Worked Examples

Problem 7.1 Determine the soil pressure distribution under the footing.

Elevation

Plan e

M 180   1.5 ft P 120

(a) B= L= 8 ft L e  1.5 ft   1.33 ft 6

1

q2 

6 (P e) 180 6 (180) P     4.92 kip/ft 2 2 3 BL BL 8(8) 8

q1 

6 P e 180 6 (180) P     0.7 kip/ft 2 2 3 BL BL 8(8) 8



(b) L=10 ft B =5 ft L e  1.5 ft   1.66 ft 6 6 (P e) 6 (180) P 180 q2      5.76 kip/ft 2 2 2 BL BL 5(10) (5)(10)  6Pe 6 (180) P 180 q1      1.42 kip/ft 2 2 BL BL 5(10) (5)(10) 2 Problem 7.2 Determine the soil pressure distribution under the footing. Use a factor of 1.2 for DL and 1.6 for LL.

Elevation

Plan P u  1.2 PD  1.6 PL  1.2(200)  1.6(250)  640 kN M

2

u

 1.2 M D  1.6 M L  0  1.6(64)  64 kN-m

e

Mu 64 L   0.1 m   0.5 m Pu 640 6

q u1 

Pu 6 (Pu e) 640 6 (64)     85.32 kN/m 2 2 3 BL BL 3(3) 3

qu 2 

Pu 6 Pu e 640 6 (64)     56.88 kN/m 2 2 3 BL BL 3(3) 3



Problem 7.3 The effective soil pressure is 4 kip/ft 2 . Determine the maximum allowable value for L.

Elevation

Plan

M 48   .3478 ft P 138 6 M 138 6 (48) L P   2  4 Assume e  2 6 L L3 L L3 e

4Lreq 3  138Lreq  288  0 3

L  6.72 ft

Problem 7.4 The allowable soil pressure is qallowable  250 kN/m2 , γsoil  18 kN/m3 , γconc  24 kN/m3 , PD = 1000 kN and PL = 1400 kN.

Plan

Elevation

P D+L  1000  1400  2400 kN qeffective  qallowable  γconc t  γsoil (h-t)  250  24(.4) 18(1  .4)  229.6 kN/m 2

P u  1.2 PD  1.6 PL  1.2(1000)  1.6(1400)  3440 kN qu 

Pu L1 L 2

Vu ( x)  L 2 q u (

L1 a  ) 2 2

L q L a M u ( x)  2 u  1   2  2 2

4

2

(a) A square footing (L1= L2 = L) P D+L 2

L

 q effective  229.6 kN/m 2

L

2400  3.23 m 229.6

qu 

Pu 3440   326 kN/m2 2 L 3.25(3.25)

use L  3.25 m

L a Vu max  L q u (  )  (3.25)(326)(1.625  .225)  1483 kN 2 2 2

M u max

L q u  L a  (3.25)(326)  (1.625  .225)2  1038 kN-m     2  2 2 2

(b)A rectangular footing with L2 = 2.5 m. P D+L  q effective  229.6 kN/m 2 L1 (L 2 ) L1 

5

2400  4.18 m 2.5(229.6)

use L1  4.2 m

qu 

Pu 3440   327.6 kN/m2 L1 L2 2.5(4.2)

Vu max  L 2 q u (

L1 a  )  (2.5)(327.6)(2.1  .225)  1536 kN 2 2 2

M u max

L q  L a  (2.5)(327.6)  2 u 1   (2.1  .225) 2  1440 kN-m 2  2 2 2

Problem 7.5 A 350 mm x 350 mm column is to be supported on a shallow foundation. Determine the dimensions (either square or rectangular) for the following conditions. The effective soil pressure is qeffective  180 kN/m2 . (a) The centerline of the column coincides with the centerline of the footing.

Plan

Elevation q 

6 (P e) P   q effective L1 L 2 L 2 L12

6 (550) 240   180 2.5L1 2.5L12 180L12  220L1  576  0

6



L1  2.5 m

(b) The center line of the column is 0.75meter from the property line.

Plan q 

Elevation


 L  6  ( 1  .75)(550)  240  550  2   180  2 2.5L1 2.5L1

180 L12  880 L1  652.5  0



L1  3.97 m

use L1  4 m

(c) The center line of the column is 0.5 meter from the centroid of the footing.

7

Plan

q 

Elevation


6 (515) 550   180 2.5L1 2.5L12 180 L12  220 L1  1236  0



L1  3.3 m

Problem 7.6 A combined footing supports two square columns: Column A is 14 inches x 14 inches and carries a dead load of 140 kips and a live load of 220 kips. Column B is 16 inches x 16 inches and carries a dead load of 260 kips and a live load of 300 kips. The effective soil pressure is qe  4.5 k/ft 2 . Assume the soil pressure distribution is uniform. Determine the footing 8

dimensions for the following geometric configurations. Establish the shear and moment diagrams corresponding to the factored loading, Pu =1.2 PD +1.6 PL.

9

10

11

12

13

14

15

16

17

18

19

Problem 7.7 Column A is 350 mm x 350 mm and carries a dead load of 1300 kN and a live load of 450 kN. Column B is 450 mm x 450 mm and carries a dead load of 1400 kN and a live load of 800 kN. The combined footing shown below is used to support these columns. Determine the soil pressure distribution and the shear and bending moment distributions along the longitudinal direction corresponding to the factored loading, Pu =1.2 PD +1.6 PL.

Elevation

20

Plan

21

22

23

Problem 7.8 Dimension a strap footing for the situation shown. The exterior column A is 14 inches x 14 inches and carries a dead load of 160 kips and a live load of 130 kips; the interior column B is 18 inches x 18 inches and carries a dead load of 200 kips and a live load of 187.5 kips; the distance between the center lines of the columns is 18 ft. Assume the strap is placed such that it does not bear directly on the soil. Take the effective soil pressure as qe  4.5 k/ft 2 . Draw shear and moment diagrams using a factored load of Pu =1.2 PD +1.6 PL.

24

Plan

Elevation

25

Problem 7.9 Column A is 350 mm x 350 mm and carries a dead load of 1300 kN and a live load of 450 kN. Column B is 450 mm x 450 mm and carries a dead load of 1400 kN and a live load of 800 kN. A strap footing is used to support the columns and the center line of Column A is 0.5 meter from the property line. Assume the strap is placed such that it does not bear directly on the soil. Determine the soil pressure distribution and the shear and bending moment distributions along the longitudinal direction corresponding to the factored loading, Pu =1.2 PD +1.6 PL.

26

Elevation

Plan

27

28

Problem 7.10 An exterior 18 in x18 in column with a total vertical service load of P1 = 180 kips and an interior 20 in x20 in column with a total vertical service load of P2 = 240 kips are to be supported at each column by a pad footing connected by a strap beam. Assume the strap is placed such that it does not bear directly on the soil. (a) Determine the dimensions L1 and L 2 for the pad footings that will result in a uniform effective soil pressure not exceeding 3 k/ft 2 under each pad footing. Use ¼ ft increments. (b) Determine the soil pressure profile under the footings determined in part (a) when an additional loading, consisting of an uplift force of 80 kips at the exterior column and an uplift force of 25 kips at the interior column, is applied.

29

30

31

32

33

Problem 8.1 For the concrete retaing wall shown, determine the factors of safety against sliding and overturning and the base pressure distribution.

1 1 1 γsoil H 2 k a  (18)(3.5) 2 ( )  49 kN 2 2 3 H 3.5 M  Pa ( )  49( )  57.16 kN-m overturning 3 3

Pa 

W1  (24)(0.75)(3)(1)  54 kN 3 W2  (24)( )(1.25)(1)  45 kN 2 W3  (24)(2)(0.5)(1)  24 kN

N  N   Wi  123 kN Fmax   N  .5(123)  61.5 kN Fmax 61.5   1.25 Pa 49 2 M  W1 (1.625)  W2 ( )(1.25)  W3 (1)  149.2 kN-m resisting 3 M 149.2 resisting F.S.    2.6 overturning M 57.16

F.S. sliding 

overturning

M net  M

resisting

 Moverturning  92 kN-m

Page 1 of Chapter I

1

x

M net N

 .748

L  x=.25 m 2 N  6 e  123  6(.25)  2 q1  1   1    107.6 kN/m L  L  2  2  e

q2 

N  6 e  123  6(.25)  2 1   1    15.3 kN/m L  L  2  2 

Problem 8.2 For the concrete retaing wall shown, determine the factors of safety against sliding and overturning and the base pressure distribution. Use the Rankine theory for soil pressure computations.

Page 2 of Chapter I

2

1 1 1 γsoil H 2 k a  (18)(3.5) 2 ( )  49 kN 2 2 3 H 3.5 M  Pa ( )  49( )  57.16 kN-m overturning 3 3

Pa 

W1  (24)(0.75)(3)(1)  54 kN 3 W2  (24)( )(1.25)(1)  45 kN 2 W3  (24)(3)(0.5)  36 kN W4  (18)(3)(1)  24 kN

N   Wi  189 kN Fmax   N  .5(123)  94.5 kN Fmax 94.5   1.928 Pa 49  314.23 kN-m

F.S. sliding 

M

resisting

F.S.

overturning

M net  M M net

x e

N



resisting

M M

resisting



overturning

314.23  5.49 57.16

 M overturning  257 kN-m

 1.36 m

L  x  1.5  1.36  .14 m 2

q1 

N  6 e  2 1    45.4 kN/m L  L 

q2 

N  6 e  2 1    8.1 kN/m L  L 

Page 3 of Chapter I

3

Problem 8.3 For the concrete retaining wall shown, determine the factors of safety against sliding and overturning and the base pressure distribution. Use the Rankine theory for soil pressure computations

Page 4 of Chapter I

4

Pa 

1 1 1 k a γs H 2  ( )(.12)(14) 2  3.92 kip 2 2 3

W1 = .15 (2) (12)  3.6 kip 1 W2 = .15 ( )(4 )(12)  3.6 kip 2 W3 = .15 (2) (13)  3.9 kip W4 = .12 (4) (12)  5.76 kip

The normal and horizontal forces are N  W1  W2  W3  W4  16.86 kip Fmax   N  .5 (16.86) = 8.43 kips

Next we compute the factors of safety. F.S.sliding =

M Boverturning

Fmax 8.43   2.15 Pa 3.92 H 14  Pa ( )  3.92( )  18.24 k-ft 3 3

M Bresisting  W1 (8)+W2 (2.67)+W3 (6.5)+W4 (11)  3.6 (8)+3.6 (2.67)+3.9(6.5)+5.76(11)  127.1 k-ft

F.S.overturning 

Page 5 of Chapter I

M Bresisting M Boverturning



127.1  6.97 18.24

5

M Bnet  M Bresisting  M Boverturning  108.86 k-ft x 

M Bnet 108.86   6.45ft N 16.86

L 13  x   6.45  .05 ft 2 2 Using the above values, the peak pressures are e

q

6e N 16.86 6(.05) (1 ± ) (1  ) L L 13 13

 q1  1.38 kip/ft 2 q 2  1.27 kip/ft 2

Problem 8.4

Page 6 of Chapter I

6

F.S. sliding  ( 1.25 

μ γc b b ) ( 2 ) (1  1 ) k a γs H b2

.5 (.15) b 1 ( 2 ) (1  ) (1/ 3) (.12) 14 b2

F.S.

overturning

2 γc  b2     k a γs  H 

F.S. sliding  (

F.S.

overturning



2

b 2  8.33

2   1  b1  b1   1      2  b2  b2    

μ γc b b ) ( 2 ) (1  1 ) k a γs H b2

2(.15)  b2     (1/3) (.12)  14 

2

2  1  1 1   1       1.75  2  b2  b2   

For b2  8.33 F.S.

overturning

2(.15)  8.33     (1/3) (.12)  14 

2

2  1   1 1    1     2.95  1.75   2  8.33  8.33   

Problem 8.5 3 3 Assume: γsoil= 0.12 k/ft , γconcrete = 0.15 k/ft , µ= .5 and Φ = 300

Page 7 of Chapter I

7

1 1 1 γsoil H 2 k a  (.12)(22) 2 ( )  9.68 kip 2 2 3 1 Ps  k a ws H = ( ) (.2) (22)  1.47 3 1 1 Pp  k p γs H'2  (3)(.12)(4) 2  2.88 kip 2 2 Pa 

F

 Pa + Ps  Pp  8.27 kip

M

 Pa (

horizontal

overturning

H H 22 22 )  Ps ( )  9.68 ( )  1.47( )  87.15 3 2 3 2

W1  .15 (1) (20.5)  3.075 20.5 )  .77 2 W3  .15 (b11) (1.5)  2.475 W2  .15 (.5) (

W4  .12 (6) (20.5)  14.76 W5  .12 (2.5) (3.5)  1.05 N 

 W  22 i

Fmax   N  11

F.S.sliding =

Fmax 11   1.33  Fhorizontal 8.27

Page 8 of Chapter I

8

M Bresisting  W1 (4.5)+W2 (3.83)+W3 (5.5)+W4 (8)+W5 (1.75)+Pp (1.33)  3.075(4.5)+.768(3.83)+2.475(5.5)+14.76(8)+1.05(1.75)+2.88(1.33)  154 kip-ft M Bresisting 154 F.S.overturning    1.76 M Boverturning 87.15

M Bnet  M Bresisting  M Boverturning  66.85 kip-ft x 

e

M Bnet 66.85   3ft N 22

L 11  x   3  3.5 ft 2 2

N

q1 (3a) 2



L  1.83 6 q1 

(22)(2) =4.89 kip/ft 2 3(3)

Problem 8.6 Allowable soil pressure = 5.0 ksf , γsoil =.12 k/ft 3 and γconcrete =.15 k/ft 3

Page 9 of Chapter I

9

1 1 1 k a γs H 2  ( )(.12)(24) 2  11.52 kip 2 2 3 H 24 M B overturning  Pa ( )  11.52( )  92.16 kip-ft 3 3

Pa 

Page 10 of Chapter I

10

W1 = .15 (1) (21)  3.15 kip 1 W2 = .15 ( )(.5 )(21)  .78 kip 2 W3 = .15 (3) (13)  5.85 kip W4 = .12 (8) (21)  20.1 kip N 

 W  30 kip i

Fmax   N  .5 (30) = 15 kips F.S.sliding =

Fmax 15   1.3 Pa 11.52

M Bresisting  W1 (4.5)+W2 (3.83)+W3 (6.5)+W4 (9)  3.15(4.5)+.78(3.83)+5.85(6.5)+20.1 (9)  236 kip-ft F.S.overturning 




236  2.56 92.16

M Bnet  M Bresisting  M Boverturning  143.84 kip-ft x 

M Bnet 143.84   4.79ft N 30

e

L 13  x   4.79  1.7 ft 2 2

q

6e N 30 6(1.7) (1 ± ) (1  ) L L 13 13


 q1  4.12 kip/ft 2 q 2  0.5 kip/ft 2

11

1 p  k a γs H  ( )(.12)(24)  0.96 kip/ft 3

Problem 8.7 Suggest values for b1 and b 2 . Take the safety factors for sliding and over turning to be equal to 2. Assume: γsoil =.12 k/ft 3 , γconcrete =.15 k/ft 3 , µ = .57, and Φ = 300


12

1 1 1 γsoil H 2 k a  (.12)(16) 2 ( )  5.12 kip 2 2 3 1 Ps  k a ws H = ( ) (.1) (16)  .53 3 1 1 Pp  k p γs H'2  (3)(.12)(4) 2  2.88 kip 2 2 Pa 

F

 Pa + Ps  Pp  2.77

M

 Pa (

horizontal

overturning

H H 16 16 )  Ps ( )  5.12 ( )  .53( )  31.55 3 2 3 2

W1  .15 (3) (14)  6.3 W2  .15 (2) (b1 +b 2 +3)  .3(b1 +b 2 +3) W3  .12 (b 2 ) (20)  2.4b 2 W4  .12 (b1 ) (2)  2.4b1 N 

 W  6.3+.3(b +b +3)+2.4b i

1

2

2

 2.4b1  7.2  2.7(b1 +b2 )

Fmax   N  .57 (7.2  2.7(b1 +b 2 ))  4.1  1.54(b1 +b 2 ) F.S.sliding =



Fmax 4.1  1.54(b1 +b 2 )  2 F 2.77  horizontal



b2  0.935  b1


13

b1 +b 2  0.935

M Bresisting  W1 (1.5  b1 )+W2 (.5b1 +.5b 2 +1.5)+W3 (b1 +.5b 2 +3)+W4 (2.5)+Pp (1.33)   6.3(1.5  b1 )  .3(b1 +b 2 +3)(.5b1 +.5b 2 +1.5)  16.8(b1 +.5b2 +3)  2.4b1 (2.5)  2.88(1.33)   106.4  3.9b1 F.S.overturning 




106.4  3.9b1 2 31.55

b1  0

 b2  .935ft

Problem 8.8

Determine the minimum value of w at which soil failure occurs (i.e., the soil pressure exceeds the allowable soil pressure). Assume: qallowable  5 k/ft 2 , γsoil =.12 k/ft 3 , γconcrete =.15 k/ft 3 , µ = .57 and Φ = 300


14

1 1 1 γsoil H 2 k a  (.12)(22) 2 ( )  9.68 kip 2 2 3 1 Ps  k a ws H = ( ) ( ws ) (22)  7.33 ws 3 1 1 Pp  k p γs (H' )2  (3)(.12)(4)2  2.88 kip 2 2 Pa 

F

 Pa + Ps  Pp  9.68  7.33 ws  2.88  6.88  7.33 ws

M

 Pa (

horizontal

overturning

H H 22 )  Ps ( )  9.68( )  7.33 ws (11)  71  80.6 ws 3 2 3

W1  .15 (2) (20)  6 kip W2  .15 (2) (14)  4.2 kip W3  .12 (7) (20)  16.8 kip W4  .12 (5) (2)  1.2 kip N 

 W  28.2 kip i

Fmax   N  .57 (28.2) = 16 kip

q

x

6e N (1  )5 L L

M net



N

6e 28.2 (1  )5 14 14

e  3.46 ft

M net

28.2 M L e   x  7  net  3.46 2 28.2

M net  99.8

M Bresisting  W1 (6)+W2 (7)+W3 (10.5)+W4 (2.5)+Pp (1.33)   6(6)  4.2(7)  16.8(10.5)  1.2(2.5)  2.88(1.33)  248.6

M net  M

resisting

 Moverturning

99.8  248.6  (71  80.6 ws ) 


ws  .96kip/ft 2

15

Problem 8.9

(a) The answer is case (a). Problem 8.10 (a) Determine the factor of safety with respect to overturning and sliding. (b) Identify the tension areas in the stem, toe, and heel and show the reinforcing pattern. (c) Determine the location of the line of action of the resultant at the base of the footing


16

1 1 γsoil H 2 k a  (.1)(12) 2 (.4)  2.88 kip 2 2 H 12 M  Pa ( )+2(15)  2.88( )  30  41.52 kip-ft overturning 3 3

Pa 

W1  (1.5)(13)(3)(.15)  2.925 kip W2  (2)(12)(.15)  3.6 kip W3  (7)(2)(010)(.1)  7 kip N   W  13.525 kip Fmax   N  .58(13.525)  7.84 kip Fmax 7.84   2.72 Pa 2.88

F.S. sliding 

M

resisting

F.S.

 93.53 kip-ft

overturning




M M

resisting

overturning



93.53  2.25 41.52

17

M net  M x e

M net N

resisting



 M overturning  52 kip-ft

52  3.84 13.525

L  x  6  3.84  2.16 ft 2


L  2ft 6

18

Problem 9.1 Take E=29,000 ksi and I= 200 in 4 .

 17 wL4 17(2)(30) 4 (12)3     33.78 in  B,0 243EI 243(29000)(200)   3 8(30)3 (12)3 δ  8L   0.794 in  BB 81EI 81(29000)(200)

 B  1 in

Page 1 of Chapter 9



RB 

 B,0  R B δBB  1 in

 B,0  1 δBB



33.78  1  41.28 kip  0.794

Problem 9.2 Take E=200 GPa and I= 80(10)6 mm4 .



 B,0  R B δBB  12 mm

 B  12 mm

Page 2 of Chapter 9

 wL4 15(6) 4 (10)9     151.875 mm  B,0 8EI 8(200)(80)106  Table 3.1  3 (6)3 (10)9 δ  L   4.5 mm  BB 3EI 3(200)(80)106

RB 

 B,0  12 δBB



151.875  12  31.08 kN  4.5

Problem 9.3 kv= 60 k/in, E=29,000 ksi, I= 200 in 4

Page 3 of Chapter 9

 17 wL4 17(1.0)(20) 4 (12)3  D,0  384EI  384(29000)(200)  2.11 in   3 (20)3 (12)3 δ  L   0.0993 in  DD 24EI 24(29000)(200) FCD 

 D,0 1 δ DD  kV



2.11 1 0.0993  60

Problem 9.4 L = 5 m, E = 200 GPa, I= 170(10)6 mm4 (a)

Page 4 of Chapter 9

 18.19 kip

B,0  110.49 mm

δBB

L3 (12)3 (10)9    1.059 mm 48EI 48(200)(170)106

RB  (b)

Page 5 of Chapter 9

 B,0 δBB

 110.4 kN 

δBB

L3 (12)3 (10)9    1.059 mm 48EI 48(200)(170)106

RB 

B δBB



8.33  7.86 kN  1.059

(c)

δBB

L3 (12)3 (10)9    1.059 mm 48EI 48(200)(170)106

RB 

(d)

Page 6 of Chapter 9

B δBB



20  18.89 kN 1.059

R B  18.89 kN 

 P 40(10)9   25.946  25.946  30.52 mm  B,0 EI (200)(170)(10)6   3 (12)3 (10)9 δ  L   1.059 mm  BB 48EI 48(200)(170)(10)6 FB 

(e)

Page 7 of Chapter 9

 B,0 δ BB 

1 kV



30.52 1.059 

1 40

 28.15 kN

 57 wL4 57(20)(12) 4 (10)9     113.16 mm  D,0 6144EI 6144(200)(170)(10)6   3 3(12)3 (10)9 δ  3L   0.596 mm  DD 256EI 256(200)(170)(10)6

FD 

 D,0 δ DD 

1 kV



113.16  17.97 kN 1 0.596  40

Problem 9.5 I  400 in 4 , E  29,000 ksi , L  54 ft , w  2.1 kip/ft ,  B  1.2 in

Page 8 of Chapter 9

(i) The distributed load shown



 B,0  R B δBB  0

The deflection terms are given in Chapter 3  4wL4     14.6 in  B,0 729 EI   4L3  δ   BB 243EI  .386 in

Page 9 of Chapter 9

RB 

 B,0 δBB



14.6  37.8 kip  0.386

(ii) The support settlement at B 

 R B δBB  1.2 in

1.2 1.2   3.1 kip R B  3.1 kip  δBB 0.386 Problem 9.6 AC  1200 mm2 , L  9 m, P  40 kN and E  200 GPa . RB 

Page 10 of Chapter 9


Problem 9.7

x  x y  4h   ( ) 2  L L 

I

Io cos θ

(a) Determine the horizontal reaction at B due to the concentrated load.




(b) Utilize the results of part (a) to obtain an analytical expression for the horizontal reaction due to a distributed loading, w( x) .

(c) Specialize (b) for a uniform loading, w( x)  w0 .


(d) Suppose the horizontal support at B is replaced by a member extending from A to B. Repeat part (a).


Problem 9.8 Consider the semi-circular arch shown below. Determine the distribution of the axial and shear forces and the bending moment. The cross section properties are constant.



Problem 9.9 Take A  20,000 mm2


I  400(10)6 mm4 and E  200 GPa

Case (a):

Mmax  0 (no bending moment)

Deflected shape Case (b):


Mmax  1101 kN-m

Deflected shape

Case (c):


Mmax  529 kN-m

Deflected shape

Problem 9.10 Consider the following values for the area of the tension rod AC: 4 in 2 , 8 in 2 , 16 in 2 .

A  30 in 2

I  1000 in 4 E  29,000 ksi .

A  30 in 2


I  1000 in 4 E  29,000 ksi

Bending moment

Axial force

Bending moment


Axial force

Bending moment

Axial force Problem 9.11


A  40 in 2

I  1200 in 4 E  29,000 ksi

Case(a):

Mmax  0 (no bending moment) Case (b):


Mmax  130 kip-ft

Case (c):

Mmax  43.3 kip-ft Case (d):


Mmax  10.9 kip-ft

Problem 9.12 Determine the horizontal reaction at support D.

(See Figure 9.38)


H D1 

P  1 kip  2

(See example 9.15)

rg 

I2 I  L 16

H D2 



Problem 9.13


2

wL 12h

rc 

I1 I  h 20 2

rg rc



5 4

1 (32) 1   2.33 kip  r 2 2 g 12(20) 1  ( 5 ) 1 3 4 3 rc

HD  HD1  HD2  1  2.33  1.33 kip 

Consider h = 2 m, 4 m, 6 m.

0  x1  9

M 0 (x1 )  101.25 x1  7.5 x12

0  x2  9

M 0 (x2 )  101.25 x2


δM(x1 ) 

x1 6 9h

δM(x2 ) 

x2 6 9h

9

9

Δ B,0 

x x 1 1 1 M 0 δM dS   (101.25 x1  7.5 x12 )( 1  6) dx1   (101.25 x2 )( 2  6) dx2  EI s EI 0 9h EI 0 9h

δ BB 

x x 1 1 1 2  δM  dS   ( 1  6)2 dx1   ( 1  6) 2 dx2  EI s EI 0 9h EI 0 9h

9

Then H B  

EIΔ B,0



9

 B,0 δBB

0.0676(10)6   0.0847(10)6  6 0.1067(10)

for h  2m for h  4m for h  6m

 25.7 kN  H B  23.7 kN  21.78 kN  


EIδ BB


0.00263(10) 6   0.00357(10) 6  6 0.0049(10)


Problem 9.14 Determine the peak positive and negative moments as a function of h. Consider h = 10 ft, 20 ft, 30 ft.

See equation (9.51) and Figure 9.49


M BA  

wL2 12

1 1.2(50)2 1 250   h 1r 12 1  1 ( h ) 1 1 g 2 25 50 2 rc

 wL  2 1 ME  18  3 1 rg 1+  2 rc  2

M  max  M BA

M

 max

      2   375 1h    3(1  )   50   

208 kip-ft   178.6 kip-ft  156.2 kip-ft

166.6 kip-ft   M E  196.4 kip-ft  218.7 kip-ft

for h  10ft for h  20ft for h  30ft for h  10ft for h  20ft for h  30ft

Problem 9.15 Using a Computer software system, determine the bending moment distribution and deflected shape due to the loading shown.

Take I1  1000 in 4 , I2  2000 in 4 , E=29,000 ksi and A=20 in 2 all members


Problem 9.16 Take E=200 GPa, I  400(10)6 mm4 , A=100000 mm2 and AC  1200 mm2 , 2400 mm2 , 4800

mm2 . Use a Computer software system.


(a)



(b)




Problem 9.17 Both cases (a) and (b) will are rigid frames and will have large horizontal displacements and will behave the same. In case (b) the added member will not carry any load. In case (c) the diagonal member changes the behavior. The frame will behave like a truss. All the members only carry axial forces and lateral movement will be very small. Case (a):

Case (b):

Case (c):


Problem 9.18 E= 29,000 ksi, A= 1 in 2 all members

δF

F0

δF2 L F0 + X1 δF 14.14(12) -3.53 -.707 423.47 84.81 4.62 14.14(12) -17.67 -.707 2119.75 84.81 -9.51 10(12) 12.5 .5 750 30 6.73 10(12) 12.5 .5 750 30 6.73 10(12) 0 1 0 120 -11.53  F δF L  4043  δF2 L  349.6

member L, in ab bc cd da bd


F0

δF

F δF L

L 4043   0.139 AE 29000 349.6 2 L δ11    δF    0.012 AE 29000 1,0   F0 δF

1,0

0.139  11.53 δ11 0.012 Knowing the value of X1 , we determine the member forces and reactions by using superposition. Member forces are listed below. 1,0  δ11 X1  0



X1  



Problem 9.19 Assume the vertical reaction at d as the force redundant. E= 200 GPa, A= 660 mm2 all members, α= 12x10-6 / 0C , T  100 C


member L, mm

δF

e temp = α ΔTL

δF e temp

ab bc cd da bd

-.606 -.714 .428 .428 1

0.67884 0.67884 0 0 0

-.4073 -.4819 0 0 0

5657 5657 3000 4000 4000

e

δF2 L 2077.4 2883.9 549.5 732.7 4000

X1 δF

-6.9 -8.1 4.8 4.8 11.4

δF  0.889  L  δF  10243.5 2

temp

1,0   etemp δF  0.889 δ11    δF 

2

L 10243.5   0.0776 AE 660(200)

1,0  δ11 X1  0



X1  

1,0 δ11



0.889  11.4 0.0776

Knowing the value of X1 , we determine the member forces and reactions by using superposition. Member forces are listed below.


Problem 9.20 E=29,000 ksi and A= 1 in 2 all members.

δF2 L F0 δF L F0 + X1 δF 12(12) 0 -.8 92.16 0 7.6 9(12) -10 -.6 38.88 648 -4.3 12(12) -19.33 -.8 92.16 2226.8 -11.7 9(12) 0 -.6 32.88 0 5.7 15(12) 16.66 1 180 2998.8 7.2 15(12) 0 1 180 0 -9.5 2  δF L 616  F0 δFL 58736

member L, in ab bc cd da ac bd

F0


δF

L 5873.6   0.2025 AE 2900(1) 616 2 L δ11    δF    0.0212 AE 2900(1) 1,0   F0 δF

1,0

.2025  9.5 δ11 .0212 Knowing the value of X1 , we determine the member forces and reactions by using superposition. Member forces are listed below. 1,0  δ11 X1  0



X1  



Problem 9.21 A1  A2  A3  A4  10 in 2 , A 5 = 5 in 2 , α= 6.5x10-6 / 0F , ΔT= 600 , E= 29,000 ksi


member L, in ab bc cd da bd

28.84(12) 28.84(12) 20(12) 20(12) 12(12) 1,0   δ11  

L/AE

e temp =

0.001193 0.001193 0.000827 0.000827 0.000993

α ΔTL 0.13497 0.13497 0.0936 0.0936 0.05616

F0 δFL  AE

 δF

2

AE

L

e

temp

δF

F0 12 -6 4.16 4.16 5

-1.8 -1.8 2.5 2.5 3

etemp δF

-.2429 -.2429 0 0 0

F0 δFL AE

-.0257 .0128 .0086 .0086 .0148

 δF 

2

L

AE .00386 .00386 .00516 .00516 .00893

F0 + X1 δF -19.1 -37.1 47.3 47.3 56.8

δF  .019  0.485  0.466

 0.027

1,0

0.466  17.26 δ11 0.027 Knowing the value of X1 , we determine the member forces and reactions by using superposition. Member forces are listed below. 1,0  δ11 X1  0




X1  



Problem 9.22 Assume the force in member ac and the reaction at support f as force redundants. 2

A=1000 mm and E=200 GPa for all the members


δF1

F0

δF2

member

L, mm

F0

δF1

δF2

F0 δF1 L

F0 δF2 L

ab bc cd de ef fa bf ac cf ce

4000 3000 3000 4000 3000 3000 5000 5000 4000 5000

-20 15 30 0 15 30 -25 0 20 -25

-.5 -.375 0 0 .375 0 .625 0 .5 -.625

-.8 -.6 0 0 0 -.6 1 1 -.8 0

40000 -16875 0 0 16875 0 78125 0 40000 -78125

64000 -27000 0 0 0 -54000 -125000 0 -64000 0

 F δF L

 80000

 F δF L

 (δF ) L

 6750

 (δF ) L

0

1

2

1


0

2

2

2

(δF1 )2 L 1000 422 0 0 422 0 1953 0 1000 1953

(δF2 )2 L 2560 1080 0 0 0 1080 5000 5000 2560 0

 206000  17280

 δF δF L 1

2

 3800

δF1 δF2 L

1000 675 0 0 0 0 3125 0 -1000 0

1,0  

F0δF1L 80000   .4 mm EA 200(1000)

 2,0  

(δF1 )2 L 6750 δ11     0.03375 EA 200(1000)

δ12  δ21  

F0δF2 L 206000   -1.03 mm EA 200(1000)

(δF2 )2 L 17280 δ22     0.0864 EA 200(1000)

δFδF 3800 1 2L   0.019 EA 200(1000)

δF2

δF1 1,0  δ11 X1  δ12 X 2  0  2,0  δ21 X1  δ22 X 2  0



X1   21.2 kN  X 2  16.58 kN

Knowing the value of X1 and X 2 , we determine the member forces and reactions by using superposition.


member ab bc cd de ef fa bf ac cf ce

F0 -20 15 30 0 15 30 -25 0 20 -25

δF1 -.5 -.375 0 0 .375 0 .625 0 .5 -.625

δF2 -.8 -.6 0 0 0 -.6 1 1 -.8 0

F0 + X1 δF1 + X 2 δF2 -22.67 13.0 30 0 7 20 -21.67 16.58 -3.86 -11.75

Member forces are listed below.


Problem 10.1 Take E = 200 GPa and A  2000 mm2

(a) A1 

1 A 2

AE (2000)200   94.28 L1 3 2(1000) A1E (1000)200   66.67 L2 3(1000) AE (2000)200   66.67 L3 6(1000)

kN mm kN mm kN mm

F(1) 

AE AE cos45 u 2  sin45 v2  66.67u 2  66.67v2 L1 L1

F(2) 

A1E v2  66.67v2 L2

F(3)  -

1

AE AE cos30 u 2  sin30 v2  57.73u 2  33.33v2 L3 L3

F

0

F

0

x

y

 F(1) cos45  F(3) cos30  45  0  F(1)sin45  F(2)  F(3)sin30  90  0

0.707 F(1)

  0.86 F(3)  45

0.707 F(1)  F(2)  0.5 F(3)  90 

0.707  66.67u 2  66.67v2   0.86  57.73u 2  33.33v2   45 0.707  66.67u 2  66.67v2   66.67v2  0.5  57.73u 2  33.33v2   90  96.78 u 2  18.47 v2  45 18.27 u 2  130.47 v2  90

u 2  0.342 mm    v2  0.645 mm 

Next, we substitute for u 2 and v2 to determine member forces F(1)  66.67u 2  66.67v2  65.8 kN F(2)  66.67v2  43 kN F(3)  57.73u 2  33.33v2  1.75 kN

2

(b) A1  2A AE (2000)200 kN   94.28 L1 3 2(1000) mm A1E (4000)200 kN   266.67 L2 3(1000) mm AE (2000)200 kN   66.67 L3 6(1000) mm

F(1) 

AE AE cos45 u 2  sin45 v2  66.67u 2  66.67v2 L1 L1

F(2) 

A1E v2  266.67v2 L2

F(3)  -

AE AE cos30 u 2  sin30 v2  57.73u 2  33.33v2 L3 L3

F

0

F

0

x

y

0.707 F(1)  0.86 F(3)  45 0.707 F(1)  F(2)  0.5 F(3)  90

 96.78 u 2  18.47 v2  45

18.27 u 2  330.47 v2  90



u 2  0.417 mm   v2  0.25 mm 

Next, we substitute for u 2 and v2 to determine member forces F(1)  66.67u 2  66.67v2  44.47 kN F(2)  266.67v2  66.67 kN F(3)  57.73u 2  33.33v2  15.74 kN

3

(c) Results based on computer based analysis listed below

1 For A1  A 2

F(1)  65.62 kN  F(2)  42.79 kN  F(3)  1.62 kN

u 2  0.342 mm   v2  0.642 mm 

90.00

45.00

o FX 6.562E+01

1

2

3

FX 1.619E+00

FX 4.279E+01

For A1  2A

4

F(1)  44.37 kN  F(2)  66.48 kN  F(3)  -15.73 kN

u 2  0.416 mm   v2  0.249 mm 

90.00

o

45.00

FX 4.438E+01

1

2

3

FX -1.573E+01

FX 6.648E+01

Problem 10.2 Take A=0.1 in 2 , A1 =0.4 in 2 , and E=29,000 ksi.

cosθ  0.6

5

sinθ  0.8

AE AE (0.1)29000 kip    9.67 L1 L3 25(12) in

A1E (0.4)29000 kip   96.67 L2 10(12) in

(a) The loading shown F(1) 

AE AE cosθ u 2  sinθ v2  5.8 u 2  7.74 v2 L1 L1

F(2)   F(3)  -

A1E v2  96.67 v2 L2 AE AE cosθ u 2  sinθ v2  5.8 u 2  7.74 v2 L3 L3

F

0

F

0

x

y

 0.6 F(1)  0.6 F(3)  6  0  0.8 F(1)  F(2)  0.8 F(3)  10  0 

6.96 u 2  6 109 v2  10



u 2  0.862 in   v2  0.092 in 

Next, we substitute for u 2 and v2 to determine the member forces F(1)  5.8 u 2  7.74 v2  5.7 kip F(2)  96.67 v2   8.89 kip F(3)  5.8 u 2  7.74 v2   4.3 kip 1 1 (b) Support #1 moves as follows: u  inch  and v  inch  8 2

6

F(1) 

AE AE cosθ u 2  sinθ v2  5.8 u 2  7.74 v2 L1 L1

F(2)   F(3)  -

A1E 1 1 (v2  )  96.67 (v2  ) L2 2 2 AE AE cosθ u 2  sinθ v2  5.8 u 2  7.74 v2 L3 L3

F

0

F

0

x

y

F(1)  F(3)



u2  0

0.8 (F(1)  F(3) )  F(2)  0 1 0.8 (15.48 v2 )  96.67 (v2  )  0 2 



v2  0.443 in 

F(1)  F(3)  3.4 kip F(2)  5.5 kip

For the following beams and frames defined in Problems 10.3- 10.18, determine the member end moments using the slope-deflection equations. Problem 10.3 Assume E=29,000 ksi, I  200 in 4 , L  30 ft ,  C  .6 in  and w  1.2 kip/ft .

7

Let

ρ AB  ρ BC

k

EI  2k L

k member AB 

k member BC 

EI k 2L

B  A

0 L   B 0.6 1  C   2L 60(12) 1200 EI 29000(200) 1   671.3 kip-ft 2L 2(30) (144)

The slope-deflection equations take the form:

 θ   90 4k 2θ   90

M AB  4k M BA 

B

B

    M BC  2k 2θ B - 3( C B )   2k 2θ B - 3ρ BC  2L       M CB  2k θ B - 3( C B )   2k θ B - 3ρ BC  2L  

Enforce moment equilibrium at node B: M BA  M BC  0  8kθ B  90  2k 2θ B - 3ρ BC   0  12kθ B  6k ρ BC  90  θB 

8

90  6(671.3)( 12(671.3)

1 ) 1200  0.01076 rad

counterclockwise

The corresponding end moments are: M AB  4(671.3)



0.01076  90  118.89 kip-ft

M BA  4(671.3) 2(0.01076)



 90  32.21 kip-ft

1   M BC  2(671.3) 2(0.01076) - 3()   32.25 kip-ft 1200   1   M CB  2(671.3) 0.01076 - 3()   17.80 kip-ft 1200  

Problem 10.4 Assume E= 200 GPa, I2  80(10)6 mm4 and L2  6 m . (a) Let

M

F AB

M FBC

k member AB 

22(6)2   66 kN-m 12 45(6)   33.75 kN-m 8

EI2 L2

k

k member BC 

EI2 k L2

22(6)2 M   66 kN-m 12 45(6) F M CB   33.75 kN-m 8 F BA


M AB  2k θ B  66 M BA  4k θ B  66 M BC  4k θ B  33.75 M CB  2k θ B  33.75

Enforce moment equilibrium at node B: MBA  MBC  0

9

4k θB  66  4k θB  33.75  0



k θB  4.03

The corresponding end moments are: M AB  2k θ B  66  74.06 kN-m M BA  4k θ B  66  -49.9 kN-m M BC  4k θ B  33.75  49.9 kN-m M CB  2k θ B  33.75  25.7 kN-m

(b) Let

k member AB 

E(2I2 )  2k L2

k member BC 

EI 2 k L2


M AB  4k θ B  66 M BA  8k θ B  66 M BC  4k θ B  33.75 M CB  2k θ B  33.75


8k θB  66  4k θB  33.75  0

The corresponding end moments are: M AB  4k θ B  66  76.7 kN-m M BA  8k θ B  66  44.5 kN-m

10



k θ B  2.68

M BC  4k θ B  33.75  44.5 kN-m M CB  2k θ B  33.75  28.4 kN-m

(c) Let

k member AB 

E(2I2 ) k (2L2 )

22(12)2  264 kN-m 12 45(6)   33.75 kN-m 8

k member BC 

EI2 k L2

22(12)2  264 kN-m 12 45(6)   33.75 kN-m 8

M FAB 

F M BA 

M FBC

F M CB

The slope-deflection equations take the form: M AB  2k θ B  264 M BA  4k θ B  264

M BC  4k θ B  33.75 M CB  2k θ B  33.75


4k θB  264  4k θB  33.75  0

The corresponding end moments are: M AB  2k θ B  264  321.5 kN-m M BA  4k θ B  264  -148.9 kN-m M BC  4k θ B  33.75  148.9 kN-m M CB  2k θ B  33.75  23.8 kN-m

11



k θ B  28.78

Problem 10.5 E=29,000 ksi and I  300 in 4 , L=20 ft

Let

k member AB  k member BC 

EI k L

1.5(20)2  50 kip-ft 12 10(20)   25 kip-ft 8

M FAB  M FBC

1.5(20)2  50 kip-ft 12 10(20)   25 kip-ft 8

F M BA 

F M CB


M AB  2k θ B  50 M BA  4k θ B  50 M BC  2k (2θ B  θC )  25 M CB  2k (θ B  2θC )  25

Enforce moment equilibrium at nodes B and C: MCB  25  0

2k (θ B  2θC )  25  25  0

M BA  M BC  0 

k θB 

4k θ B  50  2k (2θ B  θ C )  25  0 25 7

k θC  

25 14

The corresponding end moments are: M AB  2k θ B  50  57.1 kip-ft M BA  4k θ B  50  35.7 kip-ft M BC  2k (2θ B  θC )  25  35.7 kip-ft M CB  2k (θ B  2θC )  25  25 kip-ft

12



2θ C  θ B 

8k θ B  2kθ C  25

Problem 10.6 E= 200 GPa, I  80(10)6 mm4 , P=45kN, h =3 m and L  9 m .

The slope-deflection equations take the form: M AB  0

M BAmodified 

3EI θB  L

M CB  0 3EI θB  L Enforce moment equilibrium at node B: M BCmodified 

M BAmodified  M BCmodified  135  6EI θ B   135 L  EI θ B  22.5 L The corresponding end moments are:

13

M BA =67.5 kN-m

counterclockwise

M BC =67.5 kN-m

counterclockwise

Problem 10.7 E=29,000 ksi, I  200 in 4 , L  18 ft , and w  1.2 k/ft .

Because of symmetry θA  θD and θB  θC EI Let k  L EI EI k member BC   0.5 k k member AB  k member CD  k 2L L

The slope-deflection equations take the form: M AB  0 M BAmodified  M BA modified  3k θB 

M BC  k 2θ B  θC



F  M BC  k 2θ B  θ B   129.6  k θ B  129.6

M CB   M BC Enforce moment equilibrium at node B:

14

M BAmodified  M BC  0  4kθ B  129.6  0  kθ B  32.4

The corresponding end moments are: M BA  3k θ B  97.2



M BA  97.2 kip-ft

clockwise

M BC  k θ B  129.6  97.2 kip-ft counterclockwise Because of symmetry MCB   M BC  97.2 kip-ft clockwise MCD  M BA  97.2 kip-ft

counterclockwise

Problem 10.8 Assume E= 200 GPa and I  80(10)6 mm4

Let

k member AB  k member CD 

30(12) 2  144 kN-m 30 45(18) M FBC   101.25 kN-m 8 30(12) 2 F M CD   216 kN-m 20 M FAB 

EI  1.5k 12

EI k 18

30(12) 2  216 kN-m 20 45(18) F M CB   101.25 kN-m 8 30(12) 2 M FDC    144 kN-m 30

F M BA 

Because of symmetry θB  θC , MBC  MCB 15

k member BC 

The slope-deflection equations take the form: M AB  M DC  2(1.5k) θ B  144 M BA  M CD  4(1.5k) θ B  216 MBC  MCB  2k (2θB  θC )  101.25  2k θB  101.25

Enforce moment equilibrium at either node B or C: MBA  MBC  0

4(1.5k) θB  216  2k θB  101.25  0

k θ B  14.34

The corresponding end moments are: M AB  M DC  2(1.5k) θ B  144  187 kN-m

M BA  M CD  4(1.5k) θ B  216  129.9 kN-m M BC  M CB  2k θ B  101.25  129.9 kN-m Problem 10.9 E=29,000 ksi, I  400 in 4 .

Let

16

k member AB  (

EI 3 )  k 20 2

k member BC 

EI k 30

k member CD  (

EI 5 )  k 18 3

The slope-deflection equations take the form: M AB  0 3 M BA modified  3( k) θ B   75 2

M BC  2k 2θ B  θC M CB 

M CD 

  90 2k θ  2θ   90 5 2( k) 2θ   36 3 B

C

C

5 M DC  2( k) θ C   36 3 Enforce moment equilibrium at nodes B and C: M BA  M BC  0 M CB  M CD  0  3 3( k) θ B   75  2k 2θ B  θ C   90  0 2 5 2k θ B  2θ C   90  2( k) 2θ C   36  0 3  8.5k θ B  2k θ C   15 2k θ B  10.667k θ C   54  k θB  3.08

The corresponding end moments are: M AB  0 3 M BA  3( k) θ B   75   88.9 kip-ft 2

M BC  2k 2θ B  θC M CB 

M CD 

  90  88.9 kip-ft 2k θ  2θ   90   73.6 kip-ft 5 2( k) 2θ   36  73.6 kip-ft 3 B

C

C

5 M DC  2( k) θ C   36   17.2 kip-ft 3

17

k θC   5.64

Problem 10.10 Assume E= 200 GPa, and I  100(10)6 mm4 .

Let

k member AB  k member BC 

EI k L

90(6)(3) 2  60 kN-m 92 30(9) 2   202.5 kN-m 12

90(6) 2 (3)  120 kN-m 92 10(20)   202.5 kN-m 8

M FAB 

F M BA 

M FBC

F M CB


M AB  2k θ B  60 M BA  4k θ B  120 M BC  2k (2θ B  θC )  202.5 M CB  2k (θ B  2θC )  202.5

Enforce moment equilibrium at nodes B and C: MCB  135  0 M BA  M BC  0 

k θB  16.6

2k (θ B  2θC )  202.5  135  0

4k θ B  120  2k (2θ B  θC )  202.5  0 k θC  25.18

The corresponding end moments are: M AB  2k θ B  60  126.8 kN-m M BA  4k θ B  120  186.4 kN-m

18



k (θ B  2θ C )  33.75 

8k θ B  2kθ C  82.5

M BC  2k (2θ B  θC )  202.5  186.4 kN-m M CB  2k (θ B  2θC )  202.5  135 kN-m

Problem 10.11 Assume E  29,000ksi and I  100 in 4 .

Let

k AB  k BC 

E(4I) EI )   2k 20 5

k BD modified 

The modified slope-deflection equations take the form: M BA modified  M BC modified  3(2k) θB M BD modified  3(k) θB

Enforce moment equilibrium at node B: M BA modified  M BC modified  M BD modified  12  0

 3(2k) θ B  3(2k) θ B  3(k) θ B  12  k θ B  0.8 The corresponding end moments are: M BA modified  M BC modified  3(2k) θ B  4.8 kip-ft M BD modified  3(k) θ B  2.4 kip-ft

19

EI k 10

Problem 10.12 Assume E  200 GPa , Ic  120(10)6 mm4 , L  8 m , h  4 m and P=50 kN. (a) Ib  Ic (b) Ib  1.5Ic

20

21

(a) Ib  Ic  120(10)6 mm4 Ic  120(10)6 mm4 θA  θB 

B 

M AD  M AB 

(b) Ic  120(10)6 mm4 , Ib  1.5Ic  180(10)6 mm4 22


For no axial deformation, bending moment is zero throughout the structure. Problem 10.14 Assume E  200 GPa , and I  80(10)6 mm4 .

23

24

Problem 10.15 I  600 in 4 A  6 in 2 E  29, 000 k/in 2

25

Problem 10.16 Assume E  200 GPa , and I  120(10)6 mm4 . Neglecting axial deformation. HC  HD  22.5 , θ A  θC

k AC 

EI  2k 6

k BC 

E(2I) k 24

M CA  4k (θ A  3 ) M AC  4k (2θ A  3 ) MAB  2k (2θA  θC )  6k θA

M AB  M BA  50.6 kN-m M AC  M BD  50.6 kN-m MCA  MDB  84.4 kN-m

26


For no axial deformation, bending moment is zero throughout the structure.

Problem 10.18 Assume E  200 GPa , and I  80(10)6 mm4 .

27

28


29

Deflection profile

Moment Diagram

For the following beams and frames defined in Problems 10.20- 10.34, determine the member end moments using moment distribution. Problem 10.20 E = 29,000 ksi, I = 300 in 4

30

M

F AB

M F BA

2(45) 2 10(60)   135 kip-ft M F BC   75 kip-ft 30 8 2(45) 2 F    202.5 kip-ft M CB   75 kip-ft 20

DFDC  DFAB  0

 I 7I I I  ( )( )   45 60 180  L  I  4 At joints B or C DFBA  DFCD  45  7 I 7  180   4 3 DFCB  DFBC  1-  7 7 

Case(a):

31

1.0(45) 2  168.75 kip-ft 12   168.75 kip-ft

M FCD  M F DC

Case (b): M F AB set.  M F BA set. 

6EIδB 6(29000)(300)(.5)   7.45 kip-ft L2 (45) 2 (12)3

M F BC set.  M FCB set. 

32

6(29000)(300)(.5)  4.19 kip-ft (60) 2 (12)3

Case (c):

Moment diagram-loading kip-ft

Moment diagram –settlement kip-ft

Moment diagram -( loading + settlement) kip-ft

33

Problem 10.21 Check your results with Computer analysis. Assume E  200 GPa , and I  75(10)6 mm4 .

34

Shear diagram

Moment diagram

Problem 10.22 Assume EI constant. (a)

35

(b)

36

Problem 10.23 Determine the bending moment distribution. Assume EI is constant.

37

Problem 10.24 Determine the bending moment distribution. Assume I1  1.4I2 .

38

Problem 10.25 Determine the bending moment distribution.

39

Problem 10.26 Solve for the bending moments. δB  .4 in  ,E=29,000 ksi and I  240 in 4 .

40

Problem 10.27 Determine the bending moment distribution and the deflected shape. E=29,000 ksi (a)

41

Take I1  I2  1000 in 4

Moment diagram

Deflected shape

(b)

42

Take I2  1.5 I1 . Use Computer analysis.

Moment diagram

Deflected shape Case (a) has larger moment and deflection compare to case(b).

43

Problem 10.28 Determine the axial, shear, and bending moment distributions. Take Ig =2Ic

Axial Force

44

Shear diagram

Moment Diagram (kN-m) Problem 10.29 Determine the member forces and the reactions. a) Consider only the uniform load shown b) Consider only the support settlement of joint D ( δ D = .5 in↓) c) Consider only the temperature increase of ΔT  80/0 F for member BC. E = 29,000 ksi, IAB  ICD  100 in 4 , IBC  400 in 4 , α = 6.5x10-6 / 0 F

45

(a)

(b)

46

(c)

Problem 10.30 Determine the bending moment distribution for the following loadings. Take Ig  5Ic .

(a)

47

Moment diagram (kN-m)

(b)

Symmetrical loading- no sway

Moment diagram (kN-m)

48

(c)


Moment diagram (kN-m (d)

49

Moment diagram (kN-m) Problem 10.31 Solve for the bending moments.

50

Moment diagram Problem 10.32 Determine the bending moment distribution.

51

Moment diagram (kip-ft) Problem 10.33 Solve for the bending moments.


52

Moment diagram Problem 10.34 For the frame shown, determine the end moments and the reactions. Assume E  200 GPa , and

I  40(10)6 mm4 .

M F AB  0 M F BA  

24(6) 2  108 kN-m 8

 I 5I 3 4I 3 4I 3 I  ( ) ( ) ( )   4 6 4 6 4 3 4  L  I Joint B  DF  DF  2  0.4 DFDB  .2 CB  BA 5I  4 53

Problem 10.35 Determine analytic expression for the rotation and end moments at B. Take I  1000 in 4 , A=20

in 2 for all members, and α =1.0, 2.0, 5.0. Is there a upper limit for the end moment, M BD ?

54

M BD is independent of α. MBD  240 kip-ft

55

k AB  k BC 

EI k 20

k BD 

αI E  αk 20

M BA modified  3(k) θ B M BC modified  3(k) θ B M BD modified  3(αk) θ B  3(αk)

MBA modified  3(k) θB  180



k θB  60

M0

joint B

- 3(αk) θ B  3(αk)  6(k) θ B  120  0 MBD modified  3(α)(-60)  3(αk)  240

k

56

EI 29000(1000) 1   10069.4 k-ft 20 20 (12)2

k 

80  60α α



Δ 20

3.33 in (80  60α)(20)12  Δ  L   2.38 in αk 1.81 in 

for α  1.0 for α  2.0 for α  5.0

Problem 10.36 Compare the end moments and horizontal displacement at A for the rigid frames shown below. Check your results for parts (c) and (d) with a computer based analysis. Take E  200 GPa , and I  120(10)6 mm4 . A=10000 mm2 for all members.

Case (a)

57

Moment diagram kN-m Case (b)

Moment diagram kN-m

Case (c)

58

Moment diagram kN-m

Case (d)

59

Moment diagram kN-m

60

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