Solution€¦ · 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 E y

26

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2–1. The steel framework is used to support the reinforced stone concrete slab that is used for an office. The slab is 200 mm thick. Sketch the loading that acts along members BE and FED. Take a = 2 m, b = 5 m. Hint: See Tables 1.2 and 1.4.

Ans.Live load for office: 14.24 kN>m

A

B

C

D

E

Fb a

a

Solution

Beam BE. Since ba=

5 m2 m

= 2.5, the concrete slab will behave as a one-way slab.

Thus, the tributary area for this beam is rectangular, as shown in Fig. a, and the inten-sity 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

Live load for office: (2.40 kN>m2)(2 m) = 480 kN>m

14.24 kN>m Ans.

Due to symmetry, the vertical reactions 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.

27


A

B

C

D

E

Fb a

a

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

Ans.Live load for office: 21.36 kN>m

Live load for office:

3.60 kN>m

10.68 kN>m

Solution

Beam BE. Since ba=

43

6 2, the concrete slab will behave as a two-way slab.

Thus, 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

Live load for office: [(2.40 kN>m2)(3 m)] = 7.20 kN>m 21.36 kN>m Ans.


By = Ey =2 c 1

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

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

Live load for office: (2.40 kN>m2)(1.5 m) =3.60 kN>m

10.68 kN>m Ans.

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

28


2–3. The floor system used in a school classroom consists of a 4-in. reinforced stone concrete slab. Sketch the loading that acts along the joist BF and side girder ABCDE. Set a = 10 ft, b = 30 ft. Hint: See Tables 1.2 and 1.4.

A

E b

a

a

a

a

B

C

D

F

SolutionJoist BF. Since

ba=

30 ft10 ft

= 3, the concrete slab will behave as a one-way slab.

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 412

ftb(10 ft) = 0.5 k>ft

Live load for classroom: (0.04 k>ft2)(10 ft) = 0.4 k>ft

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

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, and 6.75-k end loads from the joists at A and E. The loading diagram for this girder is shown in Fig. c.

By = 13.5 k

Ans.Live load for classroom: 0.9 k>ft

Ans.

Ans.

29


Solution

Joist BF. Since ba=

15 ft10 ft

= 1.5 6 2, the concrete slab will behave as a two-way

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@ in.@thick reinforced stone concrete slab: (0.15 k>ft3)a 412

ftb(10 ft2 = 0.5 k>ft

Live load for classroom: (0.04 k>ft2)(10 ft) = 0.4 k>ft

0.9 k>ft Ans.

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

By = Fy =2 c 1

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

2= 4.50 k Ans.

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

Girder ABCDE. The loadings that are supported by this girder are the vertical reac-tions of the joist at B, C and D, which are By = Cy = Dy = 4.50 k, the 2.25-k end loads from the joists at A and E, and the triangular distributed load shown in Fig. a. Its maxi-mum intensity is

4-in.-thick reinforced stone concrete slab:

(0.15 k>ft3)a 412

ftb(5 ft) = 0.25 k>ft

Live load for classroom: (0.04 k>ft2)(5 ft) =0.20 k>ft

0.45 k>ft Ans.

The loading diagram for the girder ABCDE is shown in Fig. c.

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

A

E b

a

a

a

a

B

C

D

F

Ans.Live load for classroom: 0.9 k>ft

By = 4.50 k

Live load for classroom: 0.20 k>ft

0.45 k>ft

30


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

Solution

Beam BF. Since ba=

20 ft7.5 ft

= 2.7 7 2, the concrete slab will behave as a one-way

slab. Thus, the tributary area for this beam is a rectangle, as shown in Fig. a, and the intensity of the distributed load is:


ftb(7.5 ft) = 0.375 k>ft

Live load from classroom: (0.04k>ft2)(7.5 ft) = 0.300 k>ft

0.675 k>ft

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

By = Fy =(0.675 k>ft)(20 ft)

2= 6.75 k

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

Beam ABCD. The loading diagram for this beam is shown in Fig. c.

A

E b

a

a

a

a

B

C

D

F

Ans.Live load from classroom: 0.675 k>ft

By = 6.75 k

Ans.

Ans.

31


Ans.Reaction at B: 750 lbcReaction at A: 1.5 kc

2–6. The frame is used to support the wood deck in a residential dwelling where the live load is 40 lb>ft2. Sketch the loading that acts along members BG and ABCD. Set b = 10 ft, a = 5 ft.

SolutionFrom Table 1–4,LL = 40 psf

L2

L1=

ba=

105

= 2 … 2

b

a

aa

A

B

CD

HG

EF

Two-way slab:Reaction at B: 750 lbcReaction at A: 1.5 kc

Ans.Ans.

32


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

Ans.Reaction at B: 640 lbcReaction at A: 1920 lbc

b

a

aa

A

B

CD

HG

EF

SolutionFrom Table 1–4,LL = 40 psf

L2

L1=

ba=

88= 1 … 2

Two-way slab:

Reaction at B: 640 lbc Ans.

Reaction at A: 1920 lbc Ans.

33


Solution

*2–8. Solve Prob. 2–6 if b = 15 ft, a = 10 ft.

b

a

aa

A

B

CD

HG

EF

Ans.Reaction at B: 2 kcReaction at A: 4.5 kc

Reaction at B: 2 kcReaction at A: 4.5 kc

From Table 1.4,

LL = 40 psfb>a = 15>10 = 1.5 … 2

Two-way slab:

Ans.Ans.

34


Solution

Beam BE. Since ba=

12 ft9 ft

=43

6 2, the concrete slab will behave as a two-way

slab. Thus, the tributary area for this beam is the shaded octagonal area shown in Fig. a, and the maximum intensity of the trapezoidal distributed load is:


ftb(9 ft) = 0.45 k>ft

Floor live load: (0.4 k>ft2)(9 ft) =3.60 k>ft

4.05 k>ft


By = Ey =

12

(4.05 k>ft)(3 ft + 12 ft)

2= 15.2 k

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

Beam FED. The loadings that are supported by this beam are the vertical reactions of beam BE at E, which is Ey = 15.19 k and the triangular distributed load contributed by dotted triangular tributary area shown in Fig. a. Its maximum intensity is

4@in.@thick concrete slab: (0.15 k>ft3)a 412

ftb(4.5 ft) = 0.225 k>ft

Floor live load: (0.4 k>ft2)(4.5 ft) =1.800 k>ft

2.025 k>ft Ans.

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

2–9. The steel framework is used to support the 4-in. reinforced stone concrete slab that carries a uniform live loading of 400 lb>ft2. Sketch the loading that acts along members BE and FED. Set a = 9 ft, b = 12 ft. Hint: See Table 1.2.

A

B

C

D

E

F

ba

a

Ans.

Floor live load: 3.60 k>ft

4.05 k>ft

By = Ey = 15.2 k


2.025 k>ft

Beam BE. w max = 4.05 k>ft

Beam FED. 15.2 k at E, w max = 2.025 k>ft

Ans.

Ans.

35


2–10. Solve Prob. 2–9, with a = 6 ft, b = 18 ft.

Solution

Beam BE. Since ba=

18 ft6 ft

= 3 7 2, the concrete slab will behave as a one-way

slab. Thus, the tributary area for this beam is the shaded rectangular area shown in Fig. a, and the intensity of the uniform distributed load is:


ftb(6 ft) = 0.30 k>ft

Floor live load: (0.4 k>ft2)(6 ft) =2.40 k>ft

2.70 k>ft


By = Ey =(2.70 k>ft)(18 ft)

2= 24.3 k

The loading diagram of this 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 = 24.3 k.

The loading diagram of beam FED is shown in Fig. c.

A

B

C

D

E

F

ba

a

Ans.


2.70 k>ft

By = Ey = 24.3 k

Ans.

Ans.

36


(a)

(b)

(c)

(e)

(d)

2–11. Classify each of the structures as statically determinate, statically indeterminate, or unstable. If indeterminate, specify the degree of indeterminacy.

Solution(a) r = 2 3n = 3(1) = 3

r 6 3n

Unstable (b) r = 4 3n = 3(1) = 3

r - 3n = 4 - 1 = 1Stable and statically indeterminate to the first degree

(c) r = 9 3n = 3(3) = 9r = 3n

Stable and statically determinate (d) r = 8 3n = 3(2) = 6

r - 3n = 8 - 6 = 2Stable and statically indeterminate to the second degree

(e) r = 7 3n = 3(2) = 6r - 3n = 7 - 6 = 1Stable and statically indeterminate to the first degree

Ans.(a) Unstable(b) Stable and statically indeterminate to the first degree(c) Stable and statically determinate(d) Stable and statically indeterminate to the second degree(e) Stable and statically indeterminate to the first degree

Ans.

Ans.

Ans.

Ans.

Ans.

37


Solution(a) r 7 3n

4 7 3(1)

Statically indeterminate to the first degree.

(b) Parallel reactions

Unstable.

(c) r 7 3n

6 7 3(1)

Statically indeterminate to the third degree.

(d) Parallel reactions

Unstable.

(e)

(f)

(g)

(h)

*2–12. Classify each of the structures as statically determinate, statically indeterminate, or unstable. If indeterminate, specify the degree of indeterminacy.

(a) (b)

(c) (d)

Ans.(a) Statically indeterminate to the first degree(b) Unstable(c) Statically indeterminate to the first degree(d) Unstable

Ans.

Ans.

Ans.

Ans.

38


Ans.(a) Stable and statically indeterminate to first degree(b) Stable and statically determinate(c) Stable and statically indeterminate to first degree

2–13. 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. fixedpin pin

(a)

fixed

(b)

pin

(c)

pinroller roller

roller

Solution(a) r = 7 3n = 3(2) = 6 r - 3n = 7 - 6 = 1

Stable and statically indeterminate to first degree.

(b) r = 6 3n = 3(2) = 6 r = 3n

Stable and statically determinate

(c) r = 4 3n = 3(1) = 3 r - 3n = 4 - 3 = 1

Stable and statically indeterminate to first degree

Ans.

Ans.

Ans.

39


Ans.(a) Unstable(b) Stable and statically determinate(c) Stable and statically indeterminate to the second degree

Solution(a) r = 5 3n = 3(2) = 6

r 6 3n

Unstable

(b) r = 9 3n = 3(3) = 9

r = 3n


(c) r = 8 3n = 3(2) = 6

r - 3n = 8 - 6 = 2

Stable and statically indeterminate to the second degree

2–14. 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.

rocker

fixed

(a)

pin

pin

(b)

fixedroller roller pinpin

fixed

(c)

fixed fixed

pin

Ans.

Ans.

Ans.

40


Ans.(a) Unstable(b) Stable and statically determinate(c) Stable and statically determinate

(a)

(b)

(c)

Solution(a) Since the lines of action of the reactive forces are concur-

rent, the structure is unstable.

(b) r = 6 3n = 3(2) = 6

r = 3n


(c) r = 3 3n = 3(1) = 3

r = 3n



Ans.

Ans.

Ans.

41


(a)

(b)

(c)

(d)

Solution(a) r = 6 3n = 3(1) = 3

r - 3n = 6 - 3 = 3

Stable and statically indeterminate to the third degree

(b) r = 5 3n = 3(1) = 3

r - 3n = 5 - 3 = 2


(c) r = 5 3n = 3(1) = 3

r - 3n = 5 - 3 = 2


(d) r = 6 3n = 3(2) = 6

r = 3n

Stable and statically determinate.

Ans.(a) Stable and statically indeterminate to the third degree(b) Stable and statically indeterminate to the second degree(c) Stable and statically indeterminate to the second degree(d) Stable and statically determinate

*2–16. Classify each of the structures as statically determinate, statically indeterminate, or unstable. If indeterminate, specify the degree of indeterminacy.

Ans.

Ans.

Ans.

Ans.

42


Ans.(a) Unstable(b) Statically indeterminate to the sixth degree(c) Stable and statically determinate(d) Unstable since the lines of action of the reactive force

components are concurrent

Solution(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 = 6Statically indeterminate to the sixth degree.

(c) r = 6 3n = 3(2) = 6r = 3n

Stable and statically determinate.(d) Unstable since the lines of action of the reactive force

components are concurrent.

(a)

(b)

(c)

(d)


Ans.

Ans.

Ans.Ans.

43


Ans.NA = 18.0 kBy = 48.0 kBx = 0

SolutionEquations of Equilibrium. Referring to the FBD of the beam shown in Fig. a, NA and By can be determined directly by writ-ing the moment equations of equilibrium about points B and A, respectively.

a +ΣMB = 0; 12

(3)(12)(4) + 3(12)(6) + 12(12) - NA(24) = 0

NA = 18.0 k

a +ΣMA = 0; By(24) - 12(12) - 3(12)(18) -12

(3)(12)(20) = 0

By = 48.0 k

Write the force equation of equilibrium along the x-axis.+S ΣFx = 0, bx = 0

12 ft12 ft

6 k>ft

3 k>ft

12 k

A B

2–18. Determine the reactions on the beam.

Ans.

Ans.

Ans.

44


Ans.NB = 31.5 kAy = 9.00 kAx = 0

SolutionEquations of Equilibrium. Referring to the FBD of the beam shown in Fig. a, NB and Ay can be determined directly by writ-ing the moment equations of equilibrium about points A and B, respectively.

a +ΣMA = 0; NB(18) -12

(3)(18)(9) -12

(3)(9)(24) = 0

NB = 31.5 k

a +ΣMB = 0; 12

(3)(18)(9) -12

(3)(9)(6) - Ay(18) = 0

Ay = 9.00 k

Write the force equation of equilibrium along the x axis.+S ΣFx = 0; Ax = 0

9 ft9 ft

AB

3 k>ft 3 k>ft

9 ft

2–19. Determine the reactions at the supports.

Ans.

Ans.

Ans.

45


Ans.NA = 21.5 kBx = 10.2 kBy = 24.5 k

SolutionEquations of Equilibrium. Referring to the FBD of the beam shown in Fig. a, NA can be obtained directly by writing the moment equation of equilibrium about point B.

a +ΣMB = 0; 10(8.667) + B(6.50) + 25(4.333) - NA(13) = 0

NA = 21.5 k

Using this result to write the force equation of equilibrium along the x and y axis,

+S ΣFx = 0; Bx + 21.5a 513

b - (10 + 13 + 25)a 513

b = 0

Bx = 10.19 k = 10.2 k

bΣFy = 0; By + 21.5a1213

b - (10 + 13 + 25)a1213

b = 0

By = 24.46 k = 24.5 k

12 k10 k

1 k>ft

3 k>ft5 ft

4 ft 4 ft 4 ft

A

B

*2–20. Determine the reactions on the beam.

Ans.

Ans.

Ans.

46


Ans.Ay = 4.00 kN

MB = 63.0 kN # m

By = 17.0 kN

Bx = 0

AC B

4 kN>m

6 m 4.5 m

2–21. Determine the reactions at the supports A and B of the compound beam. There is a pin at C.

SolutionMember AC:

a +ΣMC = 0; -Ay(6) + 12(2) = 0

Ay = 4.00 kN

+ cΣFy = 0; Cy + 4.00 - 12 = 0

Cy = 8.00 kN+S ΣFx = 0; Cx = 0

Member CB:

a +ΣMB = 0; -MB + 8.00(4.5) + 9(3) = 0

MB = 63.0 kN # m

+ cΣFy = 0; By - 8 - 9 = 0

By = 17.0 kN

+S ΣFx = 0; Bx = 0

Ans.

Ans.

Ans.

Ans.

47


Ans.Ax = 0

Ay = 480 lb

By = 620 lb

A B

5 ft

200 lb>ft100 lb>ft

4 ft

Solution+S ΣFx = 0; Ax = 0

a +ΣMB = 0; 900(4.5) + 200(1.333) - Ay(9) = 0

Ay = 480 lb

+ cΣFy = 0; 480 - 1100 + By = 0

By = 620 lb

2–22. Determine the reactions at the supports.

Ans.

Ans.

Ans.

48


Ans.NA = 15.0 kNCx = 0Cy = 27.5 kNMC = 35.0 kN # m

SolutionEquations of Equilibrium. First consider the equilibrium of the FBD of segment AB in Fig. a. NA and By can be determined directly by writing the moment equations of equilibrium about points B and A, respectively.

a +ΣMB = 0; 12

(15)(3)(2) - NA(3) = 0 NA = 15.0 kN

a +ΣMA = 0; By(3)- 12

(15)(3)(1) = 0 By = 7.50 kN

Write the force equation of equilibrium along the x axis.+S ΣFx = 0; Bx = 0.

Then consider the equilibrium of the FBD of segment BC using the results of Bx and By.

+S ΣFx = 0; Cx = 0

+ cΣFy = 0; Cy - 20 - 7.50 = 0 Cy = 27.5 kN

a +ΣMC = 0; 7.50(2) + 20(1) - MC = 0 MC = 35.0 kN # m

15 kN>m

3 m 1 m 1 m

B CA

20 kN2–23. Determine the reactions at the supports A and C of the compound beam. Assume C is fixed, B is a pin, and A is a roller.

Ans.

Ans.

Ans.

49


SolutionEquations of Equilibrium:

a +ΣMA = 0; NB(24) - 2(12)(6) - 12

(2)(12)(16) = 0 NB = 14.0 k

a +ΣMB = 0; 12

(2)(12)(8) + 2(12)(18) - Ay(24) = 0 Ay = 22.0 k

+S ΣFx = 0; Ax = 0

12 ft 12 ft

BA

2 k>ft*2–24. Determine the reactions on the beam. The support at B can be assumed to be a roller.

Ans.NB = 14.0 kAy = 22.0 kAx = 0

50


SolutionEquations of Equilibrium. Referring to the FBD of the beam shown in Fig. a, FAB and Cy can be determined directly by writing the moment equations of equilibrium about points C and B, respectively,

a + ΣMC = 0; 12

(12)(6)(2) + 60 - (FAB sin 60°)(6) = 0

FAB = 25.40 kN

a + ΣMB = 0; Cy(6) + 60 -12

(12)(6)(4) = 0

Cy = 14.0 kN

Using the result of FAB to write the force equation of equilibrium along the x axis,+S Σ Fx = 0; 25.40 cos 60° - Cx = 0 Cx = 12.70 kN = 12.7 kN

Referring to the FBD of pin A, Fig. b, the force equations of equilibrium written along the x and y axis give+S ΣFx = 0; Ax - 25.40 cos 60° = 0 Ax = 12.70 kN = 12.7 kN

+ cΣFy = 0; Ay - 25.40 sin 60° = 0 Ay = 22.0 kN

A

B C

12 kN>m

60 kN ?m608

6 m

2–25. Determine the horizontal and vertical components of reaction at the pins A and C.

Ans.Cy = 14.0 kN

Cx = 12.7 kN

Ax = 12.7 kN

Ay = 22.0 kN

Ans.

Ans.

Ans.

Ans.

51


A B

48 ft

600 lb>ft 400 lb>ft

48 ft

20 ft

Solution

a + ΣMA = 0; By(96) + a1213

b20.8(72) - a 513

b20.8(10)

- a1213

b31.2(24) - a 513

b31.2(10) = 0

By = 5.117 k = 5.12 k

+ cΣFy = 0; Ay - 5.117 + a1213

b20.8 - a1213

b31.2 = 0

Ay = 14.7 k

+S ΣFx = 0; - Bx + a 513

b31.2 + a 513

b20.8 = 0

Bx = 20.0 k

2–26. Determine the reactions at the truss supports A and B. The distributed loading is caused by wind.

Ans.

By = 5.12 k

Ay = 14.7 k

Bx = 20.0 k

Ans.

Ans.

Ans.

52


2–27. The compound beam is fixed at E and supported by rockers at A and B. There are hinges (pins) at C and D. Determine the reactions at the supports. The 4-kN load is applied just to the right of the pin at D.

2 m 2 m 2 m

ADB EC

6 kN

6 kN>m4 kN

3 m 3 m

SolutionEquations of Equilibrium. Consider the equilibrium of the FBD of segment CD, Fig. b, first

a +ΣMC = 0; Dy(3) -12

(6)(3)(2) = 0 Dy = 6.00 kN

a +ΣMD = 0; 12

(6)(3)(1) - Cy(3) = 0 Cy = 3.00 kN

+S ΣFx = 0; Dx - Cx = 0 (1)

Then, using the result of Cy, the equilibrium of the FBD of segment ABC, Fig. a, gives

a +ΣMA = 0; NB(4) - 6(2) - 3.00(6) = 0 NB = 7.50 kN

a +ΣMB = 0; 6(2) - 3.00(2) - NA(4) = 0 NA = 150 kN+S ΣFx = 0; Cx = 0

Then from Eq (1), Dx = 0

Finally, using the results of Dx and Dy, the equilibrium of the FBD of segment DE gives+S ΣFx = 0; Ex = 0

+ c ΣFy = 0; Ey - 4 - 6.00 = 0 Ey = 10.0 kN

a +ΣME = 0; 4(3) + 6.00(3) - ME = 0 ME = 30.0 kN # m

Ans.NB = 7.50 kNNA = 150 kNEx = 0Ey = 10.0 kNME = 30.0 kN # m

Ans.

Ans.

Ans.

Ans.

53


Solutiona +ΣMA = 0; -20 kN(3 m) - 20 kN(6 m) - 20 kN(9 m) - 20 kN(12 m)

-8 kN(sin 60°)(15 m) + By(9 m) = 0

By = 78.2 kN+S ΣFx = 0; -Ax + 8 kN(cos 60°) = 0

Ax = 4 kN

+ cΣFy = 0; -20 kN - 20 kN - 20 kN - 20 kN - 8 kN(sin 60°) + 78.2 kN + Ay = 0

Ay = 8.71 kN

A

8 kN

608B

3 m3 m3 m3 m3 m

20 kN 20 kN 20 kN 20 kN*2–28. Determine the reactions on the beam.

Ans.By = 78.2 kNAx = 4 kNAy = 8.71 kN

Ans.

Ans.

Ans.

54


Ans.By = 15.0 kcAy = 5.00 kTEy = 10.0 kcFy = 5.00 kT

BA E FDC

200 ft 200 ft 150 ft 150 ft100 ft

15 k

100 ft

2–29. The construction features of a cantilever truss bridge are shown in the figure. Here it can be seen that the center truss CD is suspended by the cantilever arms ABC and DEF. C and D are pins. Determine the vertical reactions at the supports A, B, E, and F if a 15-k load is applied to the center truss.

Truss CD:a +ΣMD = 0; 15(200) - Cy(300) = 0 Cy = 10.0 k

+ cΣFy = 0; -15 + 10 + Dy = 0 Dy = 5.0 k

Truss ABC:a +ΣMA = 0; By(200) - 10(300) = 0 By = 15.0 kc Ans.

+ cΣFy = 0; 15 - 10 - Ay = 0 Ay = 5.0 kT Ans.

Truss DEF:

a +ΣMF = 0; 5(300) - Ey(150) = 0 Ey = 10.0 kc Ans.

+ cΣFy = 0; -5 + 10 - Fy = 0 Fy = 5.0 kT Ans.

Solution

55


SolutionEquations of Equilibrium. Consider the equilibrium of the FBD of segment BC, Fig. b, first.

a +ΣMB = 0; Nc(6) -12

(12)(6)(2) = 0 NC = 12.0 kN Ans.

a +ΣMC = 0; 12

(12)(6)(4) - By(6) = 0 By = 24.0 kN

+S ΣFx = 0; Bx = 0

Using the results of Bx and By, the equilibrium of the FBD of segment AB, Fig. a, gives

S+ ΣFx = 0; Ax = 0 Ans.

+ cΣFy = 0; Ay - 12(3) - 24.0 = 0 Ay = 60.0 kN Ans.

a +ΣMA = 0; MA - 12(3)(1.5) - 24.0(3) = 0

MA = 126 kN # m Ans.

12 kN>m

B CA

3 m 6 m

2–30. Determine the reactions at the supports A and C of the compound beam. Assume C is a roller, B is a pin, and A is fixed.

Ans.NC = 12.0 kNAx = 0Ay = 60.0 kNMA = 126 kN # m

56


P 2P

w2

w1

L__3

L__3

L__3

2–31. The beam is subjected to the two concentrated loads as shown. Assuming that the foundation exerts a linearly varying load distribution on its bottom, determine the load intensities w1 and w2 for equilibrium (a) in terms of the parameters shown; (b) set P = 500 lb, L = 12 ft.

SolutionEquations of Equilibrium: The load intensity w1 can be determined directly by summing moments about point A.

a +ΣMA = 0; PaL3b - w1LaL

6b = 0

w1 =2PL

Ans.

+ cΣFy = 0; 12

aw2 -2PL

bL +2PL

(L) - 3P = 0

w2 = a4PL

b Ans.

If P = 500 lb and L = 12 ft,

w1 =2(500)

12= 83.3 lb>ft Ans.

w2 =4(500)

12= 167 lb>ft Ans.

Ans.

w1 =2PL

w2 =4PL

w1 = 83.3 lb>ftw2 = 167 lb>ft

57


SolutionEquations of Equilibrium. Referring to the FBD of the beam shown in Fig. a, FBC can be obtained directly by writing the moment equation of equilibrium about point A.

a +ΣMA = 0; FBCa45b(0.3) + FBCa3

5b(4) - 3(4)(2) - 6(6) = 0

FBC = 22.73 kN

Using the result of FBC to write the force equation of equilibrium along the x and y axis,

+S ΣFx = 0; 22.73a45b - Ax = 0 Ax = 18.18 kN = 18.2 kN

+ cΣFy = 0; Ay + 22.73a35b - 3(4) - 6 = 0 Ay = 4.364 kN = 4.36 kN

Using the result of FBC, write the force equation of equilibrium along the x and y axis by referring to the FBD of pin C, Fig. b,

+S ΣFx = 0; Cx - 22.73a45b = 0 Cx = 18.18 kN = 18.2 kN

+ cΣFy = 0; Cy - 22.73a35b = 0 Cy = 13.64 kN = 13.6 kN

4 m 2 m

0.3 m

3 m

BA

C

3 kN>m6 kN*2–32. Determine the horizontal and vertical components of

reaction at the supports A and C. Assume the members are pin connected at A, B, and C.

Ans.Ax = 18.2 kNAy = 4.36 kNCx = 18.2 kNCy = 13.6 kN

Ans.

Ans.

Ans.

Ans.

58


SolutionEquations of Equilibrium. Member BC is a two-force member, which is reflected in the FBD diagram of member AB, Fig. a. FBC Ax can be determined directly by writing the moment equations of equilibrium about A and B, respectively.

a +ΣMA = 0; FBCa35b(4) - 24(2) = 0 FBC = 20.0 kN

a +ΣMB = 0; 24(2) - Ax(4) = 0 Ax = 12.0 kN Ans.

Write the force equation of equilibrium along the y axis using the result of FBC.

c +ΣFy = 0; 20.0a45b - Ay = 0 Ay = 16.0 kN Ans.

Then consider the FBD of pin at C, Fig. b.

+S ΣFx = 0; 20.0a3

5b - Cx = 0 Cx = 12.0 kN Ans.

+ cΣFy = 0; Cy - 20.0a45b = 0 Cy = 16.0 kN Ans.

A C

B

4 m

6 kN>m

3 m

2–33. Determine the horizontal and vertical components of reaction at the supports A and C.

Ans.Ax = 12.0 kNAy = 16.0 kNCx = 12.0 kNCy = 16.0 kN

59


C400 lb>ft

4 k 3 k4 ft 3 ft

C

A

B

12 ft

6 ft

SolutionMember AB:

a +ΣMA = 0; Bx(12) - 4.8(6) = 0

Bx = 2.40 k Ans.+d ΣFx = 0; Ax + 2.4 - 4.8 = 0

Ax = 2.40 k d Ans.

+ cΣFy = 0; Ay - By = 0 (1)

Member BC:

a +ΣMC = 0; -By(13) + 4(9) + 3(3) = 0

By = 3.46 k Ans.

+ cΣFy = 0; Cy + 3.46 - 4 - 3 = 0

Cy = 3.54 kc Ans.+d ΣFx = 0; Cx - 2.40 = 0

Cx = 2.40 k d Ans.

From Eq. (1),

Ay = 3.46 kc Ans.

2–34. Determine the horizontal and vertical components of force at the connections A, B, and C. Assume each of these connections is a pin.

Ans.Bx = 2.40 kAx = 2.40 k dBy = 3.46 kCy = 3.54 kcCx = 2.40 k dAy = 3.46 kc

60


Ans.Ay = 7.36 kcBy = 16.6 kcBx = 0.500 k S

A

B

5 k 7 k

0.5 k

10 k 2 k

8 ft

6 ft

8 ft 6 ft 6 ft2–35. Determine the reactions at the supports A and B of the frame. Assume that the support at A is a roller.

Solutionb +ΣMB = 0; -(0.5)(6) + (2)(6) - (7)(6) - (5)(14) + Ay(14) = 0

Ay = 7.36 kc Ans.

+ cΣFy = 0; 7.36 - 5 - 7 - 10 - 2 + By = 0

By = 16.6 kc Ans.

+S ΣFx = 0; -0.5 + Bx = 0

Bx = 0.500 k S Ans.

61


Ans.FBE = 1.53 k (C)FCD = 350 lb (T)

Solution

a + aMA = 0; -150(7)(3.5) +45

FBE(5) - FCD(7) = 0 (1)

a + aMF = 0; FCD(7) -45

FBE(2) = 0 (2)

Solving Eqs. (1) and (2) simultaneously,

FBE = 1531 lb = 1.53 k (C)

FCD = 350 lb (T)

2 ft

150 lb>ft

4 ft

5 ft

5 ft2 ft

A

F E D

B C

*2–36. Determine the resultant forces at pins B and C on member ABC of the four-member frame.

Ans.

Ans.

62


Ans.Ay = 4.00 kNCx = 17.3 kNCy = 4.00 kNAx = 0.667 kN

SolutionEquations of Equilibrium. First, consider the equilibrium of the FBD of segment AB, Fig. a.

a +ΣMA = 0; By(4) - 2(4)(2) = 0 By = 4.00 kN

a +ΣMB = 0; 2(4)(2) - Ay(4) = 0 Ay = 4.00 kN

+S ΣFx = 0; Ax - Bx = 0 (1)

Then, using the result of By, the equilibrium of the FBD of Segment BC, Fig. b, gives

a +ΣMC = 0; 12

(12)(3)(1) - 4.00(4) - Bx(3) = 0 Bx = 0.6667 kN

+S ΣFx = 0; Cx + 0.6667 -12

(12)(3) = 0 Cx = 17.33 kN = 17.3 kN

+ cΣFy = 0; Cy - 4.00 = 0 Cy = 4.00 kN

From Eq. (1),

Ax = 0.6667 kN = 0.667 kN

4 m

3 m

2 kN>m

12 kN>m

A B

C

2–37. Determine the horizontal and vertical reactions at A and C of the two-member frame.

Ans.

Ans.

Ans.

Ans.

63


Ans.T = 300 lbAy = 600 lbDx = 1275 lbDy = 1200 lbAx = 1275 lb

6 ft

D

AB

E

C

6 ft

1.5 ft

600 lb

6 ft

2–38. The frame supports a load of 600 lb. Determine the horizontal and vertical components of reaction at the pins A and D. Also, what is the force in the cable?

SolutionEquations of Equilibrium. First, consider the equilibrium of pulley E, Fig. a.

+ cΣFy = 0; 2T - 600 = 0 T = 300 lb

Then, consider the equilibrium of member ABC, Fig. b.

a +ΣMA = 0; 300(12) + 300(13.5) - 300(1.5) - By(6) = 0

By = 1200 lb

a +ΣMB = 0; 300(6) + 300(7.5) - 300(1.5) - Ay(6) = 0

Ay = 600 lb+S ΣFx = 0; 300 + Bx - Ax = 0 (1)

Finally, consider the equilibrium of member BD, Fig. c.

a +ΣND = 0; 1200(6) - 300(4.5) - Bx(6) = 0 Bx = 975 lb

+S ΣFx = 0; Dx - 300 - 975 = 0 Dx = 1275 lb

+ cΣFy = 0; Dy - 1200 = 0 Dy = 1200 lb

Substitute the result of Bx into Eq (1).

300 + 975 - Ax = 0 Ax = 1275 lb

Ans.

Ans.

Ans.

Ans.

Ans.

64


Ans.Cy = 6.00 kNCx = 0Dy = 6.00 kNDx = 2.00 kN

SolutionEquations of Equilibrium. The equilibrium of the FBD of member shown in Fig. a will be considered first.

a +ΣMC = 0; 3(3)(1.5) +12

(3)(3)(2) - FAB(3) = 0 FAB = 7.50 kN

a +ΣMA = 0; Cy(3) -12

(3)(3)(1) - 3(3)(1.5) = 0 Cy = 6.00 kN

+S ΣFx = 0; Cx = 0

Then using the result of FAB to consider the equilibrium of the FBD of member BD shown in Fig. b,

a +ΣMD = 0; 7.50(3) - 4(3)(1.5) - FBCa35b(3) = 0 FBC = 2.5 kN

a +ΣMB = 0; -Dy(3) + 4(3)(1.5) = 0 Dy = 6.00 kN

+S ΣFx = 0; 2.5a45b - Dx = 0 Dx = 2.00 kN

3 kN>m

4 kN>m

6 kN>m

2.25 m

3 m

B

A

D

C

2–39. Determine the horizontal and vertical force components that the pin supports at C and D exert on members AC and BD, respectively.

Ans.

Ans.

Ans.

Ans.

65


Ans.Dx = 0Dy = 0.75wLcAx = wL dAy = 0.75wLc

MA = awL2

2b B

SolutionMember BC:

a + aMB = 0; Cy(1.5L) - (1.5wL)a1.5L2

b = 0

Cy = 0.75 wL

+ c aFy = 0; By - 1.5wL + 0.75wL = 0

By = 0.75wL

Member CD:

a + aMD = 0; Cx = 0

+S ΣFx = 0; Dx = 0 Ans.

+ c aFy = 0; Dy - 0.75wL = 0

Dy = 0.75wLc Ans.

Member BC:+S ΣFx = 0; Bx - 0 = 0; Bx = 0

Member AB:+S ΣFx = 0; wL - Ax = 0

Ax = wL d Ans.

+ c aFy = 0; Ay - 0.75wL = 0

Ay = 0.75wLc Ans.

a + aMA = 0; MA - wLa12b = 0

MA = awL2

2b B Ans.

A

B

D

C

w

w

L

1.5L

*2–40. Determine the reactions at the supports A and D. Assume A is fixed and B, C, and D are pins.

0.75 wL

66


SolutionEquations of Equilibrium. Referring to the FBD of members AB and BC shown in Fig. a and b, respectively, we notice that Bx and By can be determined by solving simultaneously the moment equations of equilibrium written about A and C, respectively.

a +ΣMA = 0; Bx(6.5) + By(6) - (6)(142.25)a 2.5142.25b(5.25)

-(6)(142.25)a 6142.25b(3) - (2)(4)(2) = 0

6.5Bx + 6By = 202.75 (1)

a +ΣMC = 0; (6)(142.25)a 2.5142.25b(5.25) + (6)(142.25)a 6142.25

b(3)

+ By(6) - Bx(6.5) = 0

6.5Bx - 6By = 186.75 (2)

6 m 6 m

4 m

2.5 m

A

B

C

6 kN>m6 kN>m

2 kN>m

2–41. Determine the components of reaction at the pinned supports A and C of the two-member frame. Neglect the thickness of the members. Assume B is a pin.

67


2–41. (Continued)

Solving Eq (1) and (2) yields

Bx = 29.96 k By = 1.333 k

Using these results and writing the force equation of equilib-rium by referring to the FBD of member AB, Fig a,

+S ΣFx = 0; 2(4) + (6)(142.25)a 2.5142.25b - 29.96 + Ax = 0

Ax = 6.962 k = 6.96 k

+ cΣFy = 0; Ay + 1.333 - 6(142.25)a 6142.25b = 0

Ay = 34.67 k = 34.7 k

Referring to the FBD of member BC, Fig b,

+S ΣFx = 0; 29.96 - (6)(142.25)a 2.5142.25b - Cx = 0

Cx = 14.96 k = 15.0 k

+ cΣFy = 0; Cy - 1.333 - (6)(142.25)a 6142.25b = 0

Cy = 37.33 k = 37.3 k

Ans.Ax = 6.96 kAy = 34.7 kCx = 15.0 kCy = 37.3 k

Ans.

Ans.

Ans.

Ans.

68


Ans.Bx = 13.2 kAx = 4.80 kBy = 2.20 kDx = 13.2 kDy = 20.2 kAy = 2.20 k

SolutionEquations of Equilibrium. We will consider the equilibrium of the FBD of member AB shown in Fig. a first.

a +ΣMA = 0; Bx(20) - 6(20) - 12(12) = 0 Bx = 13.2 k

a +ΣMB = 0; 12(8) - Ax(20) = 0 Ax = 4.80 k

+ cΣFy = 0; By - Ay = 0 (1)

Using the result of Bx to consider the equilibrium of the FBD of member BCD shown in Fig. b,

a +ΣMD = 0; 12

(2)(6)(4) + 2(6)(9) - 13.2(12) + By(12) = 0

By = 2.20 k

+S ΣFx = 0; 13.2 - Dx = 0 Dx = 13.2 k

+ cΣFy = 0; Dy - 2(6) -12

(2)(6) - 2.20 = 0 Dy = 20.2 k

From Eq. (1),

2.20 - Ay = 0 Ay = 2.20 k

A

C

D

D

B

2 k>ft

6 k

12 k

8 ft

12 ft

12 ft

6 ft 6 ft

2–42. Determine the horizontal and vertical components of reaction at A, B, and D. Assume the frame is pin connected at A, B, and D, and there is a fixed-connected joint at C.

Ans.

Ans.

Ans.

Ans.

Ans.

Ans.

69


4 k>ft

12 ft

30 k 10 k

10 ft5 ft 5 ft

4 ft4 ft

8 ft

A

B D

E

FC

10 ft

2–43. The bridge frame consists of three segments which can be considered pinned at A, D, and E, rocker supported at C and F, and roller supported at B. Determine the horizontal and vertical components of reaction at all these supports due to the loading shown.

SolutionEquations of Equilibrium. We will consider the equilibrium of the FBD of member BD shown in Fig. b first.

a +ΣMD = 0; 30(10) - NB(18) = 0 NB = 16.67 k = 16.7 k

a +ΣMB = 0; Dy(18) - 30(8) = 0 Dy = 13.33 k = 13.3 k+S ΣFx = 0; Dx = 0

Next, using the result of NB to consider the equilibrium of the FBD of member ABC, shown in Fig. a,

a +ΣMA = 0; NC(4) - 4(10)(5) - 16.67(10) = 0 NC = 91.67 k = 91.7 k+S ΣFx = 0; Ax = 0

+ cΣFy = 0; 91.67 - 4(10) - 16.67 - Ay = 0 Ay = 35.0 k

Finally, using the result of Dx and Dy, the equilibrium of the FBD of member DEF, Fig. c, gives

a +ΣME = 0; 10(5) + 13.33(10) - NF(4) = 0 NF = 45.83 k = 45.8 k+S ΣFx = 0; Ex = 0

+ cΣFy = 0; 45.83 - 13.33 - 10 - Ey = 0 Ey = 22.5 k

Ans.NB = 16.7 kDy = 13.3 kDx = 0

NC = 91.7 kAx = 0Ay = 35.0 kNF = 45.8 kEx = 0Ey = 22.5 k

Ans.

Ans.

Ans.

Ans.

Ans.

Ans.

Ans.

Ans.

Ans.

70


6 ft

120 lb>ft

600 lb 600 lb800 lb800 lb

400 lb 400 lb

A C

6 ft 6 ft 6 ft

10 ft

5 ftD G

E F

B

Ans.

*2–44. Determine the horizontal and vertical reactions at the connections A and C of the gable frame. Assume that A, B, and C are pin connections. The purlin loads such as D and E are applied perpendicular to the center line of each girder.

SolutionMember AB:

a +ΣMA = 0; Bx(15) + By(12) - (1200)(5) - 600a1213

b(6) - 600a 513

b(12.5)

-400a1213

b(12) - 400a 513

b(15) = 0

Bx(15) + By(12) = 18,946.154 (1)

Member BC:

a +ΣMC = 0; -Bx(15) + By(12) + 600a1213

b(6) + 600a 513

b(12.5)

400a1213

b(12) + 400a 513

b(15) = 0

Bx(15) - By(12) = 12,446.15 (2)

Solving Eqs. (1) and (2),

Bx = 1063.08 lb By = 250.0 lb

Member AB:

+S ΣFx = 0; -Ax + 1200 + 1000a 513

b - 1063.08 = 0

Ax = 522 lb

+ cΣFy = 0; Ay - 800 - 1000a1213

b + 250 = 0

Ay = 1473 lb

Member BC:

+S ΣFx = 0; -Cx - 1000a 513

b + 1063.08 = 0

Cx = 678 lb

+ cΣFy = 0; Cy - 800 - 1000a1213

b - 250.0 = 0

Cy = 1973 lb Ans.Ax = 522 lbAy = 1473 lbCx = 678 lbCy = 1973 lb

Ans.

Ans.

Ans.

Ans.

71


2–1P. The railroad trestle bridge shown in the photo is supported by reinforced concrete bents. Assume the two simply supported side girders, track bed, and two rails have a weight of 0.5 k>ft and the load imposed by a train is 7.2 k>ft. Each girder is 20 ft long. Apply the load over the entire bridge and determine the compressive force in the columns of each bent. For the analysis assume all joints are pin connected and neglect the weight of the bent. Are these realistic assumptions?

SolutionMaximum reactions occur when the live load is over the entire span.

Load = 7.2 + 0.5 = 7.7 k>ft

R = 7.7(10) = 77 k

Then P =2(77)

2= 77 k

All members are two-force members.

a +ΣMB = 0; -77(8) + F sin 75°(8) = 0

F = 79.7 k

It is not reasonable to assume the members are pin connected, since such a framework is unstable.

Ans.F = 79.7 kIt is not reasonable to assume the members are pin connected, since such a framework is unstable

P P8 ft

18 ft

A D

B C

758 758

Ans.

Ans.

Documents

Solution€¦ · 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 E y