CH2SM

© 2009 by R.C. Hibbeler. Published by Prentice Hall, Pearson Education South Asia Pte Ltd. All rights reserved. This material is protected under copyright laws as they currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.

2–1. The frame is used to support a wood deck (not shown) that is to be subjected to a uniform load of 6.5 kN/m2. Sketch the loading that acts along members BG and ABCD. Take b 3 m, a 1.5 m.

b/a 3/1.5 2 > 1.5 (one-way slab)

1.5 m

1.5 m

1.5 m

3m

w 6.5(1.5) 9.75 kN/m 14.63 kN

14.63 kN 3m 7.31 kN

For BG, w = 9.75 kN/m

14.63 kN 14.63 kN 7.31 kN

Ans

21.94 kN

21.94 kN

For ABCD, reactions are 21.9 kN

Ans

1.5 m

2–2. The roof deck of the single storey building is subjected to a dead plus live load of 6 kN/m2. If the purlins are spaced 1.2 m and the bents are spaced 7.5 m apart, determine the distributed loading that acts along the purlin DF, and the loadings that act on the bent at A, B, C, D, and E.

7.5 m

A

B

7.5 m

E

1.5 m

1.5

7.5 m

D

1.5 m

C F

4@ 1.2 m = 4.8 m

10


2–3. The steel framework is used to support the 100-mm reinforced lightweight concrete slab that carries a uniform live loading of 25 kN/m2. Sketch the loading that acts along members BE and FD. Set b 3 m, a 2.4 m. Hint: See Table 1–3.

63.6 kN/m

1.2 m

0.6 m

1.2 m

C

57.24 kN

57.24 kN B

D 57.24 kN

A F b

E

31.8 kN/m 28.62 kN

28.62 kN 31.8 kN/m

a a

1.2 m 1.2 m E 1.2 m

1.2 m 95.4 kN

95.4 kN

DL 0.015(100) 1.5 kN/m2 LL 25 kN/m2 Total load 26.5 kN/m2

L2 b 3 = = = 1.25 < 1.5 L1 a 2.4

Reaction at B, 57.2 kN; Reaction at F, 95.4 kN

Two-way slab

Solve Prob. 2–3, with b 3.6 m, a 1.2 m.

*2–4.

31.8 kN/m

3.6 m 57.24 kN

57.24 kN 57.24 kN

28.62 kN

28.62 kN

57.24 kN

57.24 kN

DL 0.015(100) 1.5 kN/m2 LL 25 kN/m2 Total load = 26.5 kN/m2 L2 b 3.6 = = = 3 > 1.5 L1 a 1.2 One-way slab

2–5. The frame is used to support the wood deck in a residential dwelling. Sketch the loading that acts along members BG and ABCD. Set b 3 m, a 1.5 m. Hint: See Table 1– 4. From Table 1–4 LL 1.92 kN/m2 L2 b 3.5 = = =2 L1 a 1.5 One-way slab

Reaction at A : 6.48 kN

1.92(1.5) 2.88 kN/m

2.16 kN

4.32 kN

1.5 m

4.32 kN 1.5 m

2.16 kN 1.5 m

3m 4.32 kN

4.32 kN

6.48 kN

6.48 kN

11


2–6.

Solve Prob. 2–5, if b 2.4 m, a 2.4 m.

4.608 kN/m

1.2 m

1.2 m

2.76 kN

2.76 kN

2.76 kN

1.38 kN

From Table 1–4 LL 1.92 kN/m2 L2 b 2.4 = = =1< 2 L1 a 2.4 Two-way slab

2.30 kN/m

1.2 m 1.2 m

Reaction at A : 8.29 kN

2–7.

Ans

2.76 kN

2.30 kN/m

1.2 m

1.38 kN

2.30 kN/m

1.2 m

1.2 m 1.2 m

8.29 kN

8.29 kN

Solve Prob. 2–5, if b 4.5 m, a 3 m.

5.76 kN/m

1.5 m

From Table 1–3 LL 1.92 kN/m2 b/a = 4.5/3 = 1.5 Two-way slab

Reaction at 8.6 kN and 19.4 kN

1.5 m

1.5 m

8.64 kN

8.64 kN

4.32 kN 8.64 kN 8.64 kN 4.32 kN 2.88 kN/m 2.88 kN/m 2.88 kN/m

Ans 1.5 m

1.5 m

1.5 m

19.44 kN

1.5 m

1.5 m

1.5 m

19.44 kN

12


13


14


15


16


17


2–13. Classify each of the structures as statically determinate, statically indeterminiate, or unstable. If indeterminate, specify the degree of indeterminacy.

18


19


20


45 kN

7.5 kN/m A

B 3m

+3MA 0; By(6) – 45 cos 30°(6) – 22.5(1.5) 0 By 44.6 kN Ans + #3Fy 0; Ay + 44.6 – 22.5 – 45 cos 30° 0 Ay 16.8745 kN Ans 3Fx 0; –Ax + 45 sin 30° 0 Ax 22.5 kN

7.5 (3) 22.5 kN

1.5 m

Ans

21

30

4.5 m

45 kN

3m


36 kN

7.5 kN/m

36 kN

+3MB 0; Ay(6.3) + 36(2.1) + 36(4.2) + (7.5)(2.4)(7.5) 0 Ay 57.4 kN Ans

2.4 m

2.1 m

+ #3Fy 0; By + 57.4 – (7.5)(2.4) – 36 – 36 = 0 By 32.6 kN Ans

B

A 2.1 m

2.1 m

3Fx 0; Bx = 0

Ans

120 kN/m 75 kN/m A

B

1.8 m

2.7 m

45 kN/m 30 kN/m

30 kN/m B

A 810 kN·m 3.6 m

–270(3.6) – 810 + FB cos 60°(7.2) FB 495 kN Ans Ax – 495 sin 60° 0 Ax 428.7 kN

3.6 m

30(7.2) +

1 (15)(7.2) 270 kN 2

Ans 810 kN • m

Ay + 495 cos 60° – 270 0 Ay 22.5 kN

3.6 m

Ans

22

3.6 m

60


30 kN/m B

75 kN/m

3m

A

7.2 m

45 kN

45 kN/m D

C

E

0.9 m F

B

A

1.2 m

1.2 m

0.3 m G

1.2 m

1.2 m

23


N1 50 kN

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PP 225 mm 225 mm 125 mm PP PP

w2

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Z ² ²² Z N1P N1P $QV

Z

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Z ²Z

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24

PP

PP


30 kN/m N1 N1

B

P

P

P

P

2.5 m

10 kN/m

P

A

)% ² ² $QV )% N1 N1 6m

² ²$[ $QV $[ N1 $\² $QV $\ N1

2.16 kN

C 1.2 kN/m

3m

30° 3 m A

0.9 m

0.9 m

B 1.8 m

+3MA 0;

Cy(3 + 118 sin 60°) – 2.16(0.9) 0 Cy 0.426 kN Ans

+ 3F 0;

Ay – 0.426 sin 30° 0 Ay 0.213 kN

+#3Fy 0;

25

Ans

Ay + 0.426 cos 30° – 2.16 0 Ay 1.791 kN Ans


2 kN/m C

B 3m 1.5 m

(–6 1.5) + (FA)(cos 60°)(3) – (FA)(sin 60°)(1.5) 0 FA 44.78 kN ≈ 44.8 kN Ans Bx – 44.78(sin 60°) 0 Bx 38.78 kN ≈ 38.8 kN Ans –By – 6 + 44.78(cos 60°) 0 By 16.39 kN ≈ 16.4 kN Ans

A

6 kN 1.5 m

60

1.5 m

1.5 m

*2–28. The cantilever footing is used to support a wall near its edge A so that it causes a uniform soil pressure under the footing. Determine the uniform distribution loads, wA and wB, measured in kN/m at pads A and B, necessary to support the wall forces of 40 kN and 100 kN.

100 kN

40 kN 0.45 m

0.075 m

3.15 m

100 kN

0.3 m

0.45 m

(wA)(0.6)

40 kN

0.075 m

0.45 m

–40(3.15) + wB(0.9)(3.15) + 100(0.225) 0 wB 36.5 kN/m

B

A wA 0.6 m

wB(9)

2.4 m

Ans

36.5(0.9) – 140 + wA(0.6) 0

wB 0.9 m

wA 178.6 kN/m

7.5 kN/m

Ans

5 kN/m 5m

A

B

12 m

26

12 m


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2–31. Determine the reactions at the supports A and B of the frame. Assume that the support at A is a roller. 35 kN

25 kN 4m

50 kN 3m

10 kN

N1 N1

N1

3m

N1 P

A

P

P

P N1

4m

P

2.5 kN 3m

² ² ² $\ $QV $\ N1 ²²²²%\ %\ N1 $QV ²%[ $QV %[ N1

B

27


10.5 kN/m 6m 7.5 kN/m

B

9m A 14.4 m

14.4 m

N1P

N1PDWP N1 N1PDWP N1

P P

N1 P

)[

%\ N1 N1

$QV

²$[ $[ N1

$QV

P

P

)\

$\%\²

40 kN

30 kN

20 kN

12 kNm

4m

7m

6m

%\ ² ² ²

P

8m

28

$\ N1

$QV


80 kN/m 180 kN 25 kN

C

A

B 25 kN

3m 5

D 1.5 m

3m

3m

4

3

3m

3m

3m

1m

E

From FBD (b) ⎛ 4⎞ +3ME 0; F(4) – 25⎜⎝ ⎟(1) 0 5⎠

1m

+#3Fy 0;

+

3Fx 0;

F 5 kN

⎛ 4⎞ Ey + 5 – 25 ⎜ ⎟ 0 ⎝ 5⎠

Ey 15 kN

Ans.

⎛ 4⎞ Ex + 5 – 25 ⎜ ⎟ 0 ⎝ 5⎠

Ex 15 kN

Ans.

From FBD (a) By(3) – 180(3) – 5(4.5) 0 By 187.5 kN +3MA 0; +$3Fy 0; –187.5 + 180 + 5 0 Ay 2.5 kN + 3Fx 0; Ax 0

30 kN/m

B 4m

45 kN/m

40 kN

A 4m

E

D

F 3m

1m

Ans. Ans. Ans.

C 4m

4m

180 kN

120 kN 40 kN

40 kN(1 m) – Ey(4 m) 0 Ey 10 kN 10 kN – 40 kN + Fy 0 Fy 30 kN

1m 2m

120 kN(2 m) + By(4 m) – 10 kN(8 m) 0 By 80 kN Ans Ay – 120 kN + 80 kN – 10 kN 0 Ay 50 kN Ans 180 kN(2 m) – Cy(4 m) + (8 m)(30 kN) 0 Cy 150 kN Ans –30 kN + 150 kN – 180 kN + Dy 0 Dy 60 kN Ans

29

2m

4m

4m

4m

2m

2m


A B 6m 6m

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A

C

B 60 m

30 m

30 m

60 kN 60 m

D

E 45 m

360 kN

225 kN

180 kN

12 m

E

F

3m

9m

D

C 12 m

9m

3m

P

P

%\ ² $QV %\ N1 ²²$\ $\ N1 $QV

F 45 m

P

P

P

P N1

²(\ $QV (\ N1 ²²)\ $QV )\ N1

²&\ &\ N1 ²'\ '\ N1

30


67.5 kN

30 kN/m A

C

B N1

2.4 m

P

P

D 1.8 m

P

1.2 m

1.5 m

1.8 m

P

&\ ² &\ N1 %\² %\ N1 %[ N1 P

'[ ² ² '[ N1 $\²² $\ N1

P

$[ $[ N1

B N1

P

5 kN/m

5m

P P N1 P

C

4m ²$\ $\ N1 ²&\ &\ N1 ²&[ &[ N1

31

$QV $QV

20 kN/m

A

$QV

$QV $QV $QV


70 kN/m 2.5 m

75 kN

B C

350 kN

2.5 m

75 kN 2m

2m

50 kN

50 kN

+3MA 0;

By(5) – 350(2.5) – 50(2) – 75(4) 0 By 255 kN Ans

+#3Fy 0;

Ay + 255 – 350 0 Ay 95 kN

2m

2m

+!3Fx 0; Ax – 50 – 75 = 0 Ax 125 kN

A

Ans

Ans

5m

27 kN

45 kN

B

D

6m

7.5 kN/m

A

9m

C N1P

4.5 m

P

P

N1 P

3m

P

3m

² ² )'& N1 ²%[ %[ N1 $QV %\ ² %\ N1 $QV )'&

P

N1 P

²0$ 0$ N1P $[² $[ N1 $\²² $\ N1

N1 P

P

32

$QV

$QV

$QV


25 kN/m A 6m

B

12 m

E

D

6m

8m C

F 2m

2m

N1

²%\

%\ N1 P

'\²

P

'\ N1 N1 P P

&\ ² ² &\ N1

²$\² ²

$\ N1

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N1 P P

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²(\² ²

P

(\ N1

33

%\ N1


2-43. 50 kN/m B 1.8 m D

E C

A

5.4 m

34

5.4 m

25 kN/m 3 m

© 2009 by R.C. Hibbeler. Published by Prentice Hall, Pearson Education South Asia Pte Ltd. All rights reserved. This material is protected under copyright laws as they currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher. 10 kNm

2m 4m

5 3 4

3m

1.5 m

20 kNm

4m 6m

8 kN/m

5m

35


36


37


these connections is a pin and that joint B is fixed-connected.

18 kN 1.2 m

B

C

3.6 m

A

P

C

13.5 kN 0.9 m

%[ ² %[ N1 $QV $[² $[ N1 $QV $\²%\

N1

P

6 kN/m

1.8 m

%\ ² ² %\ N1 $QV &\²² &\ N1 N1

N1

&[² &[ N1

$QV

$\ N1

$QV

P

38

P

P


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1.8 kN 1.8 kN 2.7 kN 3.6 kN

E

D

2.7 kN

F

3.6 kN

G

B

1.5 m

1.8 kN/m

3m A

C

1.8 m

1.8 m

1.8 m

1.8 m

N1

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39


20 kN

12 kN

8 kNm

3m

2m

40

2m

2m

2m


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41

CH2SM

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