Assignment Solutions-Set 01 PDF

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Summary

The subject matter is organized into 14 chapters. Chapter 1 begins with a review of the important concepts of statics, followed by a formal definition of both normal and shear stress, and a discussion of normal stress in axially loaded members and average shear stress caused by direct shear. ...

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*6–4. Express the shear and moment in terms of x for 0 6 x 6 3 m and 3 m 6 x 6 4.5 m, and then draw the shear and moment diagrams for the simply supported beam.

300 N/m

A

B

3m

1.5 m

Support Reactions: Referring to the free-body diagram of the entire beam shown in Fig. a, a+ Σ MA = 0;

B y(4.5) -

1 1 (300)(3)(2) - (300)(1.5)(3.5) = 0 2 2

By = 375 N +cΣ Fy = 0;

Ay + 375 -

1 1 (300)(3) - (300)(1.5) = 0 2 2

Ay = 300 N Shear and Moment Function: For 0 … x 6 3 m, we refer to the free-body diagram of the beam segment shown in Fig. b. F +cΣ

y

= 0;

1 3 00 - (100x)x - V = 0 2 V = {300 - 50x2} N

M = 0; a+ Σ

M +

Ans.

1 x (100x)xa b - 300x = 0 3 2

M = e 300x -

50 3 x fN # m 3

Ans.

When V = 0, from the shear function, 0 = 300 - 50x2

x = 26 m

Substituting this result into the moment equation, M|x = 26 m = 489.90 N # m For 3 m 6 x … 4.5 m, we refer to the free-body diagram of the beam segment shown in Fig. c. F +cΣ

y

= 0;

V + 375 -

1 3200(4.5 - x) 4 (4.5 - x) = 0 2

V = e 100(4.5 - x)2 - 375f N M = 0; a+ Σ

375(4.5 - x) -

Ans.

1 4.5 - x b - M = 0 [200(4.5 - x)](4.5 - x) a 3 2

M = e375(4.5 - x) -

100 (4.5 - x)3 f N # m Ans. 3

Shear and Moment Diagrams: As shown in Figs. d and e.

Ans: For 0 … x 6 3 m, V = {300 - 50x2} N, 50 3 M = e 300x x f N # m, 3 For 3 m 6 x … 4.5 m, V = e100(4.5 - x)2 - 375f N, M = e 375(4.5 - x) -

100 (4.5 - x)3 f N # m 3

6–14. Draw the shear and moment diagrams for the beam.

2 kip/ ft 30 kipft

B A 5 ft

5 ft

5 ft

Ans: For 0 … x 6 5 ft: V = { -2x} kip, M = 5 -x2 6 kip # ft , For 5 ft 6 x 6 10 ft: V = -0.5 kip, M = { -22.5 - 0.5x} kip # ft , For 10 ft 6 x … 15 ft: V = -0.5 kip, M = {7.5 - 0.5x} kip # ft

6–39. Draw the shear and moment diagrams for the double overhanging beam.

400 lb

400 lb 200 lb/ft

A

B

3 ft

6 ft

3 ft

Equations of Equilibrium: Referring to the free-body diagram of the beam shown in Fig. a, a+ Σ MA = 0;

By(6) + 400(3) - 200(6)(3) - 400(9) = 0 By = 1000 lb

Fy = 0; +cΣ

Ay + 1000 - 400 - 200(6) - 400 = 0 Ay = 1000 lb

Shear and Moment Diagram: As shown in Figs. b and c.

Ans: V (lb) 600 0

3

400 6

9

x (ft) 12

400 600

M (lbft) 0

3

6

9

300 1200

1200

12

x (ft)

Draw the shear and moment diagrams for the simply supported beam.

10 kN

10 kN 15 kN m

A

B 2m

SOLUTION

2m

2m

6–41. The compound beam is fixed at A, pin connected at B, and supported by a roller at C. Draw the shear and moment diagrams for the beam.

600 N 400 N/m

C A

B 2m

2m

2m

SOLUTION Support Reactions: Referring to the free-body diagram of segment BC shown in Fig. a, a+Σ MB = 0;

Cy(2) - 400(2)(1) = 0 Cy = 400 N

F + cΣ

y

= 0;

By + 400 - 400(2) = 0 By = 400 N

Using the result of By and referring to the free-body diagram of segment AB, Fig. b, F + cΣ

y

= 0;

Ay - 600 - 400 = 0 Ay = 1000 N

MA = 0; a+Σ

MA - 600(2) - 400(4) = 0 MA = 2800 N

Shear and Moment Diagrams: As shown in Figs. c and d.

Ans: V (N) 1000 400 0

5 2

6

4

x (m)

400 M (Nm)

0

4 800

2800

5

200

2

5

x (m) 6...