analysis of selected determinate structural...

Dr. Mohammed E. Haque, P.E.

Lecture Notes

COSC321Haque 1

PDF_B4 (Analysis of Selected Determinate Structural Systems)

Analysis of Selected Determinate Structural Systems • Planar Trusses – Method of Joints

• Planar Trusses – Method of Sections

• Pinned Frames with Multi-force Members

• Retaining Walls


Lecture Notes

COSC321Haque 2


Planar Trusses – Method of Joints

Example 1: Calculate support reactions and member forces of the truss as shown.

First Calculate the support reactions

∑Fx = 0 � Bx=3k (Left)

∑MA = 0 � By(24) –3(8) – 8(12) = 0 � By = 5k (up)

∑MB = 0 � -Ay(24) -3(8) +8(12) = 0 � Ay = 3k (up)

Proof: ∑Fy = 0 � Ay + By - 8 = ? � 5 + 3 – 8 = 0

Take Joints FBD, and solve member forces

Bx

8’

6’ 6’ 6’ 6’

8k

3k

A B C

D E

By Ay

A

Ay = 3k

AD

ADx

ADy ∑Fy = 0 � ADy =3 � AD(4/5) = 3 �

AD=15/4=3.75k (C)

∑Fx = 0 � AC = ADx � AC= 3.75(3/5) = 2.25k (T) AC

FBD of

Joint A

Arrow directed towards a

joint represents Compression

member force, and arrow

directed away from the joint

represents Tension member

force.


Lecture Notes

COSC321Haque 3


D

DC AD

=3.75

DE ∑Fy = 0 � DCy = ADy � DC(4/5) = 3.75(4/5) �

DC=3.75k (T)

∑Fx = 0 � -DE+3+ADx + DCx =0 �

DE= 3+ 3.75(3/5) + 3.75(3/5) = 7.5k(C)

3k

FBD of

Joint D

B

By = 5k

BE

BEx

BEy ∑Fy = 0 � BEy =5 � BE(4/5) = 5 � BE=25/4=6.25k (C)

∑Fx = 0 � -CB + BEx – 3 = 0 � -CB +6.25(3/5) – 3 =0

� CB= 0.75(T)

CB Bx = 3k

FBD of

Joint B

C

DC=3.75

8k

CD=0.75

∑Fy = 0 � CEy +DCy –8 =0 � CE(4/5) +3.75(4/5) –8 =0 �

CE(4/5) + 3 – 8 =0 � CE = 6.25k (T)

AC=2.25

FBD of

Joint C

CE

ANSWER:

Reactions: Ay=3k (up); Bx=3k (Left), By=5k(Up)

Member Forces:

AC= 2.25(T); CB = 0.75k (T); DE=7.5k(C)

AD=3.75k(C ) ; DC=3.75k(T); CE=6.25k(T); BE=6.25k(C)


Lecture Notes

COSC321Haque 4


Example 2: Calculate support reactions and member forces, DE, DC and AC of the truss as

shown.

First Calculate the support reactions

∑Fx = 0 � Bx=3k (Left)

∑MA = 0 � By(24) –3(8) – 8(12) = 0 � By = 5k (up)

∑MB = 0 � -Ay(24) -3(8) +8(12) = 0 � Ay = 3k (up)

Proof: ∑Fy = 0 � Ay + By - 8 = ? � 5 + 3 – 8 = 0

Take a section along a-a, and solve for member forces

Bx

8’

6’ 6’ 6’ 6’

8k

3k

A B C

D E

By Ay

a

a

Ay=3k

A

D E

C

3k DE

AC

CD

∑Fy = 0 � -CDy +3 =0 � -CD(4/5) +3 = 0 �

CD=15/4=3.75k (T)

∑MC = 0 � DE(8) -3(8) - 3(12) =0 � DE= 7.5k(C )

∑Fx = 0 � AC +3+CDx-DE =0

� AC+3+ 3.75(3/5) – 7.5 =0 � AC= 2.25k (T)


Lecture Notes

COSC321Haque 5


Example 3: Calculate reactions, Ax, Ay and Bx. Find member forces, EA, EB, and DB using

section method.

C

2.5kN 2.0kN

A

D

B

E 2m

2m 2m

Ay

Ax

Bx

a

a

Find out support Reactions:

∑Fy = 0 � Ay= 2.5 + 2 = 4.5 kN (up)

∑MB = 0 � -Ax (2) +2.5(4) + 2(2) =0 � Ax= 7 kN(Right)

∑Fx = 0 � Bx= 7 kN (Left)


Lecture Notes

COSC321Haque 6


2m

C

2.5kN 2.0kN

A

D B

E

EA

a

a

EB

DB

2

1

Take a section along a-a

∑MB = 0 � -EAx(1) - EAy(2) +2.5(4) + 2(2) =0

� EA(2/√5)(1) + EA(1/√5)(2) = 14 � EA = 7.826 kN (T )

∑MC = 0 � EBx( 1) + EBy (2) – 2(2) = 0

� EB(2/√5)( 1) + EB(1/√5)(2) =4 � EB = 2.236 kN (C )

∑ME = 0 � DB(1) + 2.5(2) =0 � DB = -5 kN (C) (arrow should be

toward joint D)


Lecture Notes

COSC321Haque 7


Pinned Frames with Multi-force Members

Example 1: Calculate the support reactions at A and B. Also calculate internal pin force at C.

B

A

C

D

400 lb

200 lb

250 lb

6ft

4 ft 4 ft 4 ft

4 ft

B

A

C

D

400 lb

200 lb

250 lb

6ft

4 ft 4 ft 4 ft

4 ft

Ax

Ay

Bx

By


Lecture Notes

COSC321Haque 8


Find Support Reactions at A & B, and Pin force at C:

∑MB = 0 � Ax(6) – 200(4) – 400(4) – 250(12) = 0 � Ax=900 lb (Left)

∑Fx = 0 � Bx = 900 lb (Right)

From FBD-2:

∑MC = 0 � By(8) + 250(4) – 400(4) = 0 � By = 75 lb (up)

∑FY = 0 � By –400 – 250 + Cy = 0 � 75 – 400 –250 +Cy = 0 � Cy = 575 lb

From FBD-1:

∑MC = 0 � Ax(6) – Ay(8) +200(4) = 0 � 900(6) – Ay(8) +200(4) = 0 � Ay=775 lb (up)

∑FX = 0 � Cx – Ax =0 � Cx=900 lb

A

200 lb

4 ft 4 ft 4 ft

Ax

Ay

B C

D

400 lb

250 lb

4 ft

Bx

By Cy

Cx

Cx

C

FBD-1

FBD-2

Results: Ax=900 lb (Left); Ay=775 lb (Up); Bx=900 lb (Right); By=75 lb (Up);

Cx=900 lb; Cy=575 lb


Lecture Notes

COSC321Haque 9


Example 2: A wind force of 0.12k/ft is applied to the vertical projection of a building as shown

in fig. Calculate the support reactions at A and C. Also calculate internal pin force at B.

A C

B

20’

15’

3

4

20’ 20’

0.12k/ft


Lecture Notes

COSC321Haque 10


A C

B

20’

15’

3

4

20’ 20’

F = 0.12x35

=4.2 k

35/2=17.5’

Ax

Ay

Cx

Cy

FBD-1

C

B

35’

20’

Cx

Cy=1.838k FBD-2

Bx

By

From FBD-2

∑Fy = 0 � By = 1.838k

∑MB = 0 � Cx(35) = 1.838(20) �Cx=1.05k (Left)

∑FX = 0 � Bx – Cx =0 � Bx = 1.05k

From FBD-1

∑MA = 0 � Cy(40) – 4.2(17.5) = 0 �

Cy=1.838 k (Up)

∑Fy = 0 � Ay = 1.838 k (down)

RESULTS:

Ax=3.15k (Left); Ay=1.838k (Down)

Cx=1.05k (Left); Cy=1.383k (Up)

Bx=1.05k; By=1.838k

From FBD-1

∑FX = 0 � -Ax – Cx +4.2 =0 �

-Ax –1.05 +4.2 = 0 � Ax=3.15k (Left)


Lecture Notes

COSC321Haque 11


Retaining Walls

Common Retaining Walls

Toe Heel

Gravity or Semi-gravity Retaining wall

Toe Heel

Cantilever Retaining wall

Stem

Footing

Toe Heel

Buttress Retaining wall

Stem

Footing

Buttress

Toe Heel

Counterfort Retaining wall

Stem

Footing

Counterfort

Back Fill Back Fill

Back Fill Back Fill

Ground Level

Ground Level Ground Level

Ground Level

Ground Level

Ground Level

Ground Level

Ground Level


Lecture Notes

COSC321Haque 12


First Floor

Basement Floor

Footing

Stem

Basement Wall Foundation

Ground Level


Lecture Notes

COSC321Haque 13


Bearing Plate

Bridge Girder

Back-wall

Piles

Pile Cap

Bridge Abutment

Ground Level

Back Fill


Lecture Notes

COSC321Haque 14


Lateral Soil Pressure on Retaining Walls

Typical Angle of Internal Friction for backfill soil Soil Type φ (Degree)

Gravel and coarse sandy backfill soil 33-36

Medium to fine sandy backfill soil 29-32

Silty sand 27-30

P max = Ka γsoil h

h

γsoil

Backfill

Fig. 1: Soil Pressure on the back of wall (No surcharge)

Ka = Coefficient of Active Soil Pressure

Ka = tan2 (45

0 - φ/2)

φ = Angle of Internal Friction for backfill soil γsoil = Unit weight of Soil (pcf)

P max = Ka γsoil h

P max = w’ h Where w’ = equivalent fluid pressure of soil

w’ = Ka γsoil


Lecture Notes

COSC321Haque 15


P max = Ka γsoil h

h

γsoil

Backfill

Fig. 2: Soil Pressure on the back of wall (Sloping Backfill)

β

cos β - √(cos2 β - cos2 φ)

cos β + √(cos2 β - cos2 φ)

Ka = cos β


Lecture Notes

COSC321Haque 16


P max = Ka γsoil h

h

Fig. 3: Soil Pressure on the back of wall (with uniform surcharge)

Surcharge, wsc

Psc = Ka wsc

+

γsoil

Backfill


Lecture Notes

COSC321Haque 17


Q1(a): Analyze the stability of the reinforced cantilever retaining wall as shown in Figure. Use the following values: Concrete unit weight = 150 pcf

Soil unit weight, γsoil =110 pcf

Coefficient of Active Soil Pressure, Ka = 0.33 (Neglect Coefficient of passive Soil Pressure, Kp)

Coefficient of friction between the bottom of footing and soil, µ = 0.5

10’

2’

2’

2’ 1’ 4’


Lecture Notes

COSC321Haque 18


Solution: Step 1: Calculate lateral soil pressure and overturning moment

P max = Ka γsoil h = 0.33 (110)(12) = 435.6 psf

FH = ½ Pmax h = ½ (435.6)(12)= 2613.6 lb/ft of wall Sliding Force, FH = 2613.6 lb/ft of wall Overturning Moment, MOT about toe = 2613.6 x 4 = 10454.4 lb-ft /ft of wall.

12’

2’

2’

2’ 1’ 4’

Wsoil

Wwall

WFooting

FH

Pmax

Toe

A

2.5’

3.5’

5’

1/3(12’)=4’


Lecture Notes

COSC321Haque 19


Step 2: Calculate weights, Resisting Moment, Sliding Resisting Force; Wsoil = (4 x 10)(110) = 4400 lb./ft of wall Wwall = (1 x 10)(150) = 1500 lb / ft of wall WFooting = (2 x 7)(150) = 2100 lb / ft of wall WTotal = 4400 + 1500 + 2100 = 8000 lb / ft of wall Resisting moment , MR about toe = (4400 x 5) + (1500 x 2.5) + (2100 x 3.5)

= 33100 lb-ft/ft of wall

Sliding Resisting Force, FR = µ X WTotal = 0.5 (8000) = 4000 lb Step 3: determining the Factor of Safety (FS) against overturning and sliding. FS OT = MR / MOT = 33100 / 10454.4 = 3.17 > 2.0 OK

FS Sliding = FR / FH = 4000 / 2613.6 = 1.53 > 1.5 OK


Lecture Notes

COSC321Haque 20


Soil Pressure under the footing of the Retaining Wall f = fmax = P/A = P/(Bx1)

C.L.

B

P= WTotal

c= B/2

Case I: Load, P with zero eccentricity (e=0)

fmax = P/A + (P e /Sm) fmin = P/A - (P e /Sm) where

A= 1xB Sm = 1x(B2)/6

C.L.

B

P= WTotal

c > B/3

Case II: Load, P with small eccentricity (e<B/6)

e<B/6

fmax fmin

fmax = (2/3) P/c

C.L.

B

P= WTotal

c ≤ B/3

Case III: Load, P with

large eccentricity (e≥B/6)

e≥B/6

fmax


Lecture Notes

COSC321Haque 21


Q1(b) Calculate the soil pressure under the footing of the retaining wall of Q1(a).

Determine the resultant vertical force, WTotal intersects the bottom of the footing: c = (MR – MOT)/ WTotal = (33100 - 10454.4) / 8000 = 2.831 ft. from the Toe. Eccentricity, e = B/2 – c = (7/2) – 2.831 = 0.6693’ < B/6 (=7/6=1.1667’) Therefore, Case II is applicable.

Calculate soil pressures, fmax and fmin under the footing:

A= 1xB = 7 sqft. Sm = 1x(B2)/6 = 72 / 6 = 8.1667 ft3 P= WTotal = 8000 lb. fmax = P/A + (P e /Sm) = (8000/7) +(8000x0.6693/8.1667)

=1142.86 + 655.64 = 1798.5 psf fmin = P/A - (P e /Sm) = =1142.86 - 655.64 = 487.22 psf

C.L. of Footing

c= 2.831’ 2’

fmin

WTotal

B=7’

Toe

e

fmax

analysis of selected determinate structural...

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