lecture-23 (shear force diagram & bending moment diagram)

Post on 30-Apr-2017

222 Views

Category:

Documents

3 Downloads

Preview:

Click to see full reader

TRANSCRIPT

Lecture-23 (Shear force

diagram & bending moment

diagram)

Solid Mechanics

(CE 251)

© CE 251 (Solid Mechanics) by Anil Mandariya 1

© CE 251 (Solid Mechanics) by Anil Mandariya 2

Trigonometric functions:

(1) Line function: y = mx + c

y = dependent variable

x = independent variables

m = slope of line

c = interceptor on y-axis

y = mx

c = 0

m = + ve

y = mx + c

m = +ve

c = +ve

y = mx + c

m = -ve

c = +ve

y = mx + c

m = +ve

c= -ve

y = mx + c

m = -ve

c= -ve

+ x

+y

-x

+y

© CE 251 (Solid Mechanics) by Anil Mandariya 3

(1) Standard Equations of a Parabola with Vertex at (0, 0)

y = dependent variable

x = independent variables

1.1 y2 = 4ax

Vertex: (0, 0)

Focus: (a, 0)

Directrix: x = a

Symmetric with respect to the x axis

Axis the x axis

a < 0 (open left)

a > 0 (open right)

a < 0 (open down)

a > 0 (open up)

© CE 251 (Solid Mechanics) by Anil Mandariya 4

1.2 x2 = 4ay

Vertex: (0, 0)

Focus: (0, a)

Directrix: y = a

Symmetric with respect to the y axis

Axis the y axis

© CE 251 (Solid Mechanics) by Anil Mandariya 5

1.3 Convert quadratic equation to standard parabola

a

Aa

bxX

a

c

a

byYwhere

XAY

a

bx

a

c

a

by

a

a

c

a

b

a

bxay

a

c

a

b

a

bx

a

bxay

a

cx

a

bxay

cbxaxy

14;

2

4,

4

)2

()4

(1

4)

2(

)44

(

)(

2

2

2

2

2

2

2

22

2

2

2

22

2

2

a

c

a

by

a

c

a

by

Y

a

bx

a

bx

X

2

2

2

2

2

02

0 and

2

02

0

:are Parabola ofVertex

© CE 251 (Solid Mechanics) by Anil Mandariya 6

Vertex:

Symmetric with respect to the y axis

Axis the y axis

)2

,2

(2

2

a

c

a

b

a

b

a > 0 (open up)

y

x

0

-x

© CE 251 (Solid Mechanics) by Anil Mandariya 7

1.4 y = ax3 + bx2 + cx +d

where a ≠ 0, and b, c and d (=0) are constants (can be

zero) and If a > 0 the graph of a cubic looks similar to

one of the following graphs:

yrxqxpx

))()((

IIICase

ypx

3)(

IICase

yx

3

IICase

© CE 251 (Solid Mechanics) by Anil Mandariya 8

1.4 y = ax3 + bx2 + cx +d

where a ≠ 0, and b, c and d (=0) are constants (can be

zero) and If a < 0 the graph of a cubic looks similar to

one of the following graphs:

yrxqxpx

))()((

IIICase

ypx

3)(

IICase

yx

3

IICase

Sign Conventions SF and BM:

© CE 251 (Solid Mechanics) by Anil Mandariya

9

Sign Conventions of SF & BM

Sagging

Hogging © CE 251 (Solid Mechanics) by Anil

Mandariya 10

Steps to Calculate SF and BM in Beams:

Step-1

Determine the reactions at the supports.

Step-2

Divide the beam span in different segments according

loading and support conditions.

Step-3

Consider left side end of beam as origin.

Step-4

Choose a section corresponding to each segment which

will be at ‘x’ distance from the left end of the beam

(origin). © CE 251 (Solid Mechanics) by Anil Mandariya

11

Step-5

Take a left segment of the beam from section Y-Y.

Step-6

Convert all UDL, UVL in point loads.

Step-7

Choose the sign conventions and write the static

equilibrium equations for the beam segment.

Step-8

Solve static equilibrium equations for the each beam

segment separately according to boundary condition of

segments.

© CE 251 (Solid Mechanics) by Anil Mandariya

12

Step-9

Consider a x and y axis below the beam such that length

of the x-axis will be equal to beam span.

Step-10

Mark the all points according to loading and supports

on the x-axis.

Step-11

Draw the curve separately according to SF and BM

equations for each segments with boundary values.

© CE 251 (Solid Mechanics) by Anil Mandariya

13

Example 1:

Step-1

Determine the reactions at

the supports.

RB RD

HB

ƩFx = 0 = HB= 0 ..........(1)

ƩFy = 0 = -RB - RD + 20 + 40= 0

RD + RB = 60 kN .............(2)

ƩM = 0 = ƩMB = 0

= - (3 x 40) + (5 x RD) + RB x 0

+ HA x 0 + (2.5 x 20)= 0

RD = 70/5 = 14 kN ..............(3)

From eq. (2) & (3)

RB = 46 kN © CE 251 (Solid Mechanics) by Anil Mandariya

14

Step-2

Divide the beam span in different segments according

loading and support conditions.

Segment-1: AB

Segment-2: BC

Segment-3: CD

© CE 251 (Solid Mechanics) by Anil Mandariya

15

Step-3

Consider left side end of beam as origin.

X

Y

0

Step-4

Choose a section y-y’ corresponding to each segment

which will be at ‘x’ distance from the left end of the

beam (origin).

x

y

y’

© CE 251 (Solid Mechanics) by Anil Mandariya

16

Step-5

Take a left segment of the beam from section Y-Y.

0 x

y

y’ Step-6

Convert all UDL, UVL in point loads.

There is no UDL or UVL.

© CE 251 (Solid Mechanics) by Anil Mandariya

17

Step-7 & 8

Choose the sign conventions and write the static

equilibrium equations for the beam segment & solve them.

0 x

y

y’

Segment-1: AB (0 ≤ x ≤ 2.5)

ƩFx = 0 ..........(1)

ƩFy = 0

= 20 + Vx = 0

Vx = - 20 kN (constant).............(2)

ƩM = 0

= (x * 20) + Mx = 0

Mx = -20x (line equation with –ve

slop and zero intercept on y-axis)..(3)

From eq. (2) & (3)

At x = 0

VA = - 20 kN; MA = 0 kN-m

At x = 2.5 m

VB = - 20 kN; MB = - 50 kN-m © CE 251 (Solid Mechanics) by Anil Mandariya 18

y

y’ x

RB = 46 kN

(x - 2.5)

0

Segment-2: BC (2.5 ≤ x ≤ 5.5)

ƩFx = 0 ..........(1)

ƩFy = 0

= 20 + Vx – 46 = 0

Vx = 26 kN (constant value).............(2)

ƩM = 0

= (x * 20) - (x – 2.5)*46 + Mx = 0

Mx = 26x - 115 (line equation with +ve

slope and –ve intercept on y-axis)..........(3)

From eq. (2) & (3)

At x = 2.5 m

VB = 26.0 kN; MB = - 50.0 kN-m

At x = 5.5 m

VC = 26.0 kN; MC = 28.0 kN-m © CE 251 (Solid Mechanics) by Anil Mandariya 19

Segment-3: CD (5.5 ≤ x ≤ 7.5)

X

Y

0

x

y

y’

(x-2.5-3)

RB = 46 kN

ƩFx = 0 ..........(1)

ƩFy = 0

= 20 – 46 + 40 + Vx = 0

Vx = -14.0 kN (constant value).............(2)

ƩM = 0

= (x * 20) - (x – 2.5)*46 + (x – 2.5 – 3)*40

+ Mx = 0

Mx = -14x + 105 (line eq. with –ve slop

and +ve intercept)................................(3)

From eq. (2) & (3)

At x = 5.5 m

VC = -14.0 kN; MC = 28.0 kN-m

At x = 7.5 m

VD = -14.0 kN; MD = 0 kN-m © CE 251 (Solid Mechanics) by Anil Mandariya

20

Step-9

© CE 251 (Solid Mechanics) by Anil Mandariya

21

Consider a x and y axis below the beam such that length

of the x-axis will be equal to beam span.

x

y

Step-10

© CE 251 (Solid Mechanics) by Anil Mandariya 22

Mark the all points according to loading and supports

on the x-axis.

x

y

B A

C D

Step-11

© CE 251 (Solid Mechanics) by Anil Mandariya

23

Draw the curve separately according to SF and BM

equations for each segments with boundary values.

Segment-1: AB (0 ≤ x ≤ 2.5)

Vx = - 20 kN (constant)

Mx = -20x (line equation with

–ve slop and zero intercept on

y-axis)

At x = 0

VA = - 20 kN; MA = 0 kN-m

At x = 2.5 m

VB = - 20 kN; MB = - 50 kN-m

x

y

B A

C D

-20 kN SFD

© CE 251 (Solid Mechanics) by Anil Mandariya

24

Segment-1: AB (0 ≤ x ≤ 2.5)

Vx = - 20 kN (constant)

Mx = -20x (line equation with

–ve slop and zero intercept on

y-axis)

At x = 0

VA = - 20 kN; MA = 0 kN-m

At x = 2.5 m

VB = - 20 kN; MB = - 50 kN-m

x

y

B A

C D

-20 kN

SFD

-50 kN-m

0.0 kN-m

© CE 251 (Solid Mechanics) by Anil Mandariya

25

Segment-2: BC (2.5 ≤ x ≤ 5.5)

Vx = 26 kN (constant value)

Mx = 26x - 115 (line equation

with +ve slope and –ve

intercept on y-axis)

At x = 2.5 m

VB = 26.0 kN; MB = - 50.0 kN-m

At x = 5.5 m

VC = 26.0 kN; MC = 28.0 kN-m

Mx = 0

26x – 115 = 0

x = 4.42 m

x

y

B A

C D

-20 kN

SFD

-50 kN-m

0.0 kN-m

26 kN

E x = 4.42 m 28 kN-m

© CE 251 (Solid Mechanics) by Anil Mandariya

26

Segment-2: CD (2.5 ≤ x ≤ 7.5)

Vx = -14.0 kN (constant value)

Mx = -14x + 105 (line eq. with –

ve slop and +ve intercept)

At x = 5.5 m

VC = -14.0 kN; MC = 28.0 kN-m

At x = 7.5 m

VD = -14.0 kN; MD = 0 kN-m

x

y

B A

C D

-20 kN

SFD

0.0 kN-m

-50 kN-m

26 kN

E x = 4.42 m 28 kN-m

-14 kN

BMD

Example:2

© CE 251 (Solid Mechanics) by Anil Mandariya

27

Step-1

Determine the reactions at the supports.

Couple: arm = 2 m and

force = 8 kN

Mo = 2 x 8 = 16 kN-m

UDL: 2 Nos.

q1 = q2 = q = 4 kN/m on 2 m

Resultant point load = 4 x 2 = 8 kN

Will act at centroid of the UDL

= 2/2

= 1 m from A and B © CE 251 (Solid Mechanics) by Anil Mandariya

28

1.0 m 3.0 m 1.0 m

8.0 kN 8.0 kN

3.0 m

HA

RA RB

ƩFx = 0 = HA= 0 ..........(1)

ƩFy = 0 = -RA - RB + 8.0 + 8.0 = 0

RA + RB = 16.0 kN .............(2)

ƩM = 0 = ƩMB = 0

= RB x 0 + (1 x 8) + - (Mo = 16) + (7 x 8) - (8 x RA) + HA x 0 = 0

RA = 48/8 = 6.0 kN ..............(3)

From eq. (2) & (3)

RB = 10.0 kN

So Reactions are HA= 0; RA = 6.0 kN; RB = 10.0 kN

© CE 251 (Solid Mechanics) by Anil Mandariya

29

Step-2

Divide the beam span in different segments according

loading and support conditions.

C E D

Segment-1: AC

Segment-2: CD

Segment-3: DE

Segment-4: EB © CE 251 (Solid Mechanics) by Anil Mandariya 30

Step-3

Consider left side end of beam as origin.

C E D

Y

X

O

© CE 251 (Solid Mechanics) by Anil Mandariya

31

Step-4

Choose a section y-y’ corresponding to each segment

which will be at ‘x’ distance from the left end of the

beam (origin). 1st choose for segment AC and so on.

C E D

Y

X

O

x

y

y’

© CE 251 (Solid Mechanics) by Anil Mandariya

32

Step-5

Take a left segment of the beam from section y-y'.

C E D

Y

X

O

x

y

y’

0 kN

6.0 kN

Step-6

Convert all UDL, UVL in point loads.

UDL: 1 Nos.; q = 4 kN/m & length = x m

Resultant load of UDL = 4x kN

Will act at centroid of UDL: x/2 m

4x kN

x/2 m

© CE 251 (Solid Mechanics) by Anil Mandariya

33

Step-7 & 8

Choose the sign conventions and write the static

equilibrium equations for the beam segment & solve them.

Segment-1: AC (0 ≤ x ≤ 2.0 m)

4x kN

x/2 m

ƩFx = 0 ............................................(1)

ƩFy = 0 = - 6.0 + 4x + Vx = 0

Vx = (6- 4x) kN ............................................(2)

ƩM = 0

= - 6*x + (x/2 * 4x) + Mx = 0

Mx = - 2x2 + 6x ...........................................(3)

From eq. (2) & (3)

At x = 0 m

VA = 6.0 kN; MA = 0.0 kN-m

At x = 2.0 m

VC = -2.0 kN; MC = 4.0 kN-m © CE 251 (Solid Mechanics) by Anil

Mandariya 34

Segment-2: CD (2.0 ≤ x ≤ 4.0 m)

x

y

y’

6.0 kN x

y

y’

6.0 kN

8.0 kN

1.0 m

ƩFx = 0 ..........................(1)

ƩFy = 0 = - 6.0 + 8 + Vx = 0

Vx = -2.0 kN .............................. (2)

ƩM = 0

= - 6*x + (x – 1) * 8 + Mx = 0

Mx = - 2x + 8 .............................(3)

From eq. (2) & (3)

At x = 2.0 m

VC = -2.0 kN; MC = 4.0 kN-m

At x = 4.0 m

VD = -2.0 kN; MD = 0.0 kN-m

© CE 251 (Solid Mechanics) by Anil Mandariya 35

Segment-3: DE (4.0 ≤ x ≤ 6.0 m)

x

C D

y

y’ 6.0 kN

2.0 m

8.0 kN

1.0 m

ƩFx = 0 .....................................(1)

ƩFy = 0 = - 6.0 + 8 + Vx = 0

Vx = -2.0 kN ....................................... (2)

ƩM = 0

= - 6*x + (x – 1) * 8 – Mo (= 16) + Mx = 0

Mx = - 2x + 24 ....................................(3)

From eq. (2) & (3)

At x = 4.0 m

VD = -2.0 kN; MD = 16.0 kN-m

At x = 6.0 m

VE = -2.0 kN; ME = 12.0 kN-m

© CE 251 (Solid Mechanics) by Anil Mandariya 36

Segment-4: EB (6.0 ≤ x ≤ 8.0 m)

2.0 m

x

C D

y

y’ 6.0 kN

2.0 m

8.0 kN

1.0 m

4.0 kN/m

E

© CE 251 (Solid Mechanics) by Anil Mandariya

37

ƩFx = 0 .....................................(1)

ƩFy = 0 = - 6.0 + 8 + 4*(x – 6) + Vx = 0

Vx = 22 – 4x kN ................................ (2)

ƩM = 0

= - 6*x + (x – 1) * 8 – Mo (= 16) + 4*(x – 6)*(x – 6)/2 + Mx = 0

Mx = - 2x2 + 22x - 48 ....................................(3)

x

C D

y

y’ 6.0 kN

2.0 m

8.0 kN

1.0 m

4.0 kN/m

2.0 m

4*(x – 6)

(x – 6)/2 m

E

From eq. (2) & (3)

At x = 6.0 m

VE = -2.0 kN;

ME = 12.0 kN-m

At x = 8.0 m

VB = -10.0 kN;

ME = 0.0 kN-m

© CE 251 (Solid Mechanics) by Anil Mandariya 38

© CE 251 (Solid Mechanics) by Anil Mandariya 39

C E D

x

x

y

y

A

A B D E C

C D E B

Consider a x and y axis below the beam such that length of the x-

axis will be equal to beam span and mark the all points according

to loading and supports on the x-axis.

Step-9

& 10

© CE 251 (Solid Mechanics) by Anil Mandariya 40

Draw the curve separately according to SF and BM equations for

each segments with boundary values.

Step-11

Segment-1: AC (0 ≤ x ≤ 2.0 m)

Vx = (6- 4x) kN (line eq. with –ve slop and +ve intercept on y-axis and SF =0 at

x = 1.5 m).

Mx = - 2x2 + 6x (quadratic eq. of parabola with vertex (1.5, 4.5))

At x = 0 m

VA = 6.0 kN; MA = 0.0 kN-m

At x = 2.0 m

VC = -2.0 kN; MC = 4.0 kN-m

y

x

0

-x

Select this part

of parabola for

BMD according

to boundary

value

© CE 251 (Solid Mechanics) by Anil Mandariya 41

C E D

x

x

y

y

A

A B D E C

C D E B

6 kN

-2 kN 1.5 m

4 kN-m

2 m

© CE 251 (Solid Mechanics) by Anil Mandariya 42

Segment-2: CD (2.0 ≤ x ≤ 4.0 m)

Vx = -2.0 kN (constant value, horizontal axis parallel to x-

axis)

Mx = - 2x + 8 (line eq. with –ve slop and +ve intercept on y-

axis)

At x = 2.0 m

VC = -2.0 kN; MC = 4.0 kN-m

At x = 4.0 m

VD = -2.0 kN; MD = 0.0 kN-m

© CE 251 (Solid Mechanics) by Anil Mandariya 43

6 kN

-2 kN 1.5 m

2 m C E D

x

x

y

y

A

A B D E C

C D E B

4 kN-m

-2 kN

2 m 2 m

© CE 251 (Solid Mechanics) by Anil Mandariya 44

Segment-3: DE (4.0 ≤ x ≤ 6.0 m)

Vx = -2.0 kN (constant value, horizontal axis parallel to x-

axis)

Mx = - 2x + 24 (line eq. with –ve slop and +ve intercept on

y-axis)

At x = 4.0 m

VC = -2.0 kN; MC = 16.0 kN-m

At x = 6.0 m

VD = -2.0 kN; MD = 12.0 kN-m

© CE 251 (Solid Mechanics) by Anil Mandariya 45

6 kN

-2 kN 1.5 m

2 m C E D

x

x

y

y

A

A B D E C

C D E B

4 kN-m

-2 kN

16 kN-m

12 kN-m

-2 kN

2 m 2 m 2 m

© CE 251 (Solid Mechanics) by Anil Mandariya 46

Segment-4: EB (6.0 ≤ x ≤ 8.0 m)

Vx = 22 – 4x kN (line eq. with –ve slop and +ve intercept

on y-axis)

Mx = - 2x2 + 22x - 48 (quadratic eq. of parabola with vertex

(5.5, 36.5))

At x = 6.0 m

VE = -2.0 kN;

ME = 12.0 kN-m

At x = 8.0 m

VB = -10.0 kN;

ME = 0.0 kN-m

y

x

Select this part

of parabola for

BMD according

to boundary

value

© CE 251 (Solid Mechanics) by Anil Mandariya 47

6 kN

-2 kN 1.5 m

2 m C E D

x

x

y

y

A

A B D E C

C D E B

4 kN-m

-2 kN

16 kN-m

12 kN-m

-2 kN

2 m 2 m 2 m

-10 kN

© CE 251 (Solid Mechanics) by Anil Mandariya 48

SFD & BMD for different types of beam and

loading condition:

(1) Simply supported beam:

© CE 251 (Solid Mechanics) by Anil Mandariya 49

© CE 251 (Solid Mechanics) by Anil Mandariya 50

(2) Cantilever beam:

© CE 251 (Solid Mechanics) by Anil Mandariya 51

BM will be maximum at:

1. At sections where a concentrated load acts and the

shear force changes its sign.

2. At sections where shear force = 0

3. At supports providing vertical reactions.

4. At sections where a couple are applied.

Point of contra flexure:

A point where BM changes its sign.

Anil Mandariya 52

Important points for BM:

The slop of the SFD at any section of the beam is equal

to the intensity of the distributed load over

corresponding section.

Presence of the concentrated loads are indicated by

abrupt changes in SFD. The abrupt change (change in

ordinate) of SF at a section is equal to the

concentrated load at the section.

The slop of the BMD at any section is equal to the SF

at the corresponding section.

At concentrated load point, the slop of BMD is abruptly

chnges.

Anil Mandariya 53

Properties of SFD & BMD:

Total load on a beam between two sections =

difference between the SF at these two sections

The area of SFD between two section = difference

between BM at these two sections.

Anil Mandariya 54

Thanks

© CE 251 (Solid Mechanics) by Anil Mandariya

55

top related