × 0,140662

= 5,6474 × 1 0 −5 m 4



IT2 = Ixx2 + A2h22 (h2 = 191,46 – 173,4 = 18,06)

= 28,88 × 10–6 + 3,8 × 10–3 × 0,018062

= 3,0119 × 1 0 −5

IT3 = Iyy3 + A3h32 (h3 = 265,6 – 191,46 = 74,14)

= 1,135 × 10–6 + 2,797 × 10–3 × 0,074142

= 1,6509 × 1 0 −5

IT4 = 2[Ivv4 + A4h42] (h4 = 306,5 – 191,46 = 115,04)

= 2[0,1479 × 10–6 + 1,107 × 10–3 × 0,115042]

= 2,9596 × 10–5 m4

∴ Ixx = IT1 + IT2 + IT3 + IT4

= [5,6474 + 3,0119 + 1,6509 + 2,9596]10–5

= 132,698 × 1 0 −6 m 4

IyyT = 2Ixxchannel + Iyyi-sec + 2[Iuu + A4h42]

IT4 = 2[Iuu + A4h42] h = 86 __ 2 = 43

= 2[0,5507 × 10–6 + 1,107 × 10–3 × 0,0432]

= 5,1951 × 10–6

∴ IyyT = (2 × 13,54 × 10–6) + 3,838 × 10–6 + 5,1951 × 10–6

= 36,113 × 1 0 −6 m 4

5.3 Zmax = I xx ___ y max = 132,698 × 1 0 −6 _________ 0,19146

= 693,08 × 1 0 −6 m 3

5.4 kxx = √ __

I xx __ A T

= √ __________

132,698 × 1 0 −6 _________ 11,608 × 1 0 −3

= 106,92 mm

kyy = √ __

I yg __ A T

= √ _________

36,113 × 1 0 −6 _________ 11,608 × 1 0 −3

= 55,78 mm



6. ρ = 1 100 kg/m3

590

•

•250

150

300

5

x

y yy2

xx x

76,87 = _ y

390

5 5

Y Y

1 12

h1

h2

h2

- y AT = Σ A-moments

- y (600 × 300) – (590 × 295) = (600 × 300 × 150) – (590 × 295 × 152,5)

5 950 - y = 457 375

- y = 76,87 mm

h1 = y1 – - y = 73,13

h2 = - y – y2 = 75,37

∴ Ixx = 2[I1 + A1h12] + I2 + A2h2

2

= 2 [ 0,005 × 0, 3 3 ________ 12 + 0,005 × 0,3 × 0,0731 3 2 ] + [ 0,59 × 0,00 5 3 ________ 12 + 0,59 × 0,005 × 0,0743 7 2 ] = 3,8544 × 10–5 + 1,6322 × 10–5

= 5,48663 × 10–5 m4

M = σIxx ___ y max ymax = 300 – 76,87 = 223,13

= 60M × 5,48663 × 1 0 −5 ______________ 0,22313

= 14,754 kNm

Weight of slime/m = Volρg

= A × L × ρg

= (0,59 × 0,25) × 1 × 1 100 × 9,81

= 1 591,6725 N/m

M = w L 2 ___ 8

∴ L = √ ___

8M ___ w

= √ ___________

8 × 14,754 × 1 0 3 __________ 1 591,6725

= 8,61 m



7. D = 200

d = 180

••

•

y

y

_ y

x x y2 176,1h2

h1

y1

6,1

276,1

134,8 73,05

141,3

↔

1

2

Channel:

h = 251,5

b = 146,1

7.1 gAT = Σ A-moment

A y Ay

1 3,992 × 10–3 73,05 2,9162 × 10–4

2 π _ 4 ( 0, 2 2 − 0,1 8 2 ) 5,969 × 10–3 176,1 1,051 × 10–3

AT 9,961 × 10–3 ΣM 1,3428 × 10–3

∴ - y AT = Σ A-moment

- y = 1,3428 × 1 0 −3 _________ 9,961 × 1 0 −3

= 134,8 mm

7.2 Ixx = I1 + A1h12 + I2 + A2h2

2

h1 = 134,8 – 73,05 = 61,75

h2 = 176,1 – 134,8 = 41,3

IT1 = I1yy + A1h12

= 4,476 × 10–6 + 3,992 × 10–3 × 0,061752

= 19,698 × 10–6 m4

IT2 = I2xx + A2h22

= π __ 64 (0,24 – 0,184) + π _ 4 (0,22 – 0,182) × 0,04132

= 2,701 × 10–5 + 1,018 × 10–5

= 37,19 × 10–6 m4



∴ Ixx = IT1 + IT2

= 19,698 × 10–6 + 37,19 × 10–6

Iyy = 56,888 × 1 0 −6 m 4 Iyy = Ixx1 + Iyy2

= 44,28 × 10–6 + 27,01 × 10–6

Iyy = 71,29 × 10–6 m4

7.3 kxx = √ __

I xx __ A T

= √ _________

56,888 × 1 0 −6 _________ 9,961 × 1 0 −3

= 75,57 mm

kyy = √ __

I yy __ A T

= √ ________

71,29 × 1 0 −6 ________ 9,961 × 1 0 −3

= 84,6 mm Smallest k-value = 75,57 mm

7.4 Zmax = M ___ σ max

= I xx ___ y max

= 56,888 × 1 0 −6 _________ 0,1413

= 402,604 × 10–6 m3

7.5 M __ I = σ _ y ∴ M = σIxx ___ ymax

M = 175 × 1 0 6 × 56,888 × 1 0 −6 ________________ 0,1413

= 70,46 kNm

7.6 M __ I = σ _ y

∴ I = M.y ___ σ

y = ? ∴ I = ?

∴ Z = M __ σ

= 70,46 × 1 0 3 _______ 175 × 1 0 −6

= 402,63 × 10–6 m3

∴ Z = I _ y

∴ π __ 64 ( D 4 − D 4 )

________ D __ 2 = Z

∴ π __ 64 ( D 4 − d 4 )

________ D __ 2 = 402,63 × 10–6



∴ D4 – d4 _____ D = 402,63 × 10–6 × 64 ___________ 2π

∴ 0,42 – d4 _____ 0,4 = 4,101 × 10–3

0,44 – d4 = 1,64 × 10–3

d4 = 0,44 – 1,64 × 10–3

= 0,02396

∴ d = 393,4 mm

8. Solid

Hollow

BS

Ds = 2Bs

DH

dH DH = 2dH

∴ MS = MH

∴ σZS = σZH (M = σZ)

∴ ZS = ZH (σS = σH)

∴ IS __ yS = IH __ yH

yS = DS __ 2 and yH = DH __ 2

∴ BSDS3 × 2 ______ 12DS

= π(D4 – d4) × 2 _________ 64DH

∴ BSDS2

____ 6 = π(DH4 – dH

4) ________ 32DH

Substitute and in :

∴ BS(2BS)2

_____ 6 = π[(2dH)4 – dH4] _________ 32 × 2dH

∴ 4BS3

___ 6 = π(16dH4 – dH

4) _________ 32(2dH)

∴ 2BS3

___ 3 = π × 15d3H ______ 32 × 2

∴ BS3 = 3π × 15d3

H _______ 32 × 4

∴ BS3 = 1,1045 dH

3

Take 3 √__

: ∴ BS = 1,0337 dH

Percentage saving:

A H − A S _____ A H × 100 ___ 1 ∴ ( 1 − A S __ A H ) ∴ [ 1 − 2 B 2 _______

π _ 4 (4 d 2 − d 2 ) ]

∴ [ 1 − 0,849 B 2 _____ d 2

]

Substitute in ∴ [ 1 − 0,849(1,0337d ) 2 __________ d 2

] ∴ (1 – 0,9072) × 100 ___ 1

∴ 9,28%



9.

12 kN/m2

6

6 m1,4

Pitch

σmax = 480

Acting safe stress = 450 ___ 4 = 120 MPa

Per pitch ∴ FT = W/ m 2 × A

= 12 × 103 × (1,4 × 6)

= 100,8 kN

∴ Load/m = 100,8 ____ 6

= 16,8 kN/m

∴ M = w L 2 ___ 8 = 16,8 × 6 2 ______ 8 = 75,6 kNm

∴ Z = M __ σ = 75,6 × 1 0 3 _______ 120 × 1 0 6

= 630 × 10–6 m3

Nearest 305 × 165 × 46,1 kg/m (647 × 10–6)

10.

σ _ y = E __ R ∴ y = 0,01 ___ 2 = 0,005 m and

R = 8 _ 2 + 0,01 ___ 2 = 4,005 m

∴ σ = E.y __ R

= 100 × 1 0 9 × 0,005 ___________ 4,005

10

8 mR

y = 124,84 MPa

11.

D = 250

d = 150

y

2,5 kN/m weight

21 kN

5 m

∴ I = π __ 64 (0,254 – 0,154) = 1,669 × 10–4 m4

∴ Mmax = σI __ y = 40 × 1 0 6 × 1,669 × 1 0 −4 _______________ 0,125

= 53,407 kNm

Mmax = MPL + Mweight + MUDL

∴ Mmax = WL ___ 4 + w L 2 ___ 8 + w L 2 ___ 8



∴ 53 407 = 21 × 5 ____ 4 + 2,5 × 5 2 _____ 8 + w 5 2 ___ 8

= 26 250 + 7 812,5 + w25 ___ 8

∴ w25 ___ 8 = 19 344,575

w = 6,19 kN/m (safe U.D. Load)

12.

D = 400

d = 380

15 m

y = 200d

D

12.1 I = π __ 64 (0,44 – 0,384) = 233,098 × 10–6 m4

∴ M = σI __ y = 40 × 1 0 6 × 233,098 × 1 0 −6 ________________ 0,2

= 46 619,6 Nm

∴ Mmax = w L 2 ___ 8 + w L 2 ___ 8 (Mmax = Mload + Mweight)

46 619,6 = 1 × 1 0 3 × 1 5 2 ________ 8 + w1 5 2 ____ 8

∴ 46 619,6 = 28,125 + w1 5 2 ____ 8

∴ w1 5 2 ____ 8 = 18 494,6

∴ Load/m = w = 657,59 N/m

But W = Vol ρ × g 657,59 = A × 1 × 1 200 × 9,87 Water area: ∴ A = 0,05586 m 2

12.2 Percentage area full:

Area of water ___________ Inner area of pipe × 100 ___ 1

∴ 0,05586 ______ π _ 4 × 0,3 8 2

× 100 ___ 1

= 49,25%


MODULE


Knowledge assumed to be in place before starting with this module:• How to calculate a moment• Module 6 of this book• Th e diff erent stresses• Good knowledge of Modules 1 to 5

Learning outcomes

When students have completed this module, they should be able to:• Know the diff erence between a column and a strut• Calculate the buckling loads for columns and struts• Calculate the maximum safe load for a strut and column


• Make sure they understand the basic concept second moment of area• Make sure they understand how to use the eff ective length for columns and struts• Always make decent sketches• Make a good study of the module and underline important facts

7 Columns and struts




1.

D = 210

d = 170 le = L = 3 m

1.1 PE = π 2 EI ___ le = π 2 × 210G × π __ 64 (0,21 0 4 − 0,1 7 4 )

___________________ 3 2

= 12,54 MN

1.2 PR = σ c A ______ 1 + a ( le __ R ) 2

∴ le = √ __

I __ A

= 300M × π _ 4 (0,2 1 2 − 0,1 7 2 ) _______________ 1 + 1 ____ 7 500 ( 3 _____ 0,06755 ) = √ __________

π _ 4 (0,2 1 4 − 0,1 7 4 )

__________ π _ 4 (0,2 1 2 − 0,1 7 2 )

= 2,836 MN = 0,06755 m

2. 2.1 PE = π 2 EI ___ l e 2

1,2 × 106 = π 2 × 200 × 1 0 9 I __________ 5 2

∴ Iyy = 15,198 × 10–6 m4

∴ 203 × 203 × 46,6 kg/m

le = L = 5 m

2.2 PR = σ c A ______ 1 + a ( le _ k ) 2

le = √ __

I __ A = 51,2

∴ PR = 300M × 5,882 × 1 0 −3 _____________ 1 + 1 ____ 7 500 ( 2,5 _____ 0,0512 ) 2

= 1,339 MN SR = le _ k = 2,5 _____ 0,0512

= 48,83:1

3. 3.1 PR = σ c A ______ 1 + a ( le _ k ) 2

∴ 1,5 × 106 = 80 × 1 0 6 A ________ 1 + 1 ____ 6 400 ( 2,5 __ k ) 2

∴ 1,5 × 106 = 80 × 1 0 6 A _________ 1 + 9,766 × 1 0 −4 ________

k 2

∴ 1 + 9,766 × 1 0 −4 ________ k 2

= 53,333 A

But k2 = I __ A π( D 4 − d 4 ) 4 _________ 64π( D 2 − d 2 )

= ( D 2 − d 2 )( D 2 + d 2 ) ____________ 16( D 2 − d 2 )

= D 2 + d 2 _____ 16 ∴ k2 = 0,0625 (D2 + d2)

le = 0,56 = 2,5 m


107Module 7 • Columns and struts

Substitute in ∴ 1 + 9,766 × 1 0 −4 __________ 0,0625( D 2 + d 2 )

= 53,333 × π _ 4 (D2 – d2)

∴ 1 + 0,01562 ______ ( D 2 + d 2 )

= 41,888 (D2 – d2)

× (D2 + d2): ∴ (D2 + d2) + 0,01562 = 41,888 (D4 – d4)

∴ D2 + d2 + 0,01562 = 41,888 D4 – 41,888 d4

∴ 0,262 + d2 + 0,01562 = 41,888 (0,264) – 41,888 d4

∴ d2 + 0,08322 = 0,1914 – 41,888 d4

∴ 41,888 d4 + d2 – 0,1082 = 0

41,888 d4 + d2 – 0,1082 = 0 d4 – x2 d = x ∴ 41,888 x2 + x – 0,1082 = 0

∴ x = −1 ± √ ____________________

1 2 − 4(41,888)(−0,1082) ___________________ 2(41,888)

= − 1 ± 4,374 ________ 83,776

= 3,374 _____ 83,776

x = d2 = 0,04027 ∴ d = 200,67 mm

3.2 PE = π 2 EI ___ le = π 2 85 × 1 0 9 ( π __ 64 [ 0,2 6 4 − 0,2006 7 4 ] )

____________________ 2, 5 2

= 19,424 MN

4.

y

y y

y

_ y x x

y2 = 206,05

y1 = 101,51

2

254 × 146 × 31,6

Taper203 × 152 × 52,1

- y AT = Σ A-moments

No. A y Ay

1 6,641 × 10–3 0,101,5 6,740615 × 10–4

2 3,992 × 10–3 0,20605 8,225516 × 10–4

AT 0,010633 Σ A-mom 1,4966131 × 10–3

(D2 + d2)(D2 – d2) = (D4 – d4)



4.1 ∴ - y .0,010633 = 1,4966131 × 10–3

- y = 140,75 mm

Ixx = Ixx1 + A1h12 + Iyy2 + A2h2

2

= (47,76 × 10–6 + 6,641 × 10–3 × 0,039252) + 4,476 × 10–6 + 3,992 × 10–3 × 0,06532

= 79,46 × 10–6 m4

Iyy = Ixx2 + Iyy2

= 44,28 × 10–6 + 8,098 × 10–6

= 52,378 × 1 0 −6 m 4

4.2 S.R. k = √ __

I yg __ A T = √

_________ 52,378 × 1 0 −6 _________ 0,010633 = 0,07019 m

le _ k = 0,707 × 4,5 _______ 0,07019 = 45,33:1

4.3 PR = σA ______ 1 − a ( le _ k ) 2

= 280 × 1 0 6 × 0,010633 _____________ 1 + 1 ____ 7 500 ( 0,707 × 4,5 _______ 0,07019 ) 2

= 2,34 MN

5. 5.1 Iuu = 2(45,29 × 10–6)

= 90,58 × 10–6 m4

Ivv = 2[I + Ah2]

= 2[11,72 × 10–6 + 7,635 × 10–3 × 0,07152]

= 101,5 × 10–6 m4

u

u

x x

xxh

y

y

y

y

v

v

v

v

••

•X X

Y

Y

Ixx = Iyy = 2[I + Ah2]

= 2[28,71 × 10–6 + 7,635 × 10–3 × 0,04322]

= 85,917 × 1 0 −6 m 4

∴ Bending about xx-axis or yy-axis is the weaker axis.

∴ k = √ __

I __ A = √ ___________

85,917 × 1 0 −6 __________ 2 × 7,635 × 1 0 −3

= 0,075 m

5.2 PR = σ c A ______ 1 + a ( le _ k ) 2

2 × 106 = σ c 2 × 7,635 × 1 0 −3 ____________

1 + 1 ____ 7500 ( 0,5 × 4 _____ 0,075 ) 2

= σc 0,01395

σc = 143,39 MPa

Must check all possible axes for bending to find the smallest.



5.3 σsafe = σC ___ FOS = 143,39 _____ 4 = 35,847 MPa

5.4 PE = PR

π 2 EI ___ l e

2 = σ c A ______

1 + a ( le _ k ) 2

∴ π 2 × 200G × A k 2 ___________ l e

2 = σ c A _______

1 + a ( SR ) 2

÷ A: π 2 × 200G _______ (SR ) 2 = 143,39 × 1 0 6 ________

1 + S R 2 ____ 7 500

∴ 143,39 × 106 (SR)2 = 1,974 × 1012[1 + 1,33 × 10–4(SR)2]

= 1,974 × 1012 + 2,632 × 108(SR)2

1,198 × 108(SR)2 = 1,974 × 1012

∴ SR = √ _________

1 6477,538

= 128,36

6. le = 0,707 × 6 = 4,242 PE = π 2 EI ___

l e 2

3,5 × 106 = π 2 × 200G × I _________ 4,24 2 2

I = 31,91 × 1 0 −6 m 4

w w

y y

203 × 203 × 53,5 kg/m

150 150

Iyy = Iyy × 2 [ w × 0,1 5 3 ______ 12 ] 31,91 × 10–6 = 16,78 × 10–6 + 2(2,8125 × 10–4w)

= 2(2,8125 × 10–4w)

w = 26,89 mm

7. L = 4 m D = 80 PE = 150 kN E = 200 GPa

7.1 PE = π 2 EI ___ l e

2 ∴ 150k = π 2 EI ___

4 2

∴ Iyy = 1,216 × 10–6 m4



Iyy = π __ 64 (D4 – d4) ∴ 1,216 × 10–6 m4 = π __ 64 ( 0,0 8 4 − σ _ 4 ) ∴ d4 = 0,086 – 2,477 × 10–5

= 1,619 × 10–5

d = 63,43 mm

t = D − d ____ 2 = 8,285 mm

7.2 PR = σA ______ 1 + a ( le _ k ) 2

150k = σ π _ 4 (0,0 8 2 − 0,0634 3 2 ) _____________

1 + 1 ____ 7 500 ( 4 _ k ) 2

= σ1,8667 × 1 0 −3 __________ 1 + 1 ____ 7 500 ( 4 _____ 0,0255 ) 2

k = √ __

I yg __ A

= σ1,8667 × 1 0 −3 __________ 4,275 = √ ________

1,216 × 1 0 −6 ________ 1,877 × 1 0 −3

∴ σ = 343,5 MPa = 0,0255

∴ FOS = 350 ____ 343,5 = 1,018

8.

•

•

y

y

65 h

2

2

1

20

8.1 Ixx = Ixx1 + 2[Ixx2 + A2h22]

∴ Ixx = 0,02 × 0,0 5 3 ________ 12 + 2 [ π __ 64 (0,0 8 4 − 0,0 6 4 ) + π _ 4 (0,0 8 2 − 0,0 6 2 )0,06 5 2 ] = 2,083 × 10–7 + 21,33 × 10–6

= 21,54 × 10–6 m Iyy = Iyy1 + 2Iyy2

Iyy = 0,05 × 0,0 2 3 ________ 12 + 2 [ π __ 64 (0,0 8 4 − 0,0 6 4 ) ] = 2,782 × 10–6 m

PE = π 2 EI ___ l e

2

= π 2 × 209G × 2,782 × 1 0 −6 ________________ (3,2 × 0,5 ) 2

= 2,242 MN

OD = 80ID = 60

Must check both axes to find the smallest one.



σ = F __ A = 2,242 × 1 0 6 _______________________ 2 × [ π _ 4 (0,0 8 2 − 0,0 6 2 ) ] + (0,05 × 0,02)

= 415,32 MPa

8.2 S.R = le _ k

k = √ __

Iyy __ A

= √ ________

2,782 × 1 0 −6 ________ 5,398 × 1 0 −3

= 0,0227 m S.R = le _ k = 3,2 × 0,5 ______ 0,0227

= 70,48:1

9. PR = σA ______ 1 + a ( le __ k ) 2

k = √ __

Iyy __ A = √

_______

0,09 × 0,0 3 3 ________ 12 _______ 0,09 × 0,03

= √ ____

0,0 3 3 ____ 12

= 8,66 × 10–3 m

∴ 100 × 103 = 130 × 1 0 6 × 0,09 × 0,03 ______________ 1 + a 2,1 _______

8,66 × 1 0 −3

∴ 100 × 103 = (1 + 58 800a) = 351 000

1 + 58 800a = 3,51

∴ 58 800a = 2,51

a = 1 _______ 23 426,295

= 4,269 × 1 0 −5

10.

Le = L = 12 m

σLIM = 100 MPa

Iyy = 80% Ixx

x = 1 ___ 200

Ixx = 2Ixx2 + (Iyy1 + A1h1

2)

= 2(80,28 × 10–6) + 2 ( 0,3 × 0,01 5 3 ________ 12 + 0,3 × 0,015 × 0,157 5 2 ) = 383,985 × 10–6 m4

30 y

y

X X 90

y

y

x x

300

100 15

29

15

300

•

•

•

• •h

h

1

1

2 2



∴ Iyy = 0,8 × 383,985 × 10–6

= 307,188 × 10–6 m4

A = 2(0,3 × 0,15) + 2(5,876 × 10–3)

= 0,020752 m2

∴ k = √ __

Iyy __ AT

= √ _________

307,188 × 10–6 _________ 0,020752 = 0,122

10.1 PS = σCA(1 – x[SR])

= 100M × 0,020752 (1 – 1 ___ 200 [ le __ k ])

= 2 075 200 (1 – 1 ___ 200 [ 12 ____ 0,122 ])

= 2 075 200 – 1 020 590

= 1,055 MN

10.2 PR = σCA ______ 1 + a[ le __ k ]2

= 100M × 0,020752 ___________ 1 + 1 ____ 7 500 [ 12 ____ 0,122 ]2

= 906,22 kN

10.3 PE = π2EIyy ____ le2

= π2 × 200G × 307,188 × 10–6

_________________ 122

= 4,211 MN


MODULE


Knowledge assumed to be in place before starting with this module:• Module 1 shear stress• Torque and power• Engineering Science N4• A very good knowledge of Modules 1 to 7

Learning outcomes

When students have completed this module, they should be able to:• Recognise the two limits of a shaft transmitting power and torque• Calculate the power and torque transmitted by compound shaft s• Calculate maximum safe shear stress for a shaft


• Make sure they understand the basic concept of the torque equation• Make sure they understand compound shaft s• Always make decent sketches• Make a good study of the module and underline important facts

Shafts8




1. L = 2 m D = ? T = 20 kNm G = 80 GPa τ = 60 mPa θ = 2°

1.1 T __ J = τ __ R = Gθ __ L Shear stress: T __ J = τ 2 __ D

J = TD ___ τ 2

= 20 × 1 0 3 × D ________ 60 × 1 0 6 × 2

π __ 32 D 4 = 1,667 × 10–4 D

∴ D3 = 1,6977 × 10–3

D = 119,3 mm

Angle of twist: T __ J = Gθ __ L

J = TL __ Gθ = 20 × 1 0 3 × 2 × 188 ____________ 80 × 1 0 9 × 2 × π

∴ π __ 32 D 4 = 1,432 × 10–5

D4 = 1,459 × 10–4

D = 109,9 mm

User D = 109,9 θ = 2° and τ < 60 mPa

1.2 P = 2πNT

= 2π × 900 × 20 _________ 60

= 1 884,96 kW

= 1,885 MW

2. P = 2 MW @ 400 r/min 13% D = 100 d = 90

L = 2 m G = 80 GPa

2.1 T __ J = τ __ R = Gθ __ L

Tmax = P ___ 2πN = 2,01 0 3 × 60 _______ 2π400

= 477,465 Nm

Tmax = 477,465 × 1,13

= 539,54 Nm

∴ τ = T × R ____ J = TD __________ π __ 32 ( D 4 − d 4 ) × 2

= 539,54 × 0,1 ____________ π __ 32 (0, 1 4 − 0,0 9 4 ) × 2

= 7,99 say 8 MPa


115Module 8 • Shafts

2.2 Angle of twist ∴ T __ J = Gθ __ L

∴ θ = TL __ JG

∴ θ = 539,54 × 2 _______________ π __ 32 (0, 1 4 − 0,0 9 4 )80 × 1 0 9

= 0,004 rad. = 0,229°

3. T __ J = τ __ R = Gθ __ L D = 2,5d

3.1 T __ J = τ __ R

Stress: J = TR __ τ = 20 × 103 × D __ 2 × 1 ______

75 × 1 0 6

π __ 32 D 4 = 1,331 × 10–4 D ∴ D3 = 1,358 × 10–3

∴ D 4 − d 4 _____ D = 1,358 × 10–3

∴ (2,5d ) 4 − d 4 ________ 2,5d = 1,358 × 10–3

15,225d 3 = 1,358 × 10–3

d 3 = 8,92 × 10–5

d = 44,68 mm ∴ D = 111,7 mm

θ of twist T __ J = Gθ __ L

J = TL __ Gθ

= 20 × 1 0 3 × 1,5 × 180 _____________ 80 × 1 0 9 × 2,1 × π

π __ 32 ( D 4 − d 4 _____ D ) = 1,023 × 10–5

∴ (2,5d ) 4 − d 4 ________ 2,5d = 1,04216 × 10–4

15,225d 3 = 1,04216 × 10–4

d 3 = 6,845 d = 19 mm and D = 47,5 mm Use d = 44,68 mm D = 111,7 mm ∴ τ = 75 MPa θ < 1,5°

IMPORTANT For Question 3.1, you must check both limits to see which is the strongest.



3.2 Tmean = T max ___ 1,12

= 20 × 1 0 3 ______ 1,12

= 17,86 kNm

P = 2πNTmean

= 2π 700 ___ 60 × 17,86 × 103

= 1,31 MW

4. D = 160 G = 85 GPa L = 3 m θ = 1,1° N =400 r/min

4.1 τ __ R = T __ I

∴ τ = RT __ J

T __ J = Gθ __ L

T = GθJ ___ L

= 85G ___ 3 × 1,1 × π _____ 180 × π0,1 6 4 _____ 3 2

= 34,998 kNm

∴ τ = RT __ J

= 0,8 × 34,998 ________ π __ 32 0,1 6 4

= 43,52 MPa 4.2 Power = 2πNT = 2π × 400 ___ 60 × 34,998 ` = 1,466 MW

4.3 PH = 1,15 PS

= 1,15 × 1,466 = 1,6859 MW

D = 2d

TH = P H ___ 2πN

= 1,6859 × 60 ________ 2π × 400

= 40,248 kNm T = π __ 16 ( D 4 − d 4 _____ D ) τ

Sub. in : ∴ 40 248 = π __ 16 ( (2d ) 4 − d 4 _______ 2d ) 43,52 × 106

7,5d 3 = 4,71 × 10–3

d 3 = 6,28 × 10–4

d = 85,63 mm D = 171,27 mm



5. 5.1 TB = 1,45 TS TT = T1 + T2

∴ θB = θS θ1 = θ2

∴ T B L B ___ J B G B = T S L S ___ J S G S

(LB = LS)

∴ 1,45 ___________ π __ 32 ( D 4 − 0,0 7 4 ) 32

= 1 ________ π __ 32 0,0 7 4 × 80

Bronze

Steel θ 70

∴ 1,45 × 0,074 × 80 = 32(D4 – 0,074)

8,704 × 10–5 = D4 – 0,074

D4 = 6,303 × 10–5

D = 89,1 mm

5.2 TS = π __ 16 D 3 τ

= π __ 16 × 0,073 × 90 × 106

= 6,061 kNm

TH = π __ 16 ( D 4 − d 6 _____ D ) τ = π __ 16 ( 0,0891 − 0,0 7 6 _________ 0,0891 ) 55 × 106

= 4,729 kNm

(A) TH = 1,45 TS But (B) ∴ TS = T H ___ 1,45

= 6,061 × 1,45 = 4,729 ____ 1,45

= 8,788 kNm = 3,26 Nm

∴ Ttotal = TH + TS

= 4,729 + 3,26

= 7,989 kNm

5.3 Tmean = 7,989 ____ 1,12

= 7,133 kNm

P = 2πNT ____ 60 = 2π × 900 × 7,133 ___________ 60

= 672,269 kW

6. D = 120 d = ?

T = 20 kNm θ > 2°

G = 80 GPa θT = θ1 + θ2

T1 = T2

TS = TH = 20 kNm 1,2 m1 m

d = 80



π __ 16 D 3 τ = π __ 16 D 4 − d 4 _____ D τ

803 = 12 0 4 − d 4 ______ 120

61,44 × 106 = 1204 – d4

d 4 = 1204 – 61,44 × 106

= 1,4592 × 108

d = 110 mm θT = θ1 + θ2

2 × π ____ 180 = T H L H

____ J H G H = T S L S ___ J S G S

2 × π ____ 180 = 20k × 1,2 ______ J H 80G = 20k × 1 _____ J S 80G

∴ × 80G and ÷ 20k

∴ 139 626,34 = 1,2 __ J H + 1 __ J S

139 626,34 = 1,2 _________ π __ 32 (0,1 2 4 − d 4 )

+ 1 _____ π __ 32 0,0 8 4

–109 053,2 = 1,2 × 32 ________ π(0,1 2 4 − d 4 )

–8 921,89 = 1 _______ 0,1 2 4 − d 4

∴ 0,124 – d 4 = 1 _______ − 8 921,89

= –1,121 × 10–4

∴ d 4 = 0,124 + 1,121 × 10–4

d 4 = 3,194 × 10–4

d = 133,69 mm

7. d = 0,66 D

∴ VS = VH (LS = LH) ∴ AS = AH

∴ D12 = D2 – d 2

Substitute in : ∴ DS

2 = D2 – (0,66D)2

= 0,5644 D2H

Take √ __

: ∴ DS = 0,7513 DH

TS : TH

∴ DS3 : D H 4 − d H 4

_______ D H



Substitute and in :

(0,7513 DH)3 : D H 4 − (0,66D ) 4 __________ D H

0,424 DH3 : 0,8103 DH

3

∴ 1 : 1,911

8. T = π __ 16 d 3 τ

8.1 = π __ 16 × 0,133 × 69 × 106

= 29,765 kNm

P = 2πNT = 2π × 140 ___ 60 × 29,765

= 436,381 kW 8.2 TS = TH

∴ 29 765 = π __ 16 ( D 4 − d 4 _____ D ) 84 × 106

1,805 × 10–3 = 0,1 3 4 − d 4 _______ 0,13

2,346 × 10–4 = 0,134 – d 4

d 4 = 5,1 × 10–5

d = 84,51 mm

9. 9.1 Solid T __ θ = TGJ ___ TL

= 30 × 1 0 9 × π0,00 8 4 ____________ 0,22 × 32

= 54,84 Nm/rad.

Hollow T __ θ = TGJ ___ TL

= 40 × 1 0 9 × π(0,01 2 4 − 0,0 1 4 ) _________________ 0,22 × 32

= 191,64 Nm/rad.

9.2 θS = θH TT = T1 + T2

But θS = T S __ θ S = 54,84 θ1 = θ2

∴ θS = T S ____ 54,84

T H __ θ H = 191,64

Bronze

Al

∴ θH = T H _____ 191,64

Substitute and in

∴ T S ____ 54,84 = T H _____ 191,64

∴ TH = 3,495 TS

TT = TH + TS



∴ 15 = 3,495 TS + TS

TS = 3,34 Nm

∴ TH = 11,66 Nm

9.3 TH = π __ 16 ( D 4 − d 4 _____ D ) τ 11,66 = π __ 16 ( 0,01 2 4 − 0,0 1 4 _________ 0,012 ) τ τHd = 66,38 MPa TS = π __ 16 D 3 τ 3,34 = π __ 16 0,00 8 3 τ ∴ τsd = 33,22 MPa

9.4 θSd = T S ____ 54,84

= 3,34 ____ 54,84 = 0,061 rad.

θHd = 11,66 _____ 191,64 = 0,061 rad.

0,061 × 180 ___ π =3,5°10. Gs = 2,5 GH TT = T1 + T2

TH = 3TS θ = θ

10.1 θS = θH

∴ T S L S ___ J S G S = T H L H

____ J H G H

θ 60

÷ TS × GH: ∴ T S _____ J S 2,5 G H = 3 T S ____ J H G H

1 ________ π __ 32 0,0 6 4 × 2,5 = 3 _________

π __ 32 ( D 4 − 0,0 6 4 )

3 × 0,064 × 2,5 = D4 – 0,064

D4 = 1,1016 × 10–4

D = 102,45 mm

10.2 TH = π __ 16 ( D 4 − d 4 _____ D ) τ = π __ 16 ( 0,1024 5 4 − 0,0 6 4 ___________ 0,10245 ) 49M

= 9,129 kNm

TS = π __ 16 D 3 τ

= π __ 16 × 0,063 × 84M

= 3,563 kNm

TH = 3TS

= 3 × 3,563 = 10,689

10,689 > 9,129

Value of the ratio TH is bigger than the allowable value of 9,129 kNm.

Therefore, 10,689 kNm cannot be used and will damage the shaft.



∴ TS = T H __ 3 = 9,129 ____ 3

= 3,043 kNm

∴ TT = TH + TS

= 9,129 + 3,043

= 12,172 kNm

10.3 P = 2πNT ____ 60

= 2π × 400 ___ 60 × 12,172

= 509,86 kW

11. d = 30 D = 60 mm T __ S = τ __ R = Gθ __ L

11.1 T = π __ 16 ( D 4 − d 4 _____ D ) τ ∴ 5 × 103 = π __ 16 ( 0,0 6 4 − 0,0 3 4 ________ 0,06 ) τ ∴ τ = 125,75 MPa

11.2 Shear strain δ

δ = θR __ L

= 1 × π × 0,03 ________ 180 × 1

= 5,236 × 10–4

11.3 T __ J = Gθ __ L = τ __ R

Any one

∴ G = τL __ Rθ

= 125,75 × 1 0 6 × 1 × 180 ______________ 0,03 × 1 × π

= 240,16 GPa

11.4 Tmax = 5 __ 1,2 = 4,167 kNm

∴ P = 2πNT

= 2π 200 × 4,167 _________ 60

= 87,27 kW

12. P = 7,5 MW @ 100 r/min D = 2d d = 200

θ = 3° 80 GPa

12.1 P = 2πNT

T = 7,5 × 1 0 6 × 60 _________ 2π × 100

= 509,86 kW



T __ J = τ __ R = 716,197 MNm

∴ T = π __ 16 ( D 4 − d 4 _____ D ) τ 716,197 = π __ 16 ( 0, 4 4 − 0, 2 4 _______ 0,4 ) τ = 60,79 MPa

12.2 T __ J = τ __ R = Gθ __ L

Any one in this case:

∴ Gθ __ L = τ __ R

∴ L = GθR ___ τ

= 80 × 1 0 9 × 3 × π × 0,2 ______________ 60,79 × 1 0 6 × 180

= 13,78 m

OR Gθ __ L = T __ J

L = GθJ ___ T

= 80 × 1 0 9 × π × 3π( D 4 − d 4 ) _________________ 716 197 × 180 × 32

= 13,78 m

13. dS = 30 mm D = 45 mm ValS = ValH A S Lρg = A H Lρg ∴ AS = AH

∴ dS2 = D2 – dH

2

∴ 302 = 452 – dH2

∴ dH2 = 1 125

dH = 33,54 m

C H __ C S = G J H

___ L × L ___ G J S

= J H __ J S

= D 4 − d 4 _____ D 4 S

= 4 5 4 − 33,5 4 4 ________ 3 0 4

∴ CS = 0,286 CH

14. T = P ___ 2πN = 500k × 60 ______ 2π120

= 39,789 kNm

14.1 Tmax = 39,789 × 1,14

= 45,359 kNm

T __ J = τ __ R = Gθ __ L

All values are available, so any combination of the bending equation can be used to find L.



D = 2d

Stress: T = Jτ __ R

45 359 = J × 55 × 1 0 6 × 2 __________ D

D = 2 425,0891 J

= 2 425,0891 × π __ 32 (D4 – d 4)

∴ 1 = 238,083 ( D 4 − d 4 _____ D ) ÷ 238,083 ∴ 4,2002 × 10–3 = ( 2 d 4 − d 4 ______ 2d ) = 7,5 d3

d 3 = 5,6 × 10–4

d = 82,43 mm D = 164,86 mm

T __ J = Gθ __ L

J = TL __ Gθ

= 45,359 × 4 × 180 ___________ 80 × 1 0 9 × 1,1 × π

π __ 32 ( D 4 − d 4 ) = 1,181 × 10–4

(2d)4 – d 4 = 1,203 × 10–3

d 4 = 8,022 × 10–5

d = 94,64 mm D = 189,28 mm (usae this shaft)

14.2 δ = θR __ L

= 1,1 × π × 0,18928 ___________ 180 × 4 × 2

= 4,542 × 10–4

15. T = 120 kNm @ 150 r/min

τ = 75 MPa

15.1 Tmax = 120 × 1,1 = 132 kNm T __ J = τ __ R ∴ T = π __ 16 ( D 3 ) τ

∴ 132k = π __ 16 D3 × 75 × 106

D3 = 8,964 × 10–3

D = 207,73 mm

15.2 d = 0,65 D

Must check both, as each is a limit in the shaft: τ shear stress θ angle of twist



D3 = D 4 − d 4 _____ D

∴ 8,964 × 10–3 = D 4 − 0,6 5 4 D 4 _________ D

= 0,8215 D3

D3 = 0,0109 D = 221,8 mm d = 144,17 mm

15.3 A S − A H _____ A S

= 207,7 3 2 − (221, 8 2 − 144,1 7 2 ) __________________ 207,7 3 2

= 34,16%

16. PH = 1,18 PS

2π(1,5NS)TH = (2πNSTS)1,18

∴ TH = 0,787 TS

VH = 0,75 VS

∴ AH = 0,75 AS

D2 – d2 = 0,75 × 0,142

D2 – d2 = 0,0147

∴ d2 = D2 – 0,0147

τH = τS

∴ 16 T H D _______

π( D 4 − d 4 ) = 16 T S ____

π D S 3

Substitute in : ∴ 0,787D _____ D 4 − d 4

= 1 ____ 0,1 4 3

∴ 2,1586 × 10–3 D = D4 – d 4

Square : ∴ d

4 = (D2 – 0,0147)2

= D4 – 0,0294 D2 + 2,161 × 10–4

Substitute in : ∴ 2,1586 × 10–3 D = D4 – (D4 – 0,0294 D2 + 2,161 × 10–4) = D 4 − D 4 + 0,0294 D2 – 2,161 × 10–4 ∴ 0,0294 D2 – 2,1586 × 10–3 D – 2,161 × 10–4 = 0 ÷ 0,0294 ∴ D2 – 0,0734D – 7,35 × 10–3 = 0

D = 0,0734 ± √ _______________________

0,07342 − 4(1)(−7,35 × 1 0 −3 ) ________________________ 2

= 0,0734 ± 0,1865 __________ 2 D = 129,96 mm



From : D2 = d 2 + 0,0147

0,129962 = d 2 + 0,0147

∴ d = 46,8 mm

17. GB = 42 GPa Gal = 31 GPa D = 40 d = 32

17.1 CB = GJ __ L = 42G × π __ 32 0,0 3 4 _________ 0,5

= 6,68 kNm/rad. θ = θ

CAL = GJ __ L TT = T1 + T2

= 31G × π __ 32 (0,0 4 4 − 0,03 2 4 ) _______________ 0,5

θ 30 s

H

C = 500 = 9,2 kNm/rad.

∴ Total C = 15,88 kNm/rad.

17.2 TT = TB + Tal

600 = J B G B θ ____ L + J al G al θ ____ L

= θ[CB + Cal] ( JG __ L = C ) = θ × 15,88k

θ = 3,778 × 10–2 rad. = 2,16°

17.3 τB = RGθ ___ L = 0,015 × 42G × 0,03778 ______________ 0,5 = 47,6 MPa

τal = RGθ ___ L = 0,02 × 31G × 0,03778 _____________ 0,5 = 46,85 MPa

17.4 P = 2πNT = ωT = 82 × 600 = 49,2 kW

18. θT = θS + θH

TS = TH

θ = 2°

G = 82 GPa

T = 20 kNm

18.1 θT = TS + TH

= T L S ___ G J S + T L H

___ G J H

∴ 3 × π ____ 180 = 20k × 0,8 _________ 82G × π __ 32 0,0 8 4

+ 20k L H ______________

82G × π __ 32 (0, 1 4 − 0,0 7 4 )

∴ 0,05236 = 0,04852 + 3,8399 × 10–3 = 0,03269 LH

∴ LH = 117,45 mm

L = 800

ф 80

L

ф = 70ф 100



18.2 TS = TH ∴ C = T __ θ = 20k _____ 3 × π ___ 180

= 381,982 kNm (rad.)

19. G = 84 GPa T = T = T θT = θ1 + θ2 + θ3

1 2

3

150 250150

D ф 80ф 70D = 80

19.1 CH = C S 3

GJ __ L = GJ __ L

π __ 32 (0,0 8 4 − 0,0 7 4 )

___________ 0,15 = π __ 32 D 4 )

____ 0,25

∴ D4 = 0,25(0,0 8 4 − 0,0 7 4 ) ____________ 0,15 D = (2,825 × 10–5)0,25

= 72,9 mm

19.2 θT = θ1 + θ2 + θ3

= τ L ___ GR + τ L ___ GR + τ L ___ GR

= τ __ G [ L 1 __ R 1 + L 2 __ R 2

+ L 3 __ R 3 ]

= 60M ___ 84G [ 0,25 _____ 0,0729 + 0,15 ___ 0,04 + 0,15 ___ 0,04 ] = 0,0103 rad. = 0,588°

T3 = τ J 3 __ R 3 = 60M × π __ 32 0,072 9 4 × 2

_____________ 0,0729 = 4,564 kNm

T1 = τ J 1 __ R 1 = 60M × π __ 32 (0,0 8 4 − 0,0 7 4 ) × 2

_________________ 0,08 = 2,496 kNm

∴ Max allowable torque = 2,496 kNm T1 = T2 = T3

19.3 P = 2πNT = ωT = 20π × 2 496

= 156,828 kW

19.4 ω = 2πN ∴ 20π = 2π N

∴ N = 10 r/s

= 10 × 60 = 600 r/min

Note for Lecturers:Questions 13, 17, 18 and 19 are not in the syllabus and may be left out.


MODULE


Knowledge assumed to be in place before starting with this module:• Solving of simultaneous equations• Equilibrium conditions• Newton’s third law• What a vector is• What the components of a force are• Engineering Science N4

Learning outcomes

When students have completed this module, they should be able to:• Solve two unknown forces acting on an object• Calculate the equilibrants and resultant forces• Perform graphical and analytical solutions for three or more forces


• Make sure they understand the basic concept of forces in equilibrium• Make sure they know the equilibrium conditions• Always make decent sketches• Make a good study of the module and underline important facts

Forces9




1.

45°E cos θ

E sin θ

80°

80 sin 80°

15 sin 45°

15 cos 41°

80 cos 80°

40

20

θ

Σ H.C. = 0

∴ – E cos θ – 80 cos 80° + 40 + 15 cos 45° = 0

∴ – E cos θ – 13,892 + 40 + 10,607 = 0

∴ – E cos θ = –36,715 N

Σ V.C. = 0

∴ – E sin θ + 80 sin 80° – 15 sin 45° + 20 = 0

∴ – E sin θ + 78,785 – 10,607 + 20 = 0

∴ – E sin θ = –88,178 N

θ

VC

E

88,178

36,715

θ

VC

HC

∴ E = √ _______________

36,71 5 2 + 88,17 8 2

= 95,516 N

Tan θ = 88,178 _____ 36,715 ∴ θ = 67,39°

E = 95,516 N W 67,39 S

∴ Equilibrium force = 95,516 N E 67,39° S


129Module 9 • Forces

2.

50 cos 35°

150 sin 30°

150 cos 30°

E cos θ

E sin θ50 sin

90

100

35 θ

Σ H.C. = 0 ∴ – E cos θ + 50 cos 35° + 90 – 150 cos 30° = 0 ∴ – E cos θ + 40,958 + 90 – 129,904 – E cos θ = –1,054 Σ V.C. = 0 ∴ E sin θ + 100 – 150 sin 30° + 50 sin 35° = 0 ∴ E sin θ + 100 – 75 + 28,679 = 0 ∴ – E sin θ = –53,679

θ

E

53,679

1,054

E = √ ______________

53,67 9 2 + 1,05 4 2

= 53,689 N

tan θ = 53,679 _____ 1,054 ∴ θ = 88,88°

Resultant force = E

∴ E = 53,689 W 88,88°S

3.

Q

P 56°

56°120 N

120

Q

P

∴ cos 56° = P ___ 120 ∴ P = 120 cos 56°

P = 67,1 N



∴ sin 56° = Q __ 20 ∴ Q = 120 sin 56°

Q = 99,48 N

70°

P

P

200 N

Q

200

70Q

∴ sin 70° = 200 ___ P

∴ P = 200 _____ sin 70° = 212,84 N

tan 70° = 200 ___ Q

∴ Q = 200 _____ tan 70° = 72,79 N

4. For Figure 9.21

θ

E

E cos θ

E sin θ90°

20 N

30°

30°

200

46 N

60

30°

20 N 20 sin 30° 20 cos 30°

20 sin 30°

20 cos 30°

E cos θ θ

E sin θE

46 N

∴ Σ V.C. = 0

+ E sin θ + 20 sin 30° = 0

+ E sin θ = 10 N

Σ H.C. = 0

∴ – E cos θ + 46 – 20 cos 30° = 0

∴ – E cos θ = –28,69 N



∴ E = √ ___________

28,6 9 2 + 1 0 2

= 30,38 N

θ

10

E

28,69

E

θ

10

28,69

tan θ = 10 ____ 28,69 θ = 19,22°

Reaction = 30,38 N W19,22° N

For Figure 9.22

30°

85°

62 N 62 sin 30°62 cos 30°

80 cos 85°

80 sin 85°

θ

E sin θ

E sin θE

E cos θ

80 N

Σ V.C. = 0 E sin θ + 80 sin 85° – 62 sin 30° = 0 ∴ E sin θ + 79,7 – 31 = 0 ∴ E sin θ = –48,7 N

Σ H.C. = 0 ∴ E cos θ – 80 cos 85° – 62 cos 30° = 0 ∴ E cos θ – 6,97 – 53,69 = 0 E cos θ = +60,66 N

True position

E = √ ____________

60,6 6 2 + 48, 7 2

= 77,79 N

tan θ = 48,7 ____ 60,66

∴ θ = 38,76°

θ

48,7

є

60,66

∴ E = 77,79 N E 38,76° S

85°

30°

62 N

G

80 N



5. For Figure 9.23

B A

C

P45° 70°

120 N

Q

Scale: 1 cm = 20 kN

Q = bc = 46 kN

P = ca = 96 kN

See text book supply.

b

c

P

Q

120 N

a

For Figure 9.24

Scale: 1 cm = 20 kN

120 N

B A

C

Q P

P = ac = 74 kN

Q = bc = 74 kN

Q

120

a

b

c

P



6. For Figure 9.25

Scale: 1 cm = 10 N

100 N

80 N

A

B

C

D

65°F

P

F = ad = 92 N

P = cd = 25 N

b

c

da

7. For Figure 9.26

A

BC

D

EF

p

F 80 N

100 N60 N

50 N



F = fa = 116 N

P = ef = 50 N

80

60

50

100

ab

c

d

e

P

f


MODULE


Knowledge assumed to be in place before starting with this module:• Solving of simultaneous equations• Equilibrium conditions• Newton’s third law• What a vector is• What the components of a force are• Engineering Science N4• Module 9• Module 5, taking moments to work out the reactions at supports

Learning outcomes

When students have completed this module, they should be able to:• Calculate reaction in supports of frameworks• Graphically determine the forces in members of a framework• Determine whether a member is a strut or a tie


• Make sure they understand the basic concept of forces in equilibrium• Make sure they know the equilibrium conditions• Always make decent sketches• Make a good study of the module and underline important facts

10 Structural frameworks




Drawings are not to scale.

1. Figure 10.13

Scale: 1 cm = 1 m 1 cm = 10 N

40 40

40

C ML

B

AF

O

1

2 3

4

E

NK

d

60 60

Member Vector Force N Type

AB k1 65 S

BC l2 43 S

CD m3 43 S

DE n4 65 S

EF o4 24 T

FA o1 24 T

CF 23 40 T

FB 12 22 S

FD 34 22 S

Symmetrical loaded structure, therefore each support will be 60 N = 3 × 40 ____ 2 .

n

m

o

l3

2

14

k


137Module 10 • Structural frameworks

Figure 10.14

Scale: 1 cm = 1 m 1 cm = 10 N

12

3

76.9 N 93,1 N

20 N

100 N 50 N

4

AF

KB

1 2 4

3

L

O

MC

E

N

D

nm

l

o

k


AB k1 151 S

BC l3 98 S

CD m4 122 S

DE o4 98 T

EF o2 130 T

FA o1 130 T

FB 12 0 xx

EB 23 39 S

EC 34 23 T

Moment on A

∴ 13D = 100 × 5 + 50 × 9 + 20 × 13 D = 93,1 N

Moment on D

∴ 13A = 50 × 4 + 100 × 8 A = 76,9



Figure 10.15

Scale: 1 cm = 1 m 1 cm = 10 N

Scale: 1 cm = 10 N

n

m

l

o

k1

2

3

4

5

A B C DE

FG

K L M N30 N 20 N 40 N

47,5 NO O42,5 N 1 2 3 4 5


AB k1 74 S

BC l2 55 S

CD m4 59 S

DE n5 81 S

EF o5 94 T

FG o3 63 T

GA o1 85 T

GB 12 35 S

GC 23 15 S

FC 34 8 S

FD 45 46 S



Moments about A

∴ 12E = 30 × 3 + 20 × 6 + 40 × 9 E = 47,5

Moments about E

∴ 12A = 40 × 3 + 20 × 6 + 30 × 9 A = 42,5

Figure 10.16

Space diagram Scale: 1 cm = 10 N

Vector diagram

1

p

P

AO

N

50 N

40 N

C

M

1

D

40°

60° 20°

140°

B

m

n

o


AD op 112 S

DB p1 41 T

BC m1 114 T

CD n1 113 S

Reaction B = pm = 76 N Reaction A = op = 118 N In this case use only length for space diagram with correct angles. No calculations.



2.

O

E

D

40 N

58,12 N

CNM

L

K

A

B

FGH44°P

1 2 5 6

3 430 N

30 N

30 N

2,5 m

2,5 m

2 mP

2 m 2 m 2 m

Calculate reaction E moments about A ∴ 8A = 30 × 2,5 + 30 × 5 + 40 × 6 ∴ E = 58,12 N

Scale: 1 cm = 1 m 1 cm = 10 N

12

3

k

l

m

p

n

o


AB l1 51 S

AH p1 77 T

BC m3 42 S

BG 32 31 S

GH p2 77 T

BH 12 0 O



3.

Scale: 1 cm = 1 m 1 cm = 10 N


AB k1 87 S

BC l2 58 S

CD m4 58 S

DE n5 87 S

EF o5 34 T

FG p3 49 T

GA Q1 34 T

GB 12 40 T

GC 23 6 T

FC 34 6 T

FD 45 40 T

60 N

10 N 10 N

60 N

L

K

QP

O

N1

2 3 4

5

A

B

C

D

E

FG

M 30 N30 N

k

l

o

p

q

m

n

1

2

3

4

5

Symmetrical loaded structure

∴ Reactions is 60 N each:

10 + 10 + 30 + 40 + 30 ______________ 2 = 60



4.

40 N

60° 41°

3

3

Ao

3 3 3 3

60 N 30 N

50 N

B54,87 N

306,45

305,45

406,60

406,60

90 N

E cos θ

E sin θ

Moment on A

∴ 12B + (50 × 6) = (40 sin 60 × 3) + (40 cos 60 × 6) + (80 × 6) + (30 cos 45 × 6) + (30 sin 45 × 6)

= 103,92 + 120 + 480 + 127,28 + 127,28 12B + 300 = 958,48 B = 54,87 N

Σ V.C. = 0 ∴ E sin θ – 40 sin 60 – 60 – 30 sin 45 + 54,87 – 20 = 0 ∴ E sin θ = +34,64 + 60 + 21,21 – 54,87 + 20 = +80,98 N

Σ H.C. = 0 ∴ E cos θ + 40 cos 60 + 30 cos 45 – 50 = 0 ∴ E cos θ = –20 – 21,21 + 50 = 0 = +8,79 N

∴ E = √ ____________

80,9 8 2 + 8,7 9 2 = 81,5 N

80,98

8,79

E

80,98

θ

tan θ = 80,98 ____ 8,79 ∴ θ = 83,8°

E = 81,5 N E 83,8° N




BC 01 55 S

CD 02 55 S

DE n3 105 S

EF m4 85 S

FG 45 45 S

GB p1 0 xx

GE 43 22 S

GD 32 78 T

GC 12 0 xx

5.Member Vector Force N Type

DC m1 84 T

CE n1 60 S

PE 12 20 T

FE 02 60 S

FD 23 74 S

Reaction at B = 106 N E Reaction at A = 116 N W 25,5° N

50 N

20 N60 N

C

M

DL

K

θA F

O

B

E

NP5

4

3

2 1

30 N



Scale: 1 cm = 20 N

1

2

3

45

P

k

m

l

n

o

6.

Scale: 1 cm = 1 m 1 cm = 20 N

A

KL 40 N20 N

P

60 N

30 N

MC

D

E

N

OB

1

2

12

p

mk

l

n

o

n




AE kp 118 T

EC l2 82 T

CD n2 64 S

DB o1 86 S

DE 12 24 T

BE p1 40 S

Reaction at A = 118 N W Horizontal Reaction at B = 124 N E 45,5° N

7. Moments about B ∴ 6A + 30 sin 50° × 8 = 70 × 4,5 + 80 × 8 A = 315 + 640 – 183,85 = 128,53 N Reaction B = pk = 110 N W 32,5° N


AB p1 70 T

BC k1 56 T

CD l2 34 T

AD n2 80 S

AC 12 74 S

Space diagram

70 N 30 N

80 NP

K

B

32,5°L

C 50°D

M

N

21

A

108,5 N



Scale: 1 cm = 20 N

Vector diagram no. 7

n

m

l

k

2

1

p

Use own length for space diagram.

8. 60 N

30 N

E

D

1

23

4

5A

C P

B

l m

k

q

n

40 N

80 N

20 N

O

107,99 N

Space diagram



Scale 1 cm = 10N

Vector diagram

k

q

o

p

m

l

3

4

5

First calculate reaction at “E” roller support. Take each length on 2 units (1 unit = 1 cm)

Moment about E ∴ __

M = __

M

∴ 4C = (30 × √ __

3 ) + (60N) + (40 × 5) + (20 × 6)

∴ 4C = 51,96 + 60 + 200 + 120

C = 107,99N

Draw vector diagram and measure length qk for reaction at E = 51,42N (E 13,5°N)


AB n5 22 T

BC 45 69 S

CD 43 55 S

DB m4 46 T

CA p5 91 S


MODULE


Knowledge assumed to be in place before starting with this module:• Where the tension part of a beam will be• Where the tension part of a cantilever will be• Know the diff erent stresses

Learning outcomes

When students have completed this module, they should be able to:• Know the diff erent foundations and what they are used for• Have general knowledge of purpose and placing of steel


• Make sure they understand the basic concept of this module• Make a good study of the module and underline important facts

11 Concrete and foundations


149Module 11 • Concrete and foundations


1. The purpose of a foundation is to distribute the load evenly over a bigger area and to prevent an uneven setting.

2. Strip foundations, used for houses and double-storey buildings built on a stable soil.

Pad foundations, used for columns and piers and reinforced with steel to strengthen and reduce the thickness of the foundation.

Piled foundations, used to transfer the weight through the soft soil to a more rigid base beneath.

Raft foundations, used for houses built on soft clay and fillings, and in severe conditions they could be used as piles.


1. It can cause separation of the components and excessive shrinkage of the concrete when it dries. Water trapped in the concrete will form cavities when it dries and the water will decrease the quality of the concrete.

2. The water/cement ratio The degree of compaction How thoroughly the mixing was done How effectively the curing process was carried out The temperature at which the curing process takes place

3. Shuttering must be kept in position Cover floors with water, while water can be sprayed onto the walls Concrete can be covered by plastic sheets to keep the moisture in A waterproof film can be sprayed onto the concrete to avoid evaporation

4. Concrete is cast continuously to ensure a good strength. When silos are built, the concrete will be cast for 24 hours a day until the silo is completed. The reason for this is to prevent joints where cracks can develop if there was a stoppage in the casting process.

5. To use a good grade of aggregate with maximum allowable size. A more dense concrete is formed when the different parts fit in close together. The finer the aggregates are, the bigger the total area of all the small parts per unit mass which have to be covered by the cement paste, this will reduce the cover efficiency of the cement paste. The workability of the concrete will also be reduced.



Aggregates which are rounder and more uniform in shape mix easily and less energy is needed. Irregular-shaped aggregates are used when a concrete of ultimate strength is needed.

Additions will make the concrete more workable but will reduce the strength of the concrete.

Adding more water and cement will increase the workability, but the disadvantage of shrinkage will increase which will cause cracks. The cost per cubic metre will also be higher.

6. It must: • have a high tensile strength • bond well with the concrete • be compatible with the concrete, especially as far as the temperature

movements are concerned.

7. 7.1

7.2

8. The size of the reinforcement – the cover must be equal to the thickness of a bar and, when in groups of three or more, the cover must be equal to the diameter of a single bar of the equivalent area. The conditions to which the structure is subjected and the amount of heat it will be subjected to.

9. The reinforcement is always placed in the high tensile zone of a concrete beam. A simply supported beam has the main bars in the high tensile areas and binders are used to keep the main bars in position. A cantilever has main bars in the high tensile areas. The main bars are at the top. The main bars of a column with reinforced bars in place are placed near the outer surface to take up the tensile stress in a case of bending. The reinforcement must also be placed and maintained accurately in position. Ends passing each other should be tied together with No. 16 gauge annealed soft-iron wire.

Main bars Binders


strength of materials and structures lecturer guide

Documents