Vdocuments Structural Analysis Chapter 12 Edition 8th

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Vdocuments STRUCTURAL ANALYSIS CHAPTER 12

EDITION 8TH
Fundamentals of Materials Engineering (University of St. La Salle)
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2–1. The steel framework is used to support the C

reinforced stone concrete slab that is used for an office. The
slab is 200 mm thick. Sketch the loading that acts along
B D
members BE and FED. Take a = 2 m, b = 5 m. Hint: See
Tables 1–2 and 1–4. A
a
E
b a
F
b 5m
Beam BE. Since = = 2.5, the concrete slab will behave as a one way slab.
a 2m
Thus, the tributary area for this beam is rectangular shown in Fig. a and the intensity
of the uniform distributed load is
200 mm thick reinforced stone concrete slab:

(23.6 kN>m3)(0.2 m)(2 m) = 9.44 kN>m
480 kN>m
Live load for office: (2.40 kN>m2)(2 m) = Ans.
14.24 kN>m
Due to symmetry the vertical reaction at B and E are
By = Ey = (14.24 kN>m)(5)>2 = 35.6 kN
The loading diagram for beam BE is shown in Fig. b.
Beam FED. The only load this beam supports is the vertical reaction of beam BE
at E which is Ey = 35.6 kN. The loading diagram for this beam is shown in Fig. c.
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2–2. Solve Prob. 2–1 with a = 3 m, b = 4 m. C
B D
a
E
b a
F
b 4
Beam BE. Since = 6 2, the concrete slab will behave as a two way slab. Thus,
a 3
the tributary area for this beam is the hexagonal area shown in Fig. a and the
maximum intensity of the distributed load is
200 mm thick reinforced stone concrete slab: (23.6 kN>m3)(0.2 m)(3 m)

= 14.16 kN>m
720 kN>m
Live load for office: (2.40 kN>m2)(3 m) = Ans.
21.36 kN>m
Due to symmetry, the vertical reactions at B and E are
2c (21.36 kN>m)(1.5 m) d + (21.36 kN>m)(1 m)

1
2
By = Ey =
2
= 26.70 kN
The loading diagram for Beam BE is shown in Fig. b.
Beam FED. The loadings that are supported by this beam are the vertical reaction
of beam BE at E which is Ey = 26.70 kN and the triangular distributed load of which
its tributary area is the triangular area shown in Fig. a. Its maximum intensity is
200 mm thick reinforced stone concrete slab: (23.6 kN>m3)(0.2 m)(1.5 m)

= 7.08 kN>m
3.60 kN>m
Live load for office: (2.40 kN>m2)(1.5 m) = Ans.
10.68 kN>m
The loading diagram for Beam FED is shown in Fig. c.
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2–3. The floor system used in a school classroom consists

of a 4-in. reinforced stone concrete slab. Sketch the loading E
b
that acts along the joist BF and side girder ABCDE. Set a
a = 10 ft, b = 30 ft. Hint: See Tables 1–2 and 1–4. a
D
a
C
a
B
F
A
b 30 ft
Joist BF. Since = = 3, the concrete slab will behave as a one way slab.
a 10 ft
Thus, the tributary area for this joist is the rectangular area shown in Fig. a and the
intensity of the uniform distributed load is
4 in thick reinforced stone concrete slab: (0.15 k>ft3) a

4
ftb (10 ft) = 0.5 k>ft
12
0.4 k>ft
Live load for classroom: (0.04 k>ft2)(10 ft) = Ans.
0.9 k>ft
Due to symmetry, the vertical reactions at B and F are
By = Fy = (0.9 k>ft)(30 ft)>2 = 13.5 k Ans.
The loading diagram for joist BF is shown in Fig. b.
Girder ABCDE. The loads that act on this girder are the vertical reactions of the
joists at B, C, and D, which are By = Cy = Dy = 13.5 k. The loading diagram for
this girder is shown in Fig. c.
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*2–4. Solve Prob. 2–3 with a = 10 ft, b = 15 ft.

E
b
a
a
D
a
C
a
B
F
A
b 15 ft
Joist BF. Since = = 1.5 < 2, the concrete slab will behave as a two way
a 10 ft
slab. Thus, the tributary area for the joist is the hexagonal area as shown
in Fig. a and the maximum intensity of the distributed load is

4
ftb (10 ft) = 0.5 k>ft
12
0.4 k>ft
0.9 k>ft
Due to symmetry, the vertical reactions at B and G are
2 c (0.9 k>ft)(5 ft) d + (0.9 k>ft)(5 ft)

1
2
By = Fy = = 4.50 k Ans.
2
The loading diagram for beam BF is shown in Fig. b.
Girder ABCDE. The loadings that are supported by this girder are the vertical
reactions of the joist at B, C and D which are By = Cy = Dy = 4.50 k and the
triangular distributed load shown in Fig. a. Its maximum intensity is
4 in thick reinforced stone concrete slab:
(0.15 k>ft3) a ftb (5 ft) = 0.25 k>ft

4
12
0.20 k冫ft
0.45 k冫ft
The loading diagram for the girder ABCDE is shown in Fig. c.
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2–5. Solve Prob. 2–3 with a = 7.5 ft, b = 20 ft.

E
b
a
a
D
a
C
a
B
F
A
b 20 ft
Beam BF. Since = = 2.7 7 2, the concrete slab will behave as a one way
a 7.5 ft
slab. Thus, the tributary area for this beam is a rectangle shown in Fig. a and the
intensity of the distributed load is

4
ftb (7.5 ft) = 0.375 k>ft
12
0.300 k>ft
Live load from classroom: (0.04 k>ft2)(7.5 ft) = Ans.
0.675 k>ft
Due to symmetry, the vertical reactions at B and F are
(0.675 k>ft)(20 ft)
By = Fy = = 6.75 k Ans.
2
The loading diagram for beam BF is shown in Fig. b.
Beam ABCD. The loading diagram for this beam is shown in Fig. c.
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2–6. The frame is used to support a 2-in.-thick plywood D

floor of a residential dwelling. Sketch the loading that acts
along members BG and ABCD. Set a = 5 ft, b = 15 ft. Hint: C
See Tables 1–2 and 1–4. E
B
F
a
A
G
a
b H
a
b 15 ft
Beam BG. Since = = 3, the plywood platform will behave as a one way
a 5 ft
slab. Thus, the tributary area for this beam is rectangular as shown in Fig. a and the
intensity of the uniform distributed load is
2 in thick plywood platform: a36 b a

lb 2
ftb (5ft) = 30 lb>ft
ft2 12
Line load for residential dweller: a 40 b (5 ft) =

lb 200 lb>ft
Ans.
ft 2 230 lb>ft

(230 lb>ft)(15 ft)
By = Gy = = 1725 Ans.
2
The loading diagram for beam BG is shown in Fig. a.
Beam ABCD. The loads that act on this beam are the vertical reactions of beams
BG and CF at B and C which are By = Cy = 1725 lb. The loading diagram is shown
in Fig. c.
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2–7. Solve Prob. 2–6, with a = 8 ft, b = 8 ft. D
C
E
B
F
a
A
G
a
b H
a
b 8 ft
Beam BG. Since = = 1 < 2, the plywood platform will behave as a two
a 8 ft
way slab. Thus, the tributary area for this beam is the shaded square area shown in
Fig. a and the maximum intensity of the distributed load is
2 in thick plywood platform: (36 lb>ft3) a in b(8 ft) = 48 lb>ft

2
12
320 lb>ft
Live load for residential dwelling: (40 lb>ft)(8 ft) = Ans.
368 lb>ft
1
(368 lb>ft) (8 ft)
2
By = Gy = = 736 lb Ans.
2
The loading diagram for the beam BG is shown in Fig. b
Beam ABCD. The loadings that are supported by this beam are the vertical
reactions of beams BG and CF at B and C which are By = Cy = 736 lb and the
distributed load which is the triangular area shown in Fig. a. Its maximum intensity is
b(4 ft) = 24 lb>ft

2
2 in thick plywood platform: (36 lb>ft3)a
12 ft
160 lb>ft
Live load for residential dwelling: (40 lb>ft2)(4 lb>ft) = Ans.
184 lb>ft
The loading diagram for beam ABCD is shown in Fig. c.
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*2–8. Solve Prob. 2–6, with a = 9 ft, b = 15 ft. D
C
E
B
F
a
A
G
a
b H
a
b 15 ft
Beam BG. Since = = 1.67 < 2, the plywood platform will behave as a
a 9 ft
two way slab. Thus, the tributary area for this beam is the octagonal area shown in
Fig. a and the maximum intensity of the distributed load is
2 in thick plywood platform: (36 lb>ft3) a in b(9 ft) = 54 lb>ft

2
12
360 lb>ft
Live load for residential dwelling: (40 lb>ft2)(9 ft) = Ans.
414 lb>ft
2 c (414 lb>ft)(4.5 ft)d + (414 lb>ft)(6 ft)

1
2
By = Gy = = 2173.5 lb
2
The loading diagram for beam BG is shown in Fig. b.
Beam ABCD. The loading that is supported by this beam are the vertical
reactions of beams BG and CF at B and C which is By = Cy = 2173.5 lb and the
triangular distributed load shown in Fig. a. Its maximum intensity is
2 in thick plywood platform: (36 lb>ft3) a

2
ftb(4.5 ft) = 27 lb>ft
12
180 lb>ft
Live load for residential dwelling: (40 lb>ft2)(4.5 ft) = Ans.
207 lb>ft
The loading diagram for beam ABCD is shown in Fig. c.
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2–9. The steel framework is used to support the 4-in. C

reinforced stone concrete slab that carries a uniform live
loading of 500 lb>ft2. Sketch the loading that acts along
B D
members BE and FED. Set b = 10 ft, a = 7.5 ft. Hint: See
Table 1–2. A
a
E
b a
F
b 10
Beam BE. Since = < 2, the concrete slab will behave as a two way slab.
a 7.5
Thus, the tributary area for this beam is the octagonal area shown in Fig. a and the
maximum intensity of the distributed load is
4
4 in thick reinforced stone concrete slab: (0.15 k>ft3)a ftb (7.5 ft) = 0.375 k>ft
12
3.75 k>ft
Floor Live Load: (0.5 k>ft2)(7.5 ft) = Ans.
4.125 k>ft
2 c (4.125 k>ft)(3.75 ft)d + (4.125 k>ft)(2.5 ft)

1
2
By = Ey = = 12.89 k
2
The loading diagram for this beam is shown in Fig. b.
Beam FED. The loadings that are supported by this beam are the vertical reaction
of beam BE at E which is Ey = 12.89 k and the triangular distributed load shown in
Fig. a. Its maximum intensity is
4
4 in thick reinforced stone concrete slab: (0.15 k>ft3)a ftb (3.75 ft) = 0.1875 k>ft
12
1.875 k>ft
Floor live load: (0.5 k>ft2)(3.75 ft) = Ans.
2.06 k>ft
The loading diagram for this beam is shown in Fig. c.
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2–10. Solve Prob. 2–9, with b = 12 ft, a = 4 ft. C
B D
a
E
b a
F
b 12
Beam BE. Since = = 3 > 2, the concrete slab will behave as a one way
a 4
slab. Thus, the tributary area for this beam is the rectangular area shown in Fig. a and
the intensity of the distributed load is
4
4 in thick reinforced stone concrete slab: (0.15 k>ft2)a ftb(4 ft) = 0.20 k>ft
12
2.00 k>ft
Floor Live load: (0.5 k>ft2)(4 ft) = Ans.
2.20 k>ft
(2.20 k>ft)(12 ft)
By = Ey = = 13.2 k
2
The loading diagram of this beam is shown in Fig. b.
Beam FED. The only load this beam supports is the vertical reaction of beam
BE at E which is Ey = 13.2 k. Ans.
The loading diagram is shown in Fig. c.
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2–11. Classify each of the structures as statically

determinate, statically indeterminate, or unstable. If
indeterminate, specify the degree of indeterminacy. The
supports or connections are to be assumed as stated.
(a)
(b)
(c)
(d)
(e)
(a) r = 5 3n = 3(1) 6 5
Indeterminate to 2°. Ans.
(b) Parallel reactions

Unstable. Ans.
(c) r = 3 3n = 3(1) 6 3
Statically determinate. Ans.
(d) r = 6 3n = 3(2) 6 6
(e) Concurrent reactions

Unstable. Ans.
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*2–12. Classify each of the frames as statically determinate

or indeterminate. If indeterminate, specify the degree of
indeterminacy. All internal joints are fixed connected.
(a)
(b)
(c)
(d)
(a) Statically indeterminate to 5°. Ans.
(b) Statically indeterminate to 22°. Ans.
(c) Statically indeterminate to 12°. Ans.
(d) Statically indeterminate to 9°. Ans.
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determinate, statically indeterminate, stable, or unstable.
If indeterminate, specify the degree of indeterminacy.
The supports or connections are to be assumed as stated. pin roller
fixed
(a)
fixed pin pin fixed
(b)
pin pin
(a) r = 6 3n = 3(2) = 6
(c)
(b) r = 10 3n = 3(3) 6 10
Statically indeterminate to 1°. Ans.
(c) r = 4 3n = 3(1) 6 4
Statically determinate to 1°. Ans.
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2–14. Classify each of the structures as statically pin fixed

determinate, statically indeterminate, stable, or unstable. If
indeterminate, specify the degree of indeterminacy. The
supports or connections are to be assumed as stated.
(a) r = 5 3n = 3(2) = 6 rocker pin
r 6 3n
Unstable.
(a)
(b) r = 9 3n = 3(3) = 9
r = 3n
Stable and statically determinate.
roller pin roller pin fixed
(c) r = 8 3n = 3(2) = 6
r - 3n = 8 - 6 = 2 (b)
Stable and statically indeterminate to the
second degree.
fixed pin
fixed fixed
(c)
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indeterminate, specify the degree of indeterminacy.
(a) r = 5 3n = 3(2) = 6
r 6 3n
Unstable.
(b) r = 10 3n = 3(3) = 9 and r - 3n = 10 - 9 = 1 (a)
Stable and statically indeterminate to first degree.
(c) Since the rocker on the horizontal member can not resist a horizontal
force component, the structure is unstable.
(b)
(c)
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*2–16. Classify each of the structures as statically

(a) r = 6 3n = 3(1) = 3
r - 3n = 6 - 3 = 3
Stable and statically indeterminate to the third degree.

(a)
(b) r = 4 3n = 3(1) = 3
r - 3n = 4 - 3 = 1
Stable and statically indeterminate to the first degree.
(c) r = 3 3n = 3(1) = 3 r = 3n
(b)
(d) r = 6 3n = 3(2) = 6 r = 3n
(c)
(d)
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determinate, statically indeterminate, stable, or unstable. If
(a)
(a) r = 2 3n = 3(1) = 3 r 6 3n
Unstable.
(b) r = 12 3n = 3(2) = 6 r 7 3n
r - 3n = 12 - 6 = 6
Stable and statically indeterminate to the sixth degree.

(b)
(c) r = 6 3n = 3(2) = 6
r = 3n
(d) Unstable since the lines of action of the reactive force components are (c)
concurrent.
(d)
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20 kN 20 kN
2–18. Determine the reactions on the beam. Neglect the 26 kN
thickness of the beam. 13 12
5
By1152 - 20162 - 201122 - 26 a b1152 = 0

12 B
a + a MA = 0; A
13 6m 6m 3m
By = 48.0 kN Ans.
1262 = 0
12
+ c a Fy = 0; Ay + 48.0 - 20 - 20 -
13
Ay = 16.0 kN Ans.
Ax - a b26 = 0
+ 5
: a Fx = 0;
13
Ax = 10.0 kN Ans.
2–19. Determine the reactions on the beam. 3 k/ft

2 k/ft 2 k/ft
a + a MA = 0; -601122 - 600 + FB cos 60° (242 = 0

B
A
FB = 110.00 k = 110 k Ans. 60⬚
600 k · ft
+
: a Fx = 0; Ax = 110.00 sin 60º = 0
12 ft 12 ft
Ax = 95.3 k Ans.
+ c a Fy = 0; Ay = 110.00 cos 60º - 60 = 0

Ay = 5.00 k Ans.
*2–20. Determine the reactions on the beam. 2 k/ft
5 k/ft
10 ft
FB(26) – 52(13) – 39a b(26) = 0

1 24 ft
a + a MA = 0;
3
FB = 39.0 k Ans.
(39) – a b52 + a b (39.0) = 0

12 12 12
+ c a Fy = 0; Ay –
13 13 13
Ay = 48.0 k Ans.
-Ax + a b39 + a b 52 – a b39.0 = 0

+ 5 5 5
: a Fx = 0;
13 13 13
Ax = 20.0 k Ans.
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2–21. Determine the reactions at the supports A and B of 18 kN

the compound beam. Assume there is a pin at C. 4 kN/ m
A C B
6m 2m 2m
Equations of Equilibrium: First consider the FBD of segment AC in Fig. a. NA and

Cy can be determined directly by writing the moment equations of equilibrium
about C and A respectively.
a + a MC = 0; 4(6)(3) - NA(6) = 0 NA = 12 kN Ans.
a + a MA = 0; Cy(6) - 4(6)(3) = 0 Cy = 12 kN Ans.
Then,
+
: a Fx = 0 ; 0 - Cx = 0 Cx = 0
Using the FBD of segment CB, Fig. b,

+
: a Fx = 0 ; 0 + Bx = 0 Bx = 0 Ans.
+ c a Fy = 0; By - 12 - 18 = 0 By = 30 kN Ans.
a + a MB = 0; 12(4) + 18(2) - MB = 0 MB = 84 kN # m Ans.
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2–22. Determine the reactions at the supports A, B, D, 8k

2 k/ ft
and F.
C E
A F
B D
8 ft 4 ft 4 ft 4 ft 4 ft
2 ft
Equations of Equilibrium: First consider the FBD of segment EF in Fig. a. NF and

Ey can be determined directly by writing the moment equations of equilibrium
about E and F respectively.
a + a ME = 0; NF - (8) - 8(4) = 0 NF = 4.00 k Ans.
a + a MF = 0; 8(4) - Ey (8) = 0 Ey = 4.00 k
Then
+
: a Fx = 0; Ex = 0
Consider the FBD of segment CDE, Fig. b,

+
: a Fx = 0; Cx - 0 = 0 Cx = 0
a + a MC = 0; NP (4) - 4.00 (6) = 0 ND = 6.00 k Ans.
a + a MD = 0; Cy(4) - 4.00 (2) = 0 Cy = 2.00 k
Now consider the FBD of segment ABC, Fig. c.
a + a MA = 0; NB (8) + 2.00(12) - 2(12)(6) = 0 NB = 15.0 k Ans.
a + a MB = 0; 2(12)(2) + 2.00(4) - Ay (8) = 0 Ay = 7.00 k Ans.
+
: a Fx = 0; Ax - 0 = 0 Ax = 0 Ans.
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2–23. The compound beam is pin supported at C and 8k 12 k

supported by a roller at A and B. There is a hinge (pin) at
5 4
D. Determine the reactions at the supports. Neglect the 15 k · ft 3
thickness of the beam. A D B C
30⬚
6 ft 8 ft 8 ft 8 ft
4k 4 ft 2 ft
Equations of Equilibrium: Consider the FBD of segment AD, Fig. a.

+
: a Fx = 0; Dx - 4 sin 30° = 0 Dx = 2.00 k
a + a MD = 0; 8(2) + 4 cos 30°(12) - NA (6) = 0 NA = 9.59 k Ans.
a + a MA = 0; Dy (6) + 4 cos 30°(6) - 8(4) = 0 Dy = 1.869 k
Now consider the FBD of segment DBC shown in Fig. b,
Cx - 2.00 - 12 a b = 0
+ 3
: a Fx = 0; Cx = 9.20 k Ans.
5
1.869(24) + 15 + 12 a b (8) - NB (16) = 0

4
a + a MC = 0;
5
NB = 8.54 k Ans.
1.869(8) + 15 - 12 a b(8) - Cy (16) = 0

4
a + a MB = 0;
5
Cy = 2.93 k Ans.
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*2–24. Determine the reactions on the beam. The support 2 k/ft

at B can be assumed to be a roller.
A
B
12 ft 12 ft
Equations of Equilibrium:
1
a + a MA = 0; NB(24) – 2(12)(6) – (2)(12)(16) = 0 NB = 14.0 k Ans.
2
1
a + a MB = 0; (2)(12)(8) + 2(12)(18) – A y (24) = 0 A y = 22.0 k Ans.
2
+
: a Fx = 0; Ax = 0 Ans.
2–25. Determine the reactions at the smooth support C

and pinned support A. Assume the connection at B is fixed C
connected.
80 lb/ft
30⬚ 10 ft
A
B
6 ft
a + a MA = 0; Cy (10 + 6 sin 60°) - 480(3) = 0
Cy = 94.76 lb = 94.8 lb Ans.

+
: a Fx = 0; Ax – 94.76 sin 30° = 0
Ay = 47.4 lb Ans.
+ c a Fy = 0; Ay + 94.76 cos 30° - 480 = 0
Ay = 398 lb Ans.
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2–26. Determine the reactions at the truss supports

600 lb/ft
A and B. The distributed loading is caused by wind. 400 lb/ft
20 ft
A B
48 ft 48 ft
a + a MA = 0; By(96) + a b 20.8(72) - a b 20.8(10)

12 5
13 13
-a b31.2(24) - a b31.2(10) = 0
12 5
13 13
By = 5.117 kN = 5.12 kN Ans.
Ay - 5.117 + a b20.8 - a b31.2 = 0

12 12
+ c a Fy = 0;
13 13
Ay = 14.7 kN Ans.
-Bx + a b31.2 + a b20.8 = 0

+ 5 5
: a Fx = 0;
13 13
Bx = 20.0 kN Ans.
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2–27. The compound beam is fixed at A and supported by 15 kN

a rocker at B and C. There are hinges pins at D and E.
Determine the reactions at the supports.
A D B E
C
6m 6m
2m 2m 2m
Equations of Equilibrium: From FBD(a),
a + a ME = 0; Cy(6) = 0 Cy = 0 Ans.
+ c a Fy = 0; Ey - 0 = 0 Ey = 0
+
: a Fx = 0 ; Ex = 0
From FBD (b),
a + a MD = 0; By(4) - 15(2) = 0
By = 7.50 kN Ans.
+ c a Fy = 0; Dy + 7.50 - 15 = 0
Dy = 7.50 kN
+
: a Fx = 0; Dx = 0
From FBD (c),

a + a MA = 0; MA - 7.50(6) = 0
MA = 45.0 kN . m Ans.
+ c a Fy = 0; Ay - 7.50 = 0 Ay = 7.50 kN Ans.
+
: a Fx = 0; Ax = 0 Ans.
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*2–28. Determine the reactions at the supports A and B. 10 k

3 k/ft
The floor decks CD, DE, EF, and FG transmit their loads
3 ft 1 ft
to the girder on smooth supports. Assume A is a roller and D E F G
B is a pin. C
B
A
4 ft 4 ft 4 ft 4 ft
Consider the entire system.
a + a MB = 0; 10(1) + 12(10) - Ay (8) = 0
Ay = 16.25 k = 16.3 k Ans.

+
: a Fx = 0; Bx = 0 Ans.
+ c a Fy = 0; 16.25 - 12 - 10 + By = 0
By = 5.75 k Ans.
2–29. Determine the reactions at the supports A and B of 4 kN/m

the compound beam. There is a pin at C.
A
C B
Member AC:
a + a MC = 0; -Ay (6) + 12(2) = 0 6m 4.5 m
Ay = 4.00 kN Ans.
+ c a Fy = 0; Cy + 4.00 - 12 = 0
Cy = 8.00 kN
+
: a Fx = 0; Cx = 0
Member CB:
a + a MB = 0; -MB + 8.00(4.5) + 9(3) = 0
MB = 63.0 kN . m Ans.
+ c a Fy = 0; By - 8 - 9 = 0
By = 17.0 kN Ans.
+
: a Fx = 0; Bx = 0 Ans.
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2–30. Determine the reactions at the supports A and B of 2 kN/m

the compound beam. There is a pin at C.
A
C B
6m 4m
Member AC:
a + a MC = 0; -Ay (6) + 6(2) = 0; Ay = 2.00 kN Ans.
+
: a Fx = 0; Cx = 0
+ c a Fy = 0; 2.00 – 6 + Cy = 0; Cy = 4.00 kN
Member BC:
+ c a Fy = 0; -4.00 – 8 + By = 0; By = 12.0 kN Ans.
+
: a Fx = 0; 0 - Bx = 0; Bx = 0 Ans.
a + a MB = 0; -MB + 8(2) + 4.00 (4) = 0; MB = 32.0 kN . m Ans.
36
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2–31. The beam is subjected to the two concentrated loads P 2P

as shown. Assuming that the foundation exerts a linearly L
__ L
__ L
__
varying load distribution on its bottom, determine the load 3 3 3
intensities w1 and w2 for equilibrium (a) in terms of the
parameters shown; (b) set P = 500 lb, L = 12 ft.
w1
w2
Equations of Equilibrium: The load intensity w1 can be determined directly by

summing moments about point A.
b - w1La b = 0
L L
a + a MA = 0; Pa
3 6
2P
w1 = Ans.
L
a w2 - bL +
1 2P 2P
+ c a Fy = 0; (L) - 3P = 0
2 L L
w2 = a b
4P
Ans.
L
If P = 500 lb and L = 12 ft,
2(500)
w1 = = 83.3 lb>ft Ans.
12
4(500)
w2 = = 167 lb>ft Ans.
12
37
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*2–32 The cantilever footing is used to support a wall near 20 000 lb

its edge A so that it causes a uniform soil pressure under the
footing. Determine the uniform distribution loads, wA and 8000 lb
0.25 ft
wB, measured in lb>ft at pads A and B, necessary to support
the wall forces of 8000 lb and 20 000 lb. 1.5 ft
A B
wA wB
2 ft 8 ft 3 ft
a + a MA = 0; -8000(10.5) + wB (3)(10.5) + 20 000(0.75) = 0
wB = 2190.5 lb>ft = 2.19 k>ft Ans.
+ c a Fy = 0; 2190.5(3) - 28 000 + wA (2) = 0
wA = 10.7 k>ft Ans.
2–33. Determine the horizontal and vertical components B

of reaction acting at the supports A and C.
2m
4m
30 kN
2m
50 kN C
4m
A
3m 3m
1.5 m 1.5 m
Equations of Equilibrium: Referring to the FBDs of segments AB and BC
respectively shown in Fig. a,
a + a MA = 0; Bx (8) + By (6) - 50(4) = 0 (1)
a + a MC = 0; By (3) - Bx (4) + 30(2) = 0 (2)
38
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2–33. Continued
Solving,
By = 6.667 kN Bx = 20.0 kN
Segment AB,
+
: a Fx = 0; 50 - 20.0 - Ax = 0 Ax = 30.0 kN Ans.
+ c a Fy = 0; 6.667 - Ay = 0 Ay = 6.67 kN Ans.
Segment BC,
+
: a Fx = 0; Cx + 20.0 - 30 = 0 Cx - 10.0 kN Ans.
+ c a Fy = 0; Cy – 6.667 = 0 Cy = 6.67 kN Ans.
2–34. Determine the reactions at the smooth support A 150 lb/ ft

and the pin support B. The joint at C is fixed connected.
C
B
10 ft
5 ft
Equations of Equilibrium: Referring to the FBD in Fig. a.
A
a + a MB = 0; NA cos 60°(10) - NA sin 60°(5) - 150(10)(5) = 0
60⬚
NA = 11196.15 lb = 11.2 k Ans.
+
: a Fx = 0; Bx – 11196.15 sin 60° = 0
Bx = 9696.15 lb = 9.70 k Ans.
+ c a Fy = 0; 11196.15 cos 60° – 150(10) – By = 0 Ans.
By = 4098.08 lb = 4.10 k
39
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2–35. Determine the reactions at the supports A and B.

700 lb/ft
20 ft
500 lb/ ft B
30 ft
48 ft 48 ft
700 lb>ft at 52 ft = 36,400 lb or 36.4 k
500 lb>ft at 30 ft = 15,000 lb or 15.0 k
b (36.4) - 40 a b (36.4) - 15(15) = 0

48 20
a + a MA = 0; 96(By) – 24a
52 52
By = 16.58 k = 16.6 k Ans.
+ 20
: a Fx = 0; 15 + (36.4) – Ax = 0; Ax = 29.0 k Ans.
52
48
+ c a Fy = 0; Ay + By – (36.4) = 0; Ay = 17.0 k Ans.
52
*2–36. Determine the horizontal and vertical components 40 kN

30 kN
of reaction at the supports A and B. Assume the joints at C 20 kN
and D are fixed connections.
12 kN/m
C D
4m
7m
A
6m 8m
a + a MB = 0; 20(14) + 30(8) + 84(3.5) – Ay(8) = 0
Ay = 101.75 kN = 102 kN Ans.

+
: a Fx = 0; Bx – 84 = 0
Bx = 84.0 kN Ans.
+ c a Fy = 0; 101.75 - 20 - 30 - 40 - By = 0
By = 11.8 kN Ans.
40
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2–37. Determine the horizontal and vertical components 200N/

200 N/mm
force at pins A and C of the two-member frame.
BB
AA
3 3mm
Free Body Diagram: The solution for this problem will be simplified if one realizes
that member BC is a two force member. CC
Equations of Equilibrium:
3 3mm
a + a MA = 0; FBC cos 45° (3) – 600 (1.5) = 0
FBC = 424.26 N
+ c a Fy = 0; Ay + 424.26 cos 45° – 600 = 0
Ay = 300 N Ans.
+
: a Fx = 0; 424.26 sin 45° – Ax = 0 Ans.
Ax = 300 N
For pin C,
Cx = FBC sin 45° = 424.26 sin 45° = 300 N Ans.
Cy = FBC cos 45° = 424.26 cos 45° = 300 N Ans.
41
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2–38. The wall crane supports a load of 700 lb. Determine D

the horizontal and vertical components of reaction at the
pins A and D. Also, what is the force in the cable at the
winch W?
4 ft
4 ft 4 ft
C
A B
E
Pulley E: 60⬚
+ c ©Fy = 0; 2T – 700 = 0
W
T = 350 lb Ans.
700 lb
Member ABC:
a + a MA = 0; TBD sin 45° (4) – 350 sin 60°(4) – 700(8) = 0
TBD = 2409 lb
+ c a Fy = 0; Ay + 2409 sin 45° – 350 sin 60° - 700 = 0
Ay = 700 lb Ans.
+
: a Fx = 0; Ax - 2409 cos 45° - 350 cos 60° + 350 - 350 = 0
Ax = 1.88 k Ans.
At D:
Dx = 2409 cos 45° = 1703.1 lb = 1.70 k Ans.
Dy = 2409 sin 45° = 1.70 k Ans.
42
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2–39. Determine the resultant forces at pins B and C on 5 ft 2 ft

member ABC of the four-member frame.
150 lb/ft
A B C
4
a + a MF = 0; FCD(7) – F (2) = 0
5 BE
4
a + a MA = 0; -150(7)(3.5) + F (5) – FCD(7) = 0 4 ft
5 BE
FBE = 1531 lb = 1.53 k Ans.

F E D
FCD = 350 lb Ans.
2 ft 5 ft
*2–40. Determine the reactions at the supports is A and w

D. Assume A is fixed and B and C and D are pins.
w
B C
Member BC:
b = 0
1.5L
a + a MB = 0; Cy (1.5L) - (1.5wL)a
2 A D
Cy = 0.75 wL
1.5L
+ c a Fy = 0; By - 1.5wL + 0.75 wL = 0
By = 0.75 wL
Member CD:
a + a MD = 0; Cx = 0
+
: a Fx = 0; Dx = 0 Ans.
+ c a Fy = 0; Dy - 0.75wL = 0
Dy = 0.75 wL Ans.
43
lOMoARcPSD|24343972
*2–40. Continued
Member BC:
+
: a Fx = 0; Bx - 0 = 0; Bx = 0
Member AB:
+
: a Fx = 0; wL - Ax = 0 Ans.
Ax = wL Ans.
+ c a Fy = 0; Ay – 0.75 wL = 0
Ay = 0.75 wL Ans.
MA – wL a b = 0
L
a + a MA = 0;
2
wL2
MA = Ans.
2
2–41. Determine the horizontal and vertical reactions at 400 lb 400 lb

600 lb 600 lb
the connections A and C of the gable frame. Assume that A,
B, and C are pin connections. The purlin loads such as D 800 lb E F
800 lb
and E are applied perpendicular to the center line of each D G
B 5 ft
girder.
10 ft
120 lb/ft A C
6 ft 6 ft 6 ft 6 ft
Member AB:
Bx (15) + By(12) – (1200)(5) – 600 a b (16) – 600 a b (12.5)

12 5
a + a MA = 0;
13 13
- 400 a b (12) – 400a b (15) = 0

12 5
13 13
Bx(15) + By(12) = 18,946.154 (1)
Member BC:
b(6) + 600 a b(12.5)

12 5
a + a MC = 0; - (Bx)(15) + By(12) + (600)a
13 13
+ 400 a b (12) + 400 a b(15) = 0

12 5
13 13
Bx(15) - By(12) = 12.946.15 (2)
44
lOMoARcPSD|24343972
2–41. Continued
Solving Eqs. (1) and (2),

Bx = 1063.08 lb, By = 250.0 lb
Member AB:
-Ax + 1200 + 1000 a b - 1063.08 = 0

+ 5
: a Fx = 0;
13
Ax = 522 lb Ans.
Ay - 800 - 1000 a b + 250 = 0

12
+ c a Fy = 0;
13
Ay = 1473 lb = 1.47 k Ans.
Member BC:
b + 1063.08 = 0
+ 5
: a Fx = 0; -Cx - 1000a
13
Cx = 678 lb Ans.
Cy - 800 - 1000 a b + 250.0 = 0

12
+ c a Fy = 0;
13
Cy = 1973 lb = 1.97 k Ans.
2–42. Determine the horizontal and vertical components 50 kN 40 kN

of reaction at A, C, and D. Assume the frame is pin 1.5 m 2m 1.5 m
connected at A, C, and D, and there is a fixed-connected 15 kN/m
joint at B.
B C
4m
6m
Member CD:
A
a + a MD = 0; -Cx(6) + 90(3) = 0
Cx = 45.0 kN Ans.
+ D
: a Fx = 0; Dx + 45 - 90 = 0
Dx = 45.0 kN Ans.
+ c a Fy = 0 ; Dy - Cy = 0 (1)
Member ABC:
a + a MA = 0; Cy(5) + 45.0(4) - 50(1.5) - 40(3.5) = 0
Cy = 7.00 kN Ans.
45
lOMoARcPSD|24343972
2–42. Continued
+ c a Fy = 0; Ay + 7.00 – 50 – 40 = 0
Ay = 83.0 kN Ans.
+
: a Fx = 0; Ax – 45.0 = 0
Ax = 45.0 kN Ans.
From Eq. (1).
Dy = 7.00 kN Ans.
2–43. Determine the horizontal and vertical components 3 k/ft

at A, B, and C. Assume the frame is pin connected at these B
points. The joints at D and E are fixed connected. 6 ft
D E
1.5 k/ ft 10 ft
A C
18 ft 18 ft
a + a MA = 0; -18 ft (By ) + 16 ft (Bx) = 0 (1)
a + a MC = 0; 15 k (5ft) + 9 ft (56.92 k (cos 18.43°)) + 13 ft (56.92 k (sin 18.43° ))
-16 ft (Bx) - 18 ft (Bx) = 0 (2)
Solving Eq. 1 & 2
Bx = 24.84 k Ans.
By = 22.08 k Ans.
+
: a Fx = 0; Ax - 24.84 k = 0
Ax = 24.84 k
+ c a Fy = 0; Ay - 22.08 k = 0
Ay = 22.08 k
+
: a Fx = 0; Cx - 15 k - sin (18.43°) (56.92 k) + 24.84 k
Cx = 8.16 k Ans.
+ c a Fy = 0; Cy + 22.08 k - cos (18.43°)(56.92 k) = 0
Cy = 31.9 k Ans.
46
lOMoARcPSD|24343972
*2–44. Determine the reactions at the supports A and B. 10 kN/m

The joints C and D are fixed connected.
D
C
2m
4m B
5 3
4
4 3
a + a MA = 0; FB(4.5) + FB(2) - 30(1.5) = 0
5 5 A
FB = 9.375 kN = 9.38 kN Ans. 3m 1.5 m
4
+ c a Fy = 0; Ay + (9.375) - 30 = 0
5
Ay = 22.5 kN Ans.
+ 3
: a Fx = 0; Ax - (9.375) = 0
5
Ax = 5.63 kN Ans.
47

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Vdocuments STRUCTURAL ANALYSIS CHAPTER 12

Studocu is not sponsored or endorsed by any college or university

2–1. The steel framework is used to support the C

200 mm thick reinforced stone concrete slab:

Due to symmetry the vertical reaction at B and E are

By = Ey = (14.24 kN>m)(5)>2 = 35.6 kN

The loading diagram for beam BE is shown in Fig. b.

2–2. Solve Prob. 2–1 with a = 3 m, b = 4 m. C

200 mm thick reinforced stone concrete slab: (23.6 kN>m3)(0.2 m)(3 m)

Due to symmetry, the vertical reactions at B and E are

2c (21.36 kN>m)(1.5 m) d + (21.36 kN>m)(1 m)

The loading diagram for Beam BE is shown in Fig. b.

200 mm thick reinforced stone concrete slab: (23.6 kN>m3)(0.2 m)(1.5 m)

The loading diagram for Beam FED is shown in Fig. c.

2–3. The floor system used in a school classroom consists

4 in thick reinforced stone concrete slab: (0.15 k>ft3) a

By = Fy = (0.9 k>ft)(30 ft)>2 = 13.5 k Ans.

The loading diagram for joist BF is shown in Fig. b.

*2–4. Solve Prob. 2–3 with a = 10 ft, b = 15 ft.

4 in thick reinforced stone concrete slab: (0.15 k>ft3) a

2 c (0.9 k>ft)(5 ft) d + (0.9 k>ft)(5 ft)

The loading diagram for beam BF is shown in Fig. b.

4 in thick reinforced stone concrete slab:

(0.15 k>ft3) a ftb (5 ft) = 0.25 k>ft

2–5. Solve Prob. 2–3 with a = 7.5 ft, b = 20 ft.

4 in thick reinforced stone concrete slab: (0.15 k>ft3) a

2–6. The frame is used to support a 2-in.-thick plywood D

2 in thick plywood platform: a36 b a

Line load for residential dweller: a 40 b (5 ft) =

Due to symmetry, the vertical reactions at B and G are

2–7. Solve Prob. 2–6, with a = 8 ft, b = 8 ft. D

2 in thick plywood platform: (36 lb>ft3) a in b(8 ft) = 48 lb>ft

b(4 ft) = 24 lb>ft

*2–8. Solve Prob. 2–6, with a = 9 ft, b = 15 ft. D

2 in thick plywood platform: (36 lb>ft3) a in b(9 ft) = 54 lb>ft

Due to symmetry, the vertical reactions at B and G are

2 c (414 lb>ft)(4.5 ft)d + (414 lb>ft)(6 ft)

2 in thick plywood platform: (36 lb>ft3) a

2–9. The steel framework is used to support the 4-in. C

2 c (4.125 k>ft)(3.75 ft)d + (4.125 k>ft)(2.5 ft)

2–10. Solve Prob. 2–9, with b = 12 ft, a = 4 ft. C

The loading diagram of this beam is shown in Fig. b.

The loading diagram is shown in Fig. c.

2–11. Classify each of the structures as statically

(b) Parallel reactions

(e) Concurrent reactions

*2–12. Classify each of the frames as statically determinate

(a) Statically indeterminate to 5°. Ans.

(b) Statically indeterminate to 22°. Ans.

(c) Statically indeterminate to 12°. Ans.

(d) Statically indeterminate to 9°. Ans.

2–13. Classify each of the structures as statically

fixed pin pin fixed

2–14. Classify each of the structures as statically pin fixed

(a) r = 5 3n = 3(2) = 6 rocker pin

2–15. Classify each of the structures as statically

(b) r = 10 3n = 3(3) = 9 and r - 3n = 10 - 9 = 1 (a)

Stable and statically indeterminate to first degree.

*2–16. Classify each of the structures as statically