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Determine the moments at A, B , and C and then draw the moment diagram. EI is constan constant. t. Assum Assume e the the support at B is a roller and A and C are fixed. 11–1.
3k
3k
B
A 3 ft
3 ft
Fixed End Moments. Referring to the table on the inside back cover
2(3)(9) 2PL = = - 6 k # ft 9 9 2(3)(9) 2PL = = = 6 k # ft 9 9
(FEM)AB = (FEM)BA
4(20) PL = = - 10 k # ft 8 8 4(20) PL = = = 10 k # ft 8 8
(FEM)BC = (FEM)BC
Slope-Deflection Equations. Applyin Applying g Eq. 11–8, MN = 2Ek(2uN + uF - 3c) + (FEM)N
For span AB, MAB = 2E MBA = 2E
a 9 b [2(0) + - 3(0)] + ( - 6) = a 2 9 b - 6 a 9 b [2 + 0 - 3(0)] + 6 = a 4 9 b + 6 I
EI
uB
I
EI
uB
uB
uB
(1) (2)
For span BC , MBC = 2E MCB = 2E
a 20 b [2 + 0 - 3(0)] + ( - 10) = a 5 b a 20 b [2(0) + - 3(0)] + (10) = a 10 b I
EI
uB
I
EI
uB
uB - 10
(3)
uB + 10
(4)
Equilibrium. At Support B, MBA + MBC = 0
(5)
Substitute Substi tute Eq. 2 and 3 into (5),
a49 b EI
uB + 6 +
a 5 b EI
uB - 10 = 0
uB =
180 29EI
Substitute this result into Eqs. Eqs. 1 to 4, MAB = - 4.621 k # ft = - 4.62 k # ft
Ans.
MBA = 8.759 k # ft = 8.76 k # ft
Ans.
MBC = - 8.759 k # ft = - 8.76 k # ft
Ans.
MCB
Ans.
= 10.62 k # ft
= 10.6 k # ft
The Negative Signs indicate that M AB and MBC have the counterclockwise rotational sense. Using these results, results, the shear at both ends of span AB and BC are computed and shown in Fig. a and b, res respec pective tively ly.. Sub Subseq sequen uently tly,, the shear shear and and moment diagram can be plotted, Fig. c and d respectively.
4 06
4k
3 ft
C
10 ft
10 ft
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11–1. 11 –1.
Conti Co ntinue nued d
Determine the moments at A, B, an and d C , the then n draw draw the moment diagram for the beam. The moment of inertia of each span is indicated in the figure. Assume the support at B is a roller and A and C are fixed. E = 29(103) ksi. 11–2.
C
A
I AB
900 in.
24 ft
Fixed End Moments. Referring to the table on the inside back cover,
(FEM)AB = (FEM)BA =
(FEM)CB =
2(242) wL2 = = - 96 k # ft 12 12
2(242) wL2 = = 96 k # ft 12 12
(FEM)BC = -
30 k
2 k/ ft ft
30(16) PL = = - 60 k # ft 8 8
30(16) PL = = 60 k # ft 8 8 407
4
B
I BC
8 ft
1200 in. 8 ft
4
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11–2. 11– 2.
Conti Co ntinue nued d
Slope-Deflection Equations. Applyin Applying g Eq. 11–8, MN = 2Ek (2uN + uF - 3c) + (FEM)N
For span AB, 4
MAB = 2E
900 in c 24(12) d [2(0) + in
#
uB - 3(0)] + [ - 96(12) k in]
MAB = 6.25EuB – 1152
(1)
4
MBA = 2E
900 in c 24(12) d [2 in
#
uB + 0 - 3(0)] + 96(12) k in
MBA = 12.5EuB + 1152
(2)
For span BC , 4
MBC = 2E
1200 in c 16(12) d [2 in
#
uB + 0 - 3(0)] + [ - 60(12) k in]
MBC = 25EuB - 720
(3)
4
MCB = 2E
1200 in c 16(12) d [2(0) + in
#
uB - 3(0)] + 60(12) k in
MCB = 12.5EuB + 720
(4)
Equilibrium. At Support B, MBA + MBC = 0
(5)
Substitute Eqs. 3(2) and (3) into (5), 12.5EuB + 1152 + 25EuB - 720 = 0 uB = -
11.52 E
Substitute this result into Eqs. Eqs. (1) to (4), MAB = - 1224 k # in = - 102 k # ft
Ans.
MBA = 1008 k # in = 84 k # ft MBC = - 1008 k # in = - 84 k # ft MCB = 576 k # in = 48 k # ft
Ans. Ans. Ans.
The negative signs indicate that M AB and MBC have counterclockwise rotational senses.Using senses. Using these results, the shear at both ends of spans spans AB and BC are computed and shown in Fig. a and b, res respec pective tively ly.. Sub Subseq sequen uently tly,, the shear shear and and moment moment diagram can be plotted, Fig. c and d respectively.
4 08
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11–2. 11 –2.
Conti Co ntinue nued d
Determine the moments at the supports A and C , then draw the moment diagram. Assume joint B is a roller. EI is constant. 11–3.
25 kN
15 kN/ kN/ m
A
B
3m
MN = 2E
a b (2 I L
uN + uF - 3c) + (FEM)N
MAB =
(25)(6) 2EI (0 + uB) 6 8
MBA =
(25)(6) 2EI (2uB) + 6 8
MBC =
(15)(4)2 2EI (2uB) 4 12
MCB
(15)(4)2 2EI (uB) + = 4 12
Equilibrium. MBA + MBC = 0
25(6) 15(4)2 2EI 2EI (2uB) + (2uB) + = 0 6 8 4 12 uB =
0.75 EI MAB = - 18.5 kN # m
Ans.
MCB = 20.375 kN # m = 20.4 kN # m
Ans.
MBA = 19.25 kN # m
Ans.
MBC =
Ans.
- 19.25 kN # m
409
3m
C
4m
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Determine the moments at the supports, supports, then draw the moment diagram. Assume B is a roller and A and C are fixed. EI is constant. *11–4.
15 kN 15 kN 15 kN 25 kN/ m
A
B
3m 6m
11(25)(6)2
(FEM)AB = (FEM)BA
192
= - 51.5625 kN # m
5(25)(6)2 = = 23.4375 kN # m 192
(FEM)BC =
- 5(15)(8) = - 37.5 kN # m 16
(FEM)CB = 37.5 kN # m MN = 2E
a b (2 I L
MAB = 2E MAB =
a 6 b (2(0) + I
EIuB
MCB =
I
uB + 0 - 0) + 23.4375
(2)
a 8 b (2 I
EIuB
2
MCB = 2E
(1)
a 6 b (2
2EIuB + 23.4375 3
MBC = 2E MBC =
uB - 0) - 51.5625
- 51.5625
3
MBA = 2E MBA =
uN + uF - 3c) + (FEM)N
uB + 0 - 0) - 37.5
- 37.5
a 8 b (2(0) + I
EIuB
4
(3) uB - 0) + 37.5
+ 37.5
(4)
Equilibrium. MBA + MBC = 0
(5)
Solving: uB =
12.054 EI MAB = - 47.5 kN # m
Ans.
MBA = 31.5 kN # m MBC = - 31.5 kN # m MCB = 40.5 kN # m
Ans. Ans. Ans.
4 10
2m
C
2m
2m
2m
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Determine the moment at A, B, C and D, the then n draw the moment diagram for the beam. Assume the supports at A and D are fixed and B and C are rollers. EI is constant. 11–5.
20 kN/ m
A
5m
Fixed End Moments. Referring to the table on the inside back cover,
(FEM)AB = 0 (FEM)BC = (FEM)CB =
(FEM)BA = 0
(FEM)CD = 0
(FEM)DC = 0
20(32) wL2 = = - 15 kN # m 12 12
20(32) wL2 = = 15 kN # m 12 12
Slope-Deflection Equation. Applyin Applying g Eq. 11–8, MN = 2Ek(2uN + uF - 3c) + (FEM)N
For span AB, MAB = 2E MBA = 2E
a 5 b [2(0) + - 3(0)] + 0 = a 2 5 b a 5 b [2 + 0 - 3(0)] + 0 = a 4 5 b I
EI
uB
I
EI
uB
(1)
uB
(2)
uB
For span BC , MBC = 2E MCB = 2E
a 3 b [2 a 3 b [2 I
a 4 3 b + a 2 3 b - 15 4 2 b b + 15 - 3(0)] + 15 = a + a 3 3 EI
uB + uC - 3(0)] + ( - 15) =
I
uC + uB
EI
uC
EI
uB
EI
uB
uC
(3) (4)
For span CD, MCD = 2E MDC = 2E
a 5 b [2 + 0 - 3(0)] + 0 = a 4 5 b a 5 b [2(0) + - 3(0)] + 0 = a 2 5 b I
EI
uC
I
EI
uC
(5)
uC
(6)
uC
Equilibrium. At Support B, MBA + MBC = 0
a 4 5 b + a 4 3 b + a 2 3 b - 15 = 0 a 3215 b + a 2 3 b = 15 EI
EI
uB
EI
uB
uB
EI
EI
uC
(7)
uC
At Support C , MCB + MCD = 0
a 4 3 b + a 2 3 b EI
uC
EI
uB + 15 +
a 4 5 b EI
uC = 0
411
C
B
3m
D
5m
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11–5. 11 –5.
Conti Co ntinue nued d
a 2 3 b + a 3215 b EI
EI
uB
uC = - 15
(8)
Solving Eqs. Eqs. (7) and (8) uB =
225 22EI
uC = -
225 22EI
Substitute these results into Eqs. Eqs. (1) to (6), MAB = 4.091 kN # m = 4.09 kN # m
Ans.
MBA = 8.182 kN # m = 8.18 kN # m
Ans.
MBC = - 8.182 kN # m = - 8.18 kN # m MCB = 8.182 kN # m = 8.18 kN # m MCD = - 8.182 kN # m = - 8.18 kN # m MDC = - 4.091 kN # m = - 4.09 kN # m
Ans. Ans. Ans. Ans.
The negative sign indicates that MBC , MCD and MDC have counterclockwise rotational rotatio nal sense. sense. Using these results results,, the shear at both ends of spans AB, BC , and CD are computed and shown in Fig. a, b, an and d c respectively. Subsequently Subsequently,, the shear d e and moment diagram can be plotted, Fig. , an and d respectively.
4 12
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Determine the moments at A, B , C and D, then draw the moment moment diagram diagram for the the beam. Assum Assume e the supports at A and D are fixed and B and C are rollers. EI is constant. 11–6.
9k 2 k/ ft ft
A
B
15 ft
Fixed End Moments. Referring to the table on the inside back cover, 2
(FEM)AB
2(15) wL2 = = = - 37.5 k # ft 12 12
(FEM)BA
2(152) wL2 = = = 37.5 k # ft 12 12
(FEM)BC = (FEM)CB = 0 2(9)(15) - 2PL = = - 30 k # ft 9 9 2(9)(15) 2PL = = = 30 k # ft 9 9
(FEM)CD = (FEM)DC
Slope-Deflection Equation. Applyin Applying g Eq. 11–8, MN = 2Ek(2uN + uF - 3c) + (FEM)N
For span AB, MAB = 2E MBA = 2E
a 15 b [2(0) + - 3(0)] + ( - 37.5) = a 215 b - 37.5 a 15 b [2 + 0 - 3(0)] + 37.5 = a 415 b + 37.5 I
EI
uB
I
EI
uB
uB
uB
(1) (2)
For span BC , MBC = 2E MCB = 2E
a 15 b [2 a 15 b [2 I
I
a 415 b + a 215 b 4 2 b b - 3(0)] + 0 = a + a 15 15
uB + uC - 3(0)] + 0 = uC + uB
EI
EI
uB
uC
EI
EI
9k
uC
(3)
uB
(4)
413
C
15 ft
5 ft
D
5 ft
5 ft
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11–6. 11 –6.
Conti Co ntinue nued d
For span CD, MCD = 2E MDC = 2E
a 15 b [2 + 0 - 3(0)] + ( - 30) = a 415 b a 15 b [2(0) + - 3(0)] + 30 = a 215 b I
EI
uC
I
EI
uC
uC - 30
(5)
uC + 30
(6)
Equilibrium. At Support B, MBA + MBC = 0
a 415 b + 37.5 + a 415 b + a 215 b a 815 b + a 215 b = - 37.5 EI
EI
uB
EI
EI
uB
EI
uB
uC = 0
(7)
uC
At Support C , MCB + MCD = 0
a 415 b + a 215 b + a 415 b - 30 = 0 a 815 b + a 215 b = 30 EI
EI
uC
EI
EI
uB
uC
EI
uC
(8)
uB
Solving Eqs. (7) and (8), uC =
78.75 EI
uB = -
90 EI
Substitute these results into Eqs. Eqs. (1) to (6), MAB = - 49.5 k # ft
Ans.
MBA = 13.5 k # ft MBC = - 13.5 k # ft MCB = 9 k # ft MCD = - 9 k # ft MDC = 40.5 k # ft
Ans. Ans. Ans. Ans. Ans.
The negative signs indicate that M AB, MBC and MCD have counterclockwise rotational rotatio nal sense. sense. Using these results results,, the shear at both ends of spans AB, BC , and CD are computed and shown in Fig. a, b, an and d c respectively. Subsequently Subsequently,, the shear and moment diagram can be plotted, Fig. d, an and d e respectively.
4 14
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Determine the moment at B, then draw the the moment diagram for the beam. Assume the supports at A and C are pins and B is a roller. EI is constant. 11–7.
40 kN 20 kN
A
B
6m
Fixed End Moments. Referring to the table on the inside back cover,
(FEM)BA =
a b a P
L2
(FEM)BC = -
a2b ba + 2
b = a b c
2
40 82
22(6) 6 (2) + 2 2
d = 52.5 kN # m
3(20)(8) 3PL = = - 30 kN # m 16 16
Slope-Deflection Equations. Applyin Applying g Eq. 11–10 Since Since one of the end’s end’s support for spans AB and BC is a pin. MN = 3Ek(uN - c) + (FEM)N
For span AB, MBA = 3E
a 8 b( I
uB - 0) + 52.5 =
a38 b
uB - 0) + (- 30) =
a 3 8 b
EI
uB + 52.5
(1)
For span BC , MBC = 3E
a 8 b( I
EI
uB - 30
(2)
Equilibrium. At support B, MBA + MBC = 0
a 38 b EI
uB + 52.5 +
a34 b EI
a 3 8 b EI
uB - 30 = 0
uB = - 22.5
uB = -
30 EI
415
2m
C
4m
4m
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11–7. 11 –7.
Conti Co ntinue nued d
Substitute this result into Eqs. Eqs. (1) and (2) MBA = 41.25 kN # m
Ans.
MBC = - 41.25 kN # m
Ans.
The negative sign indicates that MBC has counterclockwise counterclockwise rotational sense. Using this result, the shear at both ends of spans AB and BC are computed and shown in Fig. a and b respe respectively ctively.. Subseq Subsequently uently,, the shear and Moment Moment diagram can be plotted,, Fig plotted Fig.. c and d respectively.
Determine the moments at A, B, an and d C , the then n draw draw the moment diagram. EI is constant. Assume the support at B is a roller and A and C are fixed. *11–8.
6k
0.5 k/ k/ ft ft
B
A
8 ft
(FEM)AB = (FEM)BA =
PL = - 12, 8
PL = 12, 8
(FEM)BC = (FEM)CB =
wL2 = - 13.5 12
wL2 = 13.5 12
uA = uC = cAB = cBC = 0
MN = 2E
a b (2 I L
uN + uF - 3c) + (FEM)N
MAB =
2EI (uB) - 12 16
MBA =
2EI (2uB) + 12 16
MBC =
2EI (2uB) - 13.5 18
MCB =
2EI (uB) + 13.5 18
Moment equilibrium at B: MBA + MBC = 0
2EI 2EI (2uB) + 12 + (2uB) - 13.5 = 0 16 18 uB =
3.1765 EI
4 16
8 ft
C
18 ft
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11–8 11 –8..
Conti Co ntinue nued d
Thus MAB = - 11.60 = - 11.6 k # ft
Ans.
MBA = 12.79 = 12.8 k # ft
Ans.
MBC = - 12.79 = - 12.8 k # ft
Ans.
MCB = 13.853 = 13.9 k # ft
Ans.
Left Segment c + a MA = 0;
- 11.60 + 6(8) + 12.79 - VBL(16) = 0 VBL = 3.0744 k
+ c a Fy = 0;
Ay = 2.9256 k
Right Segment c + a MB = 0;
- 12.79 + 9(9) - Cy(18) + 13.85 = 0 Cy = 4.5588 k
+ c a Fy = 0;
VBK = 4.412 k
At B By = 3.0744 + 4.4412 = 7.52 k
Determine the moments at each support, then draw the moment diagram. Assume A is fixed. EI is constant. 11–9.
B
A
20 ft
MN = 2E
a b (2 I L
uN + uF - 3c) + (FEM)N
MAB =
4(20)2 2EI (2(0) + uB - 0) 20 12
MBA =
4(20)2 2EI (2uB + 0 - 0) + 20 12
MBC =
2EI (2uB + uC - 0) + 0 15
MCB =
2EI (2uC + uB - 0) + 0 15
MN = 3E MCD =
a b( I L
12 k
4 k/ k/ ft ft
uN - c) + (FEM)N
3(12)16 3EI (uC - 0) 16 16 417
D
C
15 ft
8 ft
8 ft
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11–9. 11 –9.
Conti Co ntinue nued d
Equilibrium. MBA + MBC = 0 MCB + MCD = 0
Solving uC =
178.08 EI
uB = -
336.60 EI
MAB = - 167 k # ft
Ans.
MBA = 66.0 k # ft
Ans.
MBC = - 66.0 k # ft
Ans.
MCB = 2.61 k # ft
Ans.
MCD = - 2.61 k # ft
Ans.
Determine the moments at A and B, then draw draw the moment diagram for the beam. EI is constant. 11–10.
2400 lb
200 lb/ lb/ ft ft
A
B
30 ft
(FEM)AB = -
C
10 ft
1 1 (w)(L2) = - (200)(302) = - 15 k # ft 12 12
MAB =
2EI (0 + uB - 0) - 15 30
MBA =
2EI (2uB + 0 - 0) + 15 30
a MB = 0;
MBA = 2.4(10)
Solving, uB =
67.5 EI
MAB = - 10.5 k # ft
Ans.
MBA = 24 k # ft
Ans.
4 18
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Determine the moments at A, B, an and d C , the then n draw draw the moment diagram for the beam. Assume the support at A is fixed, B and C are rollers, rollers, and D is a pin. EI is constant. 11–11.
6k
6k 3 k/ ft ft
B
A
4 ft
4 ft
Fixed End Moments. Referring to the table on the inside back cover,
(FEM)AB = (FEM)BA =
2(6)(12) 2PL = = - 16 k # ft 9 9
2(6)(12) 2PL = = 16 k # ft 9 9
(FEM)BC = (FEM)CB = 0
(FEM)CD
3(122) wL2 = = = - 54 k # ft 8 8
Slope-Deflection Equations. Applyi Applying ng Eq. 11–8, for spans spans AB and BC . MN = 2Ek(2uN + uF - 3c) + (FEM)N
For span AB, MAB = 2E MBA = 2E
a 12 b [2(0) + - 3(0)] + ( - 16) = a 6 b - 16 a 12 b [2 + 0 - 3(0)] + 16 = a 3 b + 16 I
EI
uB
I
EI
uB
uB
uB
(1) (2)
For span BC , MBC = 2E MCB = 2E
a 12 b [2 a 12 b [2 I
a 3 b + a 6 b - 3(0)] + 0 = a b + a b 3 6
uB + uC - 3(0)] + 0 =
I
uC + uB
EI
EI
uB
uC
EI
EI
uC
(3)
uB
(4)
Applying Applyi ng Eq. 11–10 for span CD, MN = 3Ek(uN - c) + (FEM)N MCD = 3E
a 12 b ( I
uC - 0) + ( - 54) =
a 4 b EI
uC - 54
(5)
Equilibrium. At support B, MBA + MBC = 0
a 3 b + 16 + a 3 b + a 6 b a 2 3 b + a 6 b = - 16 EI
EI
EI
uB
EI
uB
uB
EI
uC = 0
(6)
uC
At support C , MCB + MCD = 0
a 3 b + a 6 b + a 4 b a 712 b + a 6 b = 54 EI
EI
uC
uC
EI
EI
uB
EI
uC - 54 = 0
(7)
uB
419
4 ft
C
12 ft
D
12 ft
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11–11. 11–1 1.
Continue Cont inued d
Solving Eqs. Eqs. (6) and (7) uC =
1392 13EI
uB = -
660 13EI
Substitute these results into Eq. (1) to (5) MAB = - 24.46 k # ft = - 24.5 k # ft
Ans.
MBA = - 0.9231 k # ft = - 0.923 k # ft MBC = 0.9231 k # ft = 0.923 k # ft MCB = 27.23 k # ft = 27.2 k # ft MCD = - 27.23 k # ft = - 27.2 k # ft
Ans. Ans. Ans. Ans.
The negative signs indicates that M AB, MBA, and MCD have counterclockwise rotational rotatio nal sense. sense. Using these results results,, the shear at both ends of spans AB, BC , and CD are computed and shown in Fig. a, b, an and d c respectively. Subsequently Subsequently,, the shear and moment diagram can be plotted, Fig. d and e respectively.
4 20
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Determine the moments acting at A and B. Assume A is fixed supported, B is a roller roller,, and C is a pin. EI is constant. *11–12.
20 kN/ kN/ m m
3m
9m
wL2 = - 54, 30
(FEM)BA =
wL2 = 81 20
(FEM)BC =
C
B
A
(FEM)AB =
80 kN
3m
3PL = - 90 16
Applying Eqs. Eqs. 11–8 and 11–10, MAB =
2EI (uB) - 54 9
MBA =
2EI (2uB) + 81 9
MBC =
3EI (uB) - 90 6
Moment equilibrium at B: MBA + MBC = 0 EI 4EI (uB) + 81 + uB - 90 = 0 9 2 uB =
9.529 EI
Thus, MAB = - 51.9 kN # m
Ans.
MBA = 85.2 kN # m
Ans.
MBC = - 85.2 kN # m
Ans.
Determine the moments at A, B, an and d C , the then n draw draw the moment diagram for each member. Assume all joints are fixed connected. EI is constant. 11–13.
4 k/ k/ ft ft B A
18 ft 9 ft
C
2
(FEM)AB =
- 4(18) = - 108 k # ft 12
(FEM)BA = 108 k # ft (FEM)BC = (FEM)CB = 0
421
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11–13. 11–1 3.
Continue Cont inued d
a b (2 + = 2 a b (2(0) + 18
MN = 2E MAB
I L
uN
I
E
uF - 3c) + (FEM)N uB - 0) - 108
MAB = 0.1111EIuB - 108 MBA = 2E
(1)
a 18 b (2 I
uB + 0 - 0) + 108
MBA = 0.2222EIuB + 108 MBC = 2E
(2)
a 9 b (2 I
uB + 0 - 0) + 0
MBC = 0.4444EIuB MCB = 2E
a 9 b (2(0) + I
(3) uB - 0) + 0
MCB = 0.2222EIuB
(4)
Equilibrium MBA + MBC = 0
(5)
Solving Eqs. Eqs. 1–5: uB =
- 162.0 EI
MAB = - 126 k # ft
Ans.
MBA = 72 k # ft
Ans.
MBC = - 72 k # ft
Ans.
MCB = - 36 k # ft
Ans.
4 22
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Determine the moments at the supports, supports, then draw the moment diagram. The members are fixed connected at the supports and at joint B. The moment moment of inertia of each member is given in the figure. Take E = 29(103) ksi . 11–14.
20 k 8 ft
8 ft B
A
I AB
800 in 4
6 ft I BC
1200 in4
15 k
6 ft
(FEM)AB
- 20(16) = = - 40 k # ft 8
C
(FEM)BA = 40 k # ft (FEM)BC =
- 15(12) = - 22.5 k # ft 8
(FEM)CB = 22.5 k # ft MN = 2E MAB
a b (2 I L
uN + uF - 3c) + (FEM)N
2(29)(103)(800) (2(0) + uB - 0) - 40 = 16(144)
MAB = 20,138.89uB - 40 MBA =
2(29)(103)(800) 16(144)
(1)
(2uB + 0 - 0) + 40
MBA = 40,277.78uB + 40 MBC =
(2)
2(29)(103)(1200) (2uB + 0 - 0) - 22.5 12(144)
MBC = 80,555.55uB - 22.5 MCB =
2(29)(103)(1200) 12(144)
(3)
(2(0) + uB - 0) + 22.5
MCB = 40,277.77uB + 22.5
(4)
Equilibrium. MBA + MBC = 0
(5)
Solving Solvin g Eqs. 1–5: uB = - 0.00014483
MAB = - 42.9 k # ft
Ans.
MBA = 34.2 k # ft MBC = - 34.2 k # ft MCB = 16.7 k # ft
Ans.
423
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Determine the moment at B, then draw the moment moment diagram for each member of the frame. frame. Assume the support at A is fixed and C is pinned. EI is constant. 11–15.
2 kN/ kN/ m
A
B
Fixed End Moments. Referring to the table on the inside back cover,
(FEM)AB
2(32) wL2 = = = - 1.50 kN # m 12 12
(FEM)BA
2(32) wL2 = = = 1.50 kN # m 12 12
3m 4m
(FEM)BC = 0 Slope-Deflection Equations. Applyin Applying g Eq. 11–8 for member AB,
C
MN = 2Ek(2uN + uF - 3c) + (FEM)N MAB = 2E MBA = 2E
a 3 b [2(0) + - 3(0)] + ( - 1.50) = a 2 3 b - 1.50 a 3 b [2 + 0 - 3(0)] + 1.50 = a 4 3 b + 1.50 I
EI
uB
I
EI
uB
uB
uB
(1) (2)
Applying Applyin g Eq. 11–10 for member BC , MN = 3Ek(uN - c) + (FEM)N MBC = 3E
a 4 b( I
uB - 0) + 0 =
a 3 4 b EI
(3)
uB
Equilibrium. At Joint B, MBA + MBC = 0
a43 b EI
uB + 1.50 + uB = -
a 3 4 b EI
uB = 0
0.72 EI
Substitute this result into Eqs. Eqs. (1) to (3) MAB = - 1.98 kN # m
Ans.
MBA = 0.540 kN # m
Ans.
MBC = - 0.540 kN # m
Ans.
The negative signs indicate that M AB and MBC have counterclockwise rotational sense.. Using these sense these results results,, the shear shear at both ends ends of member member AB and BC are computed and shown in Fig. a and b respe respectively ctively.. Subse Subsequentl quently y, the shear and moment diagram can be plotted, Fig. c and d respectively.
4 24
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Determine the moments at B and D, th then en draw draw the moment diagram. diagram. Assum Assume e A and C are pinned and B and D are fixed connected. EI is constant. *11–16.
8k 15 ft A
10 ft B
12 ft D
(FEM)BA = 0 (FEM)BC =
- 3(8)(20) = - 30 k # ft 16
(FEM)BD = (FEM)DB = 0
a b( = 3 a b( 15
MN = 3E MBA
I L
uN - c) + (FEM)N
I
E
uB - 0) + 0
MBA = 0.2EIuB MBC = 3E
(1)
a 20 b ( I
uB - 0) - 30
MBC = 0.15EIuB - 30
a b (2 = 2 a b (2 12
MN = 2E MBD
(2)
I L
I
E
uN + uF - 3c) + (FEM)N uB + 0 - 0) + 0
MBD = 0.3333EIuB MDB = 2E
a 12 b (2(0) + I
(3) uB - 0) + 0
MDB = 0.1667EIuB
(4)
Equilibrium. MBA + MBC + MBD = 0
(5)
Solving Solvin g Eqs. 1–5: uB =
43.90 EI
MBA = 8.78 k # ft
Ans.
MBC = - 23.41 k # ft
Ans.
MBD = 14.63 k # ft
Ans.
MDB = 7.32 k # ft
Ans.
425
10 ft C
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Determine the moment that each member exerts on the joint at B, then draw the moment moment diagram for each member of the frame.Assume the support at A is fixed and C is a pin. EI is constant. 11–17.
2 k/ ft ft
B C
15 ft
6 ft
10 k
6 ft
A
Fixed End Moments. Referring to the table on the inside back cover,
(FEM)AB = (FEM)BC
10(12) PL = = - 15 k # ft 8 8
10(12) PL = = 15 k # ft 8 8
(FEM)BA =
2(152) wL2 = = = - 56.25 k # ft 8 8
Slope Reflection Equations. Applyin Applying g Eq. 11–8 for member AB, MN = 2Ek(2uN + uF - 3c) + (FEM)N
a 12 b [2(0) + - 3(0)] + (–15) = a 6 b - 15 = 2 a b [2 + 0 - 3(0)] + 15 = a b + 15 12 3 I
MAB = 2E MBA
E
EI
uB
I
EI
uB
uB
uB
(1) (2)
For member BC , app applyin lying g Eq. 11–1 11–10 0 MN = 3Ek(uN - c) + (FEM)N MBC = 3E
a 15 b ( I
uB - 0) + ( - 56.25) =
a 5 b EI
uB - 56.25
(3)
Equilibrium. At joint B, MBA + MBC = 0
a 3 b EI
uB + 15 +
a 5 b
uB =
EI
uB - 56.25 = 0
77.34375 EI
Substitute this result into Eqs. Eqs. (1) to (3) MAB = - 2.109 k # ft = - 2.11 k # ft
Ans.
MBA = 40.78 k # ft = 40.8 k # ft
Ans.
MBC = - 40.78 k # ft = - 40.8 k # ft
Ans.
The negative signs indicate that M AB and MBC have counterclockwise rotational sense.. Using these sense these results results,, the shear shear at both both ends ends of member member AB and BC are computed and shown in Fig. a and b respec respectively tively.. Subseq Subsequently uently,, the shear and Moment diagram can be plotted, Fig. c and d respectively.
4 26
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11–17. 11–1 7.
Continue Cont inued d
427
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Determine the moment that each member exerts on the joint at B, then draw the moment moment diagram for each member of the frame. Assume the supports at A, C , an and dD are pins. EI is constant. 11–18.
B
D
C
6m
6m
8m 12 kN/ m
Fixed End Moments. Referring to the table on the inside back cover,
(FEM)BA =
A
2
2
12(8 ) wL = = 96 kN # m 8 8
(FEM)BC = (FEM)BD = 0
Slope-Reflection Slope-Reflec tion Equation. Equation. Since the far end of each each members are pinned, pinned, Eq. 11–10 can be applied MN = 3Ek(uN - c) + (FEM)N
For member AB, MBA = 3E
a 8 b( I
uB - 0) + 96 =
a 3 8 b EI
uB + 96
(1)
For member BC , MBC = 3E
a 6 b( I
uB - 0) + 0 =
a 2 b EI
(2)
uB
For member BD, MBD = 3E
a 6 b( I
uB - 0) + 0 =
EI u 2 B
(3)
Equilibrium. At joint B, MBA + MBC + MBD = 0
a 38 b EI
uB + 96 +
a 2 b EI
uB = -
uB +
EI uB = 0 2
768 11EI
Substitute this result into Eqs. Eqs. (1) to (3) MBA = 69.82 kN # m = 69.8 kN # m
Ans.
MBC = - 34.91 kN # m = - 34.9 kN # m
Ans.
MBD = - 34.91 kN # m = - 34.9 kN # m
Ans.
The negative signs indicate that MBC and MBD have counterclockwise rotational sense. Using these results, results, the shear at both ends of members AB, BC , an and d BD are computed and shown in Fig. a , b and c respec respectively tively.. Subse Subsequently quently,, the shear and and moment diagrams can be plotted, Fig. d and e respectively.
4 28
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11–18. 11– 18.
Continued Cont inued
429
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Determine the moment at joints D and C , then draw the moment diagram for each member of the frame. Assume the supports at A and B are pins. EI is constant. 11–19.
3 k/ ft ft C
D
12 ft
B
A
5 ft
Fixed End Moments. Referring to the table on the inside back cover,
(FEM)DC = -
3(102) wL2 = = - 25 k # ft 12 12
3(102) wL2 = = 25 k # ft 12 12
(FEM)CD =
(FEM)DA = (FEM)CB = 0 Slope-Deflection Equations. For member CD, app applyin lying g Eq. 11– 11–8 8 MN = 2Ek(2uN + uF - 3c) + (FEM)N
a 10 b [2 = 2 a b [2 10
MDC = 2E MCD
E
I
a 2 5 b + a 5 b 2 b + a 5 b - 3(0)] + 25 = a 5 EI
uD + uC - 3(0)] + (–25) =
I
EI
uC + uD
EI
uD
EI
uC
uC - 25
uD + 25
(1) (2)
For members AD and BC , app applyin lying g Eq. 11–1 11–10 0 MN = 3Ek(uN - c) + (FEM)N MDA = 3E MCB = 3E
a 13 b ( a 13 b ( I
a 313 b 3 b - 0) + 0 = a 13 EI
uD - 0) + 0 =
I
uC
EI
uD
(3)
uC
(4)
Equilibrium. At joint D, MDC + MDA = 0
a 2 5 b + a 5 b - 25 + a 313 b a 4165 b + a 5 b = 25 EI
EI
uD
EI
EI
uC
EI
uD
uD = 0
(5)
uC
At joint C , MCD + MCB = 0
a 2 5 b + a 5 b + 25 + a 313 b a 4165 b + a 5 b = –25 EI
EI
uC
EI
EI
uD
uC
EI
uC = 0
(6)
uD
Solving Eqs. (5) and (6) uD =
1625 28EI
uC = -
1625 28EI
Substitute these results into Eq. (1) to (4) MDC = - 13.39 k # ft = - 13.4 k # ft
Ans.
MCD = 13.39 k # ft = 13.4 k # ft
Ans.
MDA = 13.39 k # ft = 13.4 k # ft
Ans.
MCB = - 13.39 k # ft = - 13.4 k # ft
Ans.
4 30
10 ft
5 ft
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11–19. 11– 19.
Continued Cont inued
The negative signs indicate that MDC and MCB have counterclockwise rotational sense. Using these results, results, the shear at both ends of of members AD, CD, an and d BC are computed and shown in Fig. a, b , an and d c respec respectively tively.. Subseq Subsequently uently,, the shear and moment diagrams can be plotted, Fig. d and e respectively.
431
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Determine the moment that each member exerts on the joints at B and D, then draw draw the moment moment diagram for each member of the frame. Assume the supports at A, C , an and d E are pins. EI is constant. *11–20.
12 kN/ m
10 kN D
4m
E
16 kN/ m
15 kN B
Fixed End Moments. Referring to the table on the inside back cover,
(FEM)BA = (FEM)BD = (FEM)DB = 0
4m
2
2
(FEM)BC = -
16(3 ) wL = = - 18 kN # m 8 8
(FEM)DE = -
12(32) wL2 = = - 13.5 kN # m 8 8
A
Slope-Deflection Equations. For member AB, BC , an and d ED, app applyin lying g Eq. 11–1 11–10. 0. MN = 3Ek(uN - c) + (FEM)N
a 4 b ( - 0) + 0 = a 3 4 b = 3 a b ( - 0) + (- 18) = - 18 3 = 3 a b ( - 0) + (- 13.5) = - 13.5 3
MBA = 3E MBC MDE
I
E
I
E
I
EI
uB
uB
(1)
uB
EIuB
(2)
EIuD
(3)
uD
For member BD, applyin applying g Eq. 11–8 MN = 2Ek(2uN + uF - 3c) + (FEM)N MBD = 2E MDB = 2E
a 4 b [2 a 4 b [2 I
I
a 2 b + a b 2
uB + uD - 3(0)] + 0 = EIuB + uD + uB - 3(0)] + 0 = EIuD
EI
EI
uD
(4)
uB
(5)
Equilibrium. At Joint B, MBA + MBC + MBD = 0
a34 b EI
uB + EIuB - 18 + EIuB +
a 114 b + a 2 b EI
EI
uB
a 2 b EI
uD = 0
uD = 18
(6)
At joint D, MDB + MDE = 0 EIuD +
a 2 b EI
2EIuD +
uB + EIuD - 13.5 = 0
a 2 b EI
uB = 13.5
(7)
Solving Eqs. Eqs. (6) and (7) uB =
39 7EI
uD =
75 14EI
4 32
C
3m
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11–20. 11– 20.
Continued Cont inued
Substitute these results into Eqs. Eqs. (1) to (5), MBA = 4.179 kN # m = 4.18 kN # m
Ans.
MBC = - 12.43 kN # m = - 12.4 kN # m
Ans.
MDE = - 8.143 kN # m = - 8.14 kN # m
Ans.
MBD = 8.25 kN # m
Ans.
MDB = 8.143 kN # m = 8.14 kN # m
Ans.
The negative signs indicate that MBC and MDE have counterclockwise rotational sense. Using these results, results, the shear at both ends of members AB, BC , BD and DE are computed and shown on Fig. a, b, c and d respe respectively ctively.. Subseq Subsequently uently,, the shear and moment diagram can be be plotted, Fig. e and f .
433
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Determine the moment at joints C and D, then draw the moment diagram for each member of the frame. Assume the supports at A and B are pins. EI is constant. 11–21.
D
C
8 kN/ m
6m
Fixed End Moments. Referring to the table on the inside back cover,
(FEM)DA =
8(62) wL2 = = 36 kN # m 8 8 B
A
(FEM)DC = (FEM)CD = (FEM)CB = 0 5m
Slope-Deflection Equations. Here, cDA = cCB = c and cDC = cCD = 0
For member CD, app applyin lying g Eq. 11– 11–8, 8, MN = 2Ek (2uN + uF - 3c) + (FEM)N MDC = 2E MCD = 2E
a 5 b [2 a 5 b [2 I
a 4 5 b + a 2 5 b 4 2 b b - 3(0)] + 0 = a + a 5 5 EI
uD + uC - 3(0)] + 0 =
I
EI
uC + uD
EI
uD
EI
uC
uC
(1)
uD
(2)
For member AD and BC , app applyin lying g Eq. 11–1 11–10 0 MN = 3Ek (uN - c) + (FEM)N
a 6 b( = 3 a b( 6
MDA = 3E MCB
I
uD
I
E
a 2 b - a 2 b + 36 ) + 0 = a b - a b 2 2 EI
c) + 36 =
EI
uC - c
EI
uD
c
EI
uC
c
(3) (4)
Equilibrium. At joint D, MDA + MDC = 0
a 2 b - a 2 b EI
EI
uD
c + 36 +
a 4 5 b + a 2 5 b EI
uD
EI
uC = 0
1.3EIuD + 0.4EIuC - 0.5EIc = –36
(5)
At joint C , MCD + MCB = 0
a 4 5 b + a 2 5 b + a 2 b - a 2 b EI
EI
uC
uD
EI
uC
EI
c = 0
0.4EIuD + 1.3EIuC - 0.5EIc = 0
(6)
Consider the horizontal force equilibrium for the entire frame
+
:
a Fx = 0; 8(6) -
VA - VB = 0
Referring to the FBD of member AD and BC in Fig. a, a + a MD = 0;
8(6)(3) - MDA - VA(6) = 0 VA = 24 -
MDA
6
and a + a MC = 0;
- MCB - VB(6) = 0 VB = -
MCB
6
= 0
4 34
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11–21. 11– 21.
Continued Cont inued
Thus, 8(6) -
a 24 -
MDA
6
b - a -
MCB
6
b=0
MDA + MCB = - 144
a 2 b - a 2 b EI
uD
EI
c + 36 +
a 2 b - a 2 b EI
uC
EI
c = - 144
0.5EIuD + 0.5EIuC - EIc = - 180
(7)
Solving Solvin g of Eqs. Eqs. (5), (6) and (7) uC =
80 EI
uD =
40 EI
c =
240 EI
Substitute these results into Eqs. Eqs. (1) to (4), MDC = 64.0 kN # m
Ans.
MCD = 80.0 kN # m
Ans.
MDA = - 64.0 kN # m
Ans.
MCB = - 80.0 kN # m
Ans.
The negative signs indicate that MDA and MCB have counterclockwise rotational sense. Using these results, results, the shear at both ends of of members AD, CD, an and d BC are computed and shown in Fig. b, c, an and d d, respec respectively tively.. Subseq Subsequently uently,, the shear and moment diagram can be plotted, Fig. e and f respectively.
435
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Determine the moment at joints A, B , C , an and d D, then draw the moment diagram for each member of the frame.. Assum frame Assume e the supports supports at A and B are fixed. EI is constant. 11–22.
D
C
3m
Fixed End Moments. Referring to the table on the inside back cover, 2
2
30(3 ) wL = = = 13.5 kN # m 20 20
(FEM)AD
A
3m
30(32) wL2 = = 9 kN # m 30 30
(FEM)DA =
(FEM)DC = (FEM)CD = (FEM)CB = (FEM)BC = 0 Slope-Deflection Equations. Here, cAD = cDA = cBC = cCB = c and cCD = cDC = 0
Applying Applyin g Eq. 11–8, MN = 2Ek(2uN + uF - 3c) + (FEM)N
For member AD,
a 3 b [2(0) + - 3 ] + ( - 13.5) = a 2 3 b 4 b -2 = 2 a b (2 + 0 - 3 ) + 9 = a 3 3 I
MAD = 2E MDA
uD
I
E
EI
c
uD
EI
c
uD
uD - 2EIc - 13.5
(1)
EIc + 9
(2)
For member CD, MDC = 2E
a 3 b [2 a 3 b [2 I
MCD = 2E
a 4 3 b + a 2 3 b 4 2 b b - 3(0)] + 0 = a + a 3 3 EI
uD + uC - 3(0)] + 0 =
I
uC + uD
EI
uD
EI
EI
uC
uC
(3)
uD
(4)
For member BC,
a 3 b [2(0) + - 3 ] + 0 = a 2 3 b 4 b = 2 a b [2 + 0 - 3 ] + 0 = a 3 3
MBC = 2E MCB
E
I
uC
I
EI
c
uC
EI
c
uC - 2EIc
uC - 2EIc
(5) (6)
Equilibrium. At Joint D, MDA + MDC = 0
a43 b EI
a 4 3 b + a 2 3 b a 8 3 b + a 2 3 b - 2 = – 9
uD - 2EIc + 9 +
EI
EI
uD
EI
uD
EI
uC = 0
(7)
EIc
uC
At joint C , MCD + MCB = 0
a 4 3 b + a 2 3 b + a 4 3 b EI
uC
EI
uD
EI
B
30 kN/ m
uC - 2EIc = 0
4 36
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11–22. 11– 22.
Continued Cont inued
a 2 3 b + a 8 3 b EI
EI
uD
uC - 2EIc = 0
(8)
Consider the horizontal force equilibrium for the entire frame,
+
a Fx = 0;
:
1 (30)(3) - VA - VB = 0 2
Referring to the FBD of members AD and BC in Fig. a a + a MD = 0;
1 (30)(3)(2) - MDA - MAD - VA(3) = 0 2 VA = 30 -
MDA
3
MAD
-
3
and a + a MC = 0;
- MCB - MBC - VB(3) = 0 VB = -
MCB
-
3
MBC
3
Thus, 1 (30)(3) 2
a 30 -
MDA
3
-
MAD
3
b - a -
MCB
3
-
MBC
3
b=0
MDA + MAD + MCB + MBC = - 45
a43 b EI
a 2 3 b - 2 - 13.5 + a 4 3 b 2 b - 2 = - 45 + a 3
uD - 2EIc + 9 +
EI
EI
uC
EI
EIc
uD
uC - 2EIc
EIc
2EIuD + 2EIuC - 8EIc = - 40.5
(9)
Solving Solvin g of Eqs. Eqs. (7), (8) and (9) uC =
261 56EI
uD =
9 56EI
c =
351 56EI
Substitute these results into Eq. (1) to (6), MAD = - 25.93 kN # m = - 25.9 kN # m
Ans.
MDA = - 3.321 kN # m = - 3.32 kN # m
Ans.
MDC = 3.321 kN # m = 3.32 kN # m
Ans.
MCD = 6.321 kN # m = 6.32 kN # m
Ans.
MBC = - 9.429 kN # m = - 9.43 kN # m
Ans.
MCB = - 6.321 kN # m = - 6.32 kN # m
Ans.
The negative signs indicate that M AD, M DA, M BC and MCB have counterclockwise rotational sense.Using these results, results, the shear at both ends of members AD, CD and BC are computed and shown on Fig. b, c and d, respectively respectively.. Subsequently Subsequently,, the shear and moment diagram can be plotted, Fig. e and d respectively.
437
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11–22. 11–2 2.
Continue Cont inued d
4 38
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Determine the moments acting at the supports A and D of the battered-column frame. Take E = 29(103) ksi , I = 600 in4. 11–23.
4 k/ k/ ft ft B
C
6k
20 ft
A
D
15 ft
(FEM)BC = -
wL2 = - 1600 k # in. 12
(FEM)CB =
wL2 = 1600 k # in. 12
uA = uD = 0
¢ 25
cAB = cCD = cBC = -
1.2 ¢ 20
cBC = - 1.5cCD = - 1.5cAB c = - 1.5c
a b (2 + - 3 ) + (FEM) 600 b (0 + - 3 ) + 0 = 116,000 = 2 a 25(12) 600 b (2 + 0 - 3 ) + 0 = 232,000 = 2 a 25(12) 600 b (2 + - 3( - 1.5 )) - 1600 = 2 a 20(12)
MN = 2E MAB MBA MBC
(where c = cBC, c = cAB = cCD)
I L
uN
uF
E
c
uB
E
uB
E
uB
N
uB - 348,000c
c
uB - 348,000c
c
uC
c
= 290,000uB + 145,000uC + 652,500c - 1600 MCB = 2E
600 a 20(12) b (2
uC + uB - 3( - 1.5c)) + 1600
= 290,000uC + 145,000uB + 652,500c - 1600 MCD = 2E
600 a 20(12) b (2
uC + 0 - 3c) + 0
= 232,000uC - 348,000c MDC = 2E
600 a 25(12) b (0 +
uC - 3c) + 0
= 116,000uC - 348,000c Moment equilibrium at B and C : MBA + MBC = 0
522,000uB + 145,000uC + 304,500c = 1600
(1)
MCB + MCD = 0
145,000uB + 522,000uC + 304,500c = - 1600
(2)
439
20 ft
15 ft
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11–23. 11–2 3.
Continue Cont inued d
using the FBD of the frame, c + a M0 = 0; MAB + MDC -
a
b (41.667)(12) + –a b (41.667)(12) - 6(13.333)(12) = 0 25(12)
MBA + MAB
25(12) MDC
MCD
- 0.667MAB - 0.667MDC - 1.667MBA - 1.667MCD - 960 = 0 464,000uB + 464,000uC - 1,624,000c = - 960 Solving Eqs Eqs.. (1), (2) and (3), (3), uB = 0.004030 rad uC = - 0.004458 rad c = 0.0004687 in.
MAB = 25.4 k # ft
Ans.
MBA = 64.3 k # ft MBC = - 64.3 k # ft MCB = 99.8 k # ft MCD = - 99.8 k # ft MDC = - 56.7 k # ft
Ans.
4 40
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Wind loads are transmitted to the frame at joint E. If A, B , E , D , and F are all pin connected and C is fixed connected, conne cted, determ determine ine the moments at joint C and draw the bending moment diagrams for the girder BCE. EI is constant. *11–24.
E
12 kN B
C
8m
cBC = cCE = 0 cAB = cCD = cCF = c
A
D
F
Applying Applyi ng Eq. 11–10, MCB =
3EI (uC - 0) + 0 6
MCE =
3EI (uC - 0) + 0 4
MCD =
3EI (uC - c) + 0 8
6m
(1)
Moment equilibrium at C: MCB + MCE + MCD = 0
3EI 3EI 3EI (uC) + (uC) + (uC - c) = 0 6 4 8 c = 4.333uC
(2)
From FBDs of members AB and EF : c + a MB = 0;
VA = 0
c + a ME = 0;
VF = 0
Since AB and FE are two-force members, members, then for the entire frame:
+
:
a FE = 0;
VD - 12 = 0;
VD = 12 kN
From FBD of member CD: c + a MC = 0;
MCD - 12(8) = 0 MCD = 96 kN # m
Ans.
From Fro m Eq. (1), 96 = uC =
3 EI(uC - 4.333uC) 8
- 76.8 EI
From Fro m Eq. (2), c =
- 332.8 EI
Thus, MCB = - 38.4 kN # in
Ans.
MCE = - 57.6 kN # m
Ans.
441
4m