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
6 (P e) P q effective L1 L 2 L 2 L12
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 (P e) P q effective L1 L 2 L 2 L12
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
M Bresisting M Boverturning
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
Page 11 of Chapter I
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
Page 12 of Chapter I
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
Page 13 of Chapter I
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
M Bresisting M Boverturning
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
Page 14 of Chapter I
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 )
Page 15 of Chapter I
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
Page 16 of Chapter I
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
Page 17 of Chapter I
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
Page 18 of Chapter I
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
Page 11 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.
Page 12 of Chapter 9
Page 13 of Chapter 9
Page 14 of Chapter 9
(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 .
Page 15 of Chapter 9
(d) Suppose the horizontal support at B is replaced by a member extending from A to B. Repeat part (a).
Page 16 of Chapter 9
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.
Page 17 of Chapter 9
Page 18 of Chapter 9
Problem 9.9 Take A 20,000 mm2
Page 19 of Chapter 9
I 400(10)6 mm4 and E 200 GPa
Case (a):
Mmax 0 (no bending moment)
Deflected shape Case (b):
Page 20 of Chapter 9
Mmax 1101 kN-m
Deflected shape
Case (c):
Page 21 of Chapter 9
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
Page 22 of Chapter 9
I 1000 in 4 E 29,000 ksi
Bending moment
Axial force
Bending moment
Page 23 of Chapter 9
Axial force
Bending moment
Axial force Problem 9.11
Page 24 of Chapter 9
A 40 in 2
I 1200 in 4 E 29,000 ksi
Case(a):
Mmax 0 (no bending moment) Case (b):
Page 25 of Chapter 9
Mmax 130 kip-ft
Case (c):
Mmax 43.3 kip-ft Case (d):
Page 26 of Chapter 9
Mmax 10.9 kip-ft
Problem 9.12 Determine the horizontal reaction at support D.
(See Figure 9.38)
Page 27 of Chapter 9
H D1
P 1 kip 2
(See example 9.15)
rg
I2 I L 16
H D2
Problem 9.13
Page 28 of Chapter 9
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
Page 29 of Chapter 9
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
Page 30 of Chapter 9
EIδ BB
for h 2m for h 4m for h 6m
0.00263(10) 6 0.00357(10) 6 6 0.0049(10)
for h 2m for h 4m for h 6m
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
Page 31 of Chapter 9
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
Page 32 of Chapter 9
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.
Page 33 of Chapter 9
(a)
Page 34 of Chapter 9
Page 35 of Chapter 9
(b)
Page 36 of Chapter 9
Page 37 of Chapter 9
Page 38 of Chapter 9
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):
Page 39 of Chapter 9
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
Page 40 of Chapter 9
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
Page 41 of Chapter 9
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.
Page 42 of Chapter 9
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
Page 43 of Chapter 9
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
Page 44 of Chapter 9
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
Page 45 of Chapter 9
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
Page 46 of Chapter 9
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
Page 47 of Chapter 9
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.
Page 48 of Chapter 9
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.
Page 49 of Chapter 9
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
The slope-deflection equations take the form:
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
The slope-deflection equations take the form:
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
Enforce moment equilibrium at node B: MBA MBC 0
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
Enforce moment equilibrium at node B: MBA MBC 0
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
The slope-deflection equations take the form:
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
The slope-deflection equations take the form:
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
Problem 10.13 Assume E 29,000ksi and I 200 in 4 .
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
Problem 10.17 Assume E 29,000ksi and I 200 in 4 .
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
Problem 10.19 Assume E 29,000ksi and I 200 in 4 .
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)
Symmetrical loading- no sway
Moment diagram (kN-m (d)
49
Moment diagram (kN-m) Problem 10.31 Solve for the bending moments.
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Moment diagram Problem 10.32 Determine the bending moment distribution.
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Moment diagram (kip-ft) Problem 10.33 Solve for the bending moments.
Symmetrical loading- no sway
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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 ?
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M BD is independent of α. MBD 240 kip-ft
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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
M0
joint B
- 3(αk) θ B 3(αk) 6(k) θ B 120 0 MBD modified 3(α)(-60) 3(αk) 240
k
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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)
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Moment diagram kN-m Case (b)
Moment diagram kN-m
Case (c)
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Moment diagram kN-m
Case (d)
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Moment diagram kN-m
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