9789810682627_sm_03

7/25/2019 9789810682627_sm_03

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© 2009 by R.C. Hibbeler. Published by Prentice Hall, Pearson Education South Asia Pte Ltd. All rights reserved. This material is protected under copyright

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30 kN/m

6 @ 4 m 24 m

4 m

A

B

M L

K

J I

C D E F G

H N

3–6.

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15

22.5 19.2

47.2

28.016.0

40

15

47.2

15

25

16

21.521.5

35

15

35

35 kN (C)

+15 – 35 0 21.5 kN (T) Ans

– 21.5 8.0 kN (C) Ans

21.5 kN (C)

+ y 0; F HE sin 51.3° – 15 0 F HE 19.2 kN (T) Ans

+ x 0; 40 – 19.2 cos 51.3° – F DE = 0 F DE = 28.0 kN (T) Ans

– 2(21.5 16.0 kN (T)

16.0 16.0 kN (T)

+ 25 0

47.2

47.2 kN (C)

40.0 kN (T)

47.2 kN (C) Ans

15.0 kN (C) Ans

+ x 0; 28.0 – 16.0 – F ID cos 61.9° = 0 F ID = 25.5 kN (T) Ans

+ y 0; 25.5 sin 61.9° – F HD 0 F HD 22.5 kN (C) Ans

+ x 0; F IH cos 32.0° – 47.2 cos 32.0°

+ 19.2 sin 38.7° = 0

F IH 33.0 kN (C) Ans

2.4 m2.4 m2.4 m1.8 m1.8 m

1.5 m

1.5 m

1.5 m

4.5 m

A

J

B C D E

F

15 kN 15 kN

15 kN

15 kN

G

H

I

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2 m 2 m 2 m 2 m 2 m 2 m 2 m 2 m

2 m

2 m

3 kN 3 kN

4 kN

A

B

I

H G

C D

F

E

5 kN 3 kN 4 kN 3 kN 5 kN

2 m

2 m

2 m 4 m 2 m

5 kN 3 kN

5 kN

–5(4) (6)

7.45 kN (C)

6.0 kN (T)

–5(6) + 3(2) + F BC (4) 0

5(2) – 3(6) + F BH sin 63.43°(6) 0

1.49 kN (T)

1.49 kN (T) Check

5 – 3 – 7.45(sin 26.56°) + F BH (sin 63.43°) 0

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15 kN

15 kN

1.5 m

1.5 m

1.5 m

15 kN

2.4 m2.4 m2.4 m1.8 m1.8 m

2.25 m

2.25 m

10 kN

1.5 m

1.5 m

1.5 m

2.4 m2.4 m2.4 m1.8 m1.8 m

2.25 m

2.25 m

15 kN

15 kN

15 kN

10 kN

36.28 kN

18.67 kN

10 kN

22.67 kN4.5

1.8

2.25

1.8

28.33 kN

2.88

2.83

2.4

1.5

26.67 kN

2.83

15 kN

50.31 kN2.4

1.5

3.843

2.4

15 kN

42.67 kN

15 kN

2.41.5

3

2.4

19.21 kN

50.31 kN

30.67 kN18.67 kN

5.14.5

2.4

–F y(10.8) + 15(8.4) + 15(6) + 15(3.6) + 10(1.8) = 0F y = 26.67 kN

A y + 26.67 – 10 – 15 – 15 – 15 = 0

A y = 28.33 kN

36.28( ) – F JI ( ) = 0

F JI = 36.28 kN (C) Ans

F JB – 10 + 36.28( ) – 36.28( ) = 0

F JB = 10 kN (C) Ans

F BI ( ) – 10 = 0

F BI = 10.77 kN (T) Ans

F BC – 22.67 + 10.77( ) = 0

F BC = 18.67 kN (T) Ans

F CI = 0 Ans

F CD = 18.67 kN (T) Ans

26.67 – F FG( ) = 0

F FG = 50.31 kN (C) Ans

F FE – 50.31( ) = 0

F FE = 42.67 kN Ans

–50.31( ) + ( ) = 0

F GH = 50.31 (C) Ans

F GE – 15 – 50.31( ) + 50.31( ) = 0

F GE = 15 kN (C) Ans

42.67 – F ED – 19.21( ) = 0

F ED = 30.67 kN (T) Ans

30.67 – 18.67 – F DI ( ) = 0

F DI = 25.5 kN (T) Ans

25.5( ) – F DH = 0

F DH = 22.5 kN (C) Ans

F HI ( ) – 50.31( ) + 19.21( ) = 0

F HI = 36.10 kN (C) Ans

F EH ( ) – 15 = 0

F EH = 19.21 kN (T) Ans

F AB – 36.28( ) = 0

F AB = 22.67 kN (T) Ans

28.33 – F AJ ( ) = 0

F AJ = 36.28 kN (C) Ans

2.252.88

1.82.88

2.252.88

2.252.88

2.25

2.88

2.25

2.88

4.54.85

1.84.85

1.52.83

2.42.83

2.42.83

2.42.83

1.52.83

1.52.83

33.84

2.43.84

2.45.1

4.55.1

2.4

2.84

2.4

2.84

2.4

3.84

10 kN

22.5 kN

H

I

J

A

B

G

F

C D E

2.25 m

2.25 m

15 kN

15 kN

15 kN

1.5 m

1.5 m

1.5 m

1.8 m1.8 m 2.4 m2.4 m2.4 m

10 kN

3–15.

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3–16. Solve Prob. 3-15 assuming there is no external load

on joints J , H, and G and only the vertical load of 15 kN exists

on joint I .

15 kN

2.25 m

2.25 m

1.8 m 1.8 m 2.4 m 2.4 m 2.4 m

1.5 m

1.5 m

1.5 m

2.841.5

2.4

5 kN

9.47 kN2.4

1.5

3.84

8 kN2.4

3

2.84

9.47 kN

1.5

2.4

5.1

8 kN4.5

2.4

2.88

8 kN

10 kN

2.25

1.8

2.882.25

1.8

12.8 kN

8 kN 8 kN1.8 4.5

) Ans

Ans

1.5 m

1.5 m

1.5 m

2.4 m2.4 m2.4 m1.8 m1.8 m

2.25 m

2.25 m

15 kN

2.25 m

2.25 m

1.8

15(3.6) – 10.8F y = 0; F y = 5 kN

A y + 5 – 15 = 0; A y = 10 kN

5 – F GF ( ) = 0; F GF = 9.47 kN (C) Ans

9.47( ) – F EF = 0; F EF = 8 kN (T) Ans

F HG( ) – 9.47( ) = 0

F HG = 9.47 kN (C)

8 – F ED = 0

F ED = 8 kN (T)

F IH ( ) – 9.47( ) = 0

F IH = 9.47 kN (C)

8 – F CD = 0; F CD = 8 kN (T)

8 – F CB = 0

F CB = 8 kN (T)

–F AJ ( ) + 10 = 0F AJ = 12.8 kN (C)

F AB – 12.8( ) = 0

F AB = 8 kN (T)

–F IJ ( ) + 12.8( ) = 0

F IJ = 12.8 kN

1.5

2.84

2.42.84

2.42.84

2.42.84 2.42.84

2.25

2.88

1.82.88

2.252.88

2.252.88

2.42.84

H

I

J

A

B

G

F

C D E

2.25 m

2.25 m

15 kN

15 kN

15 kN

1.5 m

1.5 m

1.5 m

1.8 m1.8 m 2.4 m2.4 m2.4 m

10 kN

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3–22. Members AB and BC can each support amaximum compressive force of 4 kN, and members AD, DC , and BD can support a maximum tensileforce of 7.5 kN. If a = 3 m, determine the greatestload P the truss can support.

3–21.

F AB = 4 kN (C)

–4

F AD = 2.915 kN < 7.5 kN

(4) (2.915) = 0

P = 4.243 kN

–4.243 – 2.915(2)

F BD = 5.657 kN < 7.5 kN

Pmax = 4.24 kN

4 kN

2.915 kN

4.243 kN

2.915 kN

3–23. Members AB and BC can each support amaximum compressive force of 4 kN, and members AD, DC , and BD can support a maximum tensile forceof 10 kN. If a = 1.8 m, determine the greatest load Pthe truss can support.

3–22.

F AD = 2.915 kN < 7.5 kN

(4) (2.915) = 0

P = 4.243 kN

–4.243 – 2.915(2)

F BD = 5.657 kN < 10 kN

Pmax = 4.24 kN

–4

F AB = 4 kN (C)

4 kN

2.915 kN2.915 kN

4.243 kN

B

P

A

C a a

3__a

4

1__a

4D

B

P

A

C

a a

3__a

4

1__a

4D

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2 A y – 4 – 3 – 4 = 0 A y = 5.5 kN

3 kN

4 kN4 kN

3 m3 m

3 m

4 kN

4.5 tan 30° m

3 cos2 30° m

3 m1.5 m

A y = 5.5 kN

F BC (4.5 tan 30°) + 4(4.5 – 3 cos2 30°) – 5.5(4.5) = 0

F BC = 6.062 kN (T) = 6.06 kN (T)

F FBsin 60°(3) – 4(3 cos2 30°) = 0F FB = 3.464 kN (T) = 3.46 kN (T)

F GF sin 30°(3) + 4(3 – 3 cos2 30°) – 5.5(3) = 0

F GF = 9 kN (C)

A

B

G

3 m 3 m 3 mC

306030 60

F

3 kN

D

E

4 kN4 kN

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-

-

10.46 kN10.46 kN

15 kN20.93 kN 8.66 kN

6 m

15 kN18.125 kN 18.125 kN

18.13(3) – F IH (5.196) = 0; F IH = 10.46 kN (T)

18.13(9) – 15(6) – 10.46(5.196) – F JH cos 30°(6) = 0

F JH = 3.61 kN (T) Ans

F AB = = 20.93 kN (C)

F AI = 20.93(cos 60°) = 10.46 kN (T)

–F IBcos 30° + 10.46(cos 30°) + 15(cos 60°) – 10.46(cos 30°) = 0

F IB = 8.66 kN (T) Ans

8.66(cos 30°) – F BJ (cos 30°) = 0

F BJ = 8.66 kN (C) Ans

18.125

sin 60°

3 mm

15 kN

B

C D

A FH G I

J K E

3 m

3 m

3 m

3 m3 m3 m3 m

6 m

10 kN 10 kN

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A y = 10 kN

10 kN

5 m

F AC cos 33.69° – F ADcos 45° = 0

10 – F ADsin 45° + F AC sin 33.69° = 0

F AC = 36.055 kN = 36.1 kN (T) Ans

F AD = 42.426 kN = 42.4 kN (C) Ans

42.426sin 45° – 10 – F DBsin 33.69° = 0

F DB = 36.055 kN = 36.1 kN (T) Ans

42.426cos 45° – F DC + F DBcos 33.69° = 0

F DC = 60 kN (C) Ans

F CB = 42.4 kN (C) Ans

D

B A

C

5 m

2 m 1 m

10 kN 10 kN

2 m

2 m

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AB C

D

E

J K L

I H

G F

2 m

2 m 2 m4 m

4 m

2 kN 2 kN

4 m

2 m

2 m 4 m 2 m

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A

B

C

D

E

G

F

3 m 6 m

6 m 6 m

9 kN 9 kN

3 m

2.1 m

4.2 m

90 kN

3 m6 m3 m

9 kN 9 kN

90 kN

1.5 kN

2.7 m

1.8 m

0.9 m

J

I

H

G

K

A

L

B C D E F

2.4 m 2.4 m 2.4 m 2.4 m 2.4 m 2.4 m

2 kN

1 kN

2 kN

2 kN 2 kN

1 kN

2 kN

2.4 m2.4 m

1.8 m

0.9 m

1(4.8) – 5.75(4.8) + F DE (1.8) + 2(2.4) = 0F DE = 10 kN (T) Ans

F DI ( )(7.2) – 2(4.8) – 2(2.4) = 0

F DI = 3.333 = 3.33 kN (C) Ans

2(2.4) + 2(4.8) + 1(7.2) – 5.75(7.2) – F JI 3

73

(7.2) = 0

F JI = 7.832 = 7.83 kN (C) Ans

3

5

2 kN

1 kN

-

5.75 kN

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A B

C

D

E

F G

H

I

J

1.2 m

1.2 m

1.2 m

1.2 m

1.2 m

2.4 m

125 kN150 kN

A B

C

D

E

F G

H

I

J

1.2 m

1.2 m

1.2 m

1.2 m

1.2 m

2.4 m

125 kN150 kN

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-

21

82

(15.1) + F DK – 4 = 0

FDK = 0.665 kN Ans

1.5 kN 3 kN 3 kN

3 m

2 m

3 m 3 m

9.5 kN

4 kN

15.1 kN

15.1 kN

F LK = 15.1 kN (C)

F LD = 0.5 kN (T) Ans

9.5 – 1.5 – 3 – 3 –8

145F LD –

1

82F LK = 0

9.5(6) – 1.5(6) – 3(3) – F CD(2.67) = 0F CD = 14.625 = 14.6 kN (T) Ans

9

145

F LD –9

82

F LK + 14.625 = 0

4 kN

K L

M N

A

B C D E F

G

3 kN

J

3 kN3 kN

3 kN

I

1.5 kN1.5 kN

H 3 m

2 m

6 @ 3 m 18 m

F DK 0.665 kN (C)

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B A

M

L K J I H

G

P ON

4 m 4 m 4 m 4 m 4 m 4 m

3 m

3 m

6 kN

C

7.5 kN

D E F

9 kN6(20) + 7.5(16) + 9(12) – A y(24) = 0

A y = 14.5 kN

F KJ (6) + 6(4) – 14.5(8) = 0F KJ = 15.333 kN (C) = 15.3 kN (C)

F CD(6) + 6(4) – 14.5(8) = 0F CD = 15.333 kN (T) = 15.3 kN (T)

+ 15.33 – 15.33 = 0

14.5 – 6 – 7.5 – F ND

= 1.6667

0.833 kN (C)0.833 kN (T)

12 m

9 kN

4 m4 m4 m

6 kN

7.5 kN

6 m

4 m 4 m

7.5 kN6 kN A y = 14.5 kN

7.5 kN

4 m4 m

A y = 14.5 kN 6 kN

F CD = 15.33 kN

F KJ = 15.33 kN

6 m

3 m

9 m 9 m 9 m 9 m 9 m

A

J I H G

EDC B

F K L M

12 kN 12 kN20 kN 20 kN

3 m

20 kN 20 kN 35.2 kN

9 m9 m

3 m

3 m

28.8 kN 12 kN 12 kN

9 m 9 m

3 m

3 m

–68.4(6) – 20(9) + 35.2(18) – F HL(6)3

10

0

F HL = 7.59 kN (C) Ans

F IH (6) + 12(9) – 28.8(18) 0; F IH 68.4 kN (C) Ans

F CD(6) – 28.8(18) + 12(9) 0; F CD 68.4 (T) Ans

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3 kN

4 m

2 m2 m

3.866*

4.243* (T)4.243* (T)

6.866* (T)

6.156* (C)3.740* (T)

5.026* (T)

3* (C)

6* (C)

3.866*

3*

3.740 + x(0.526) = 0

x = 7.109

= 6.866 + (7.109)(–1.41)

= –3.188 kN

= 3.2 kN (C)

In a similar manner:

F AB = 2.9 kN (C) Ans

F EB = 4.1 kN (T) Ans

F BC = 2.9 kN (C) Ans

F EF = 2.4 kN (C) Ans

F CF = 2.9 kN (T) Ans

F CD = 8.0 kN (C) Ans

F ED = 5.8 kN (C) Ans

F DA = 7.1 kN (T) Ans

D

E

B

C A

F

3 kN

2 m 2 m

4 m

3030

45 45

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K J

C

E

D

F G

H I

L

B A

2 m

4 m

2 m

2 m 4 m

10 kN

7.5 kN 10 kN

7.5 (C) 8.84 (T) 5.305 (T) 1.77 (C)1.77 (T)

2.5 (C)

2.5 kN

1.77 (C)

1.77 (C)8.84 (C)7.07 (C)

5 (T)

8.84 (C)7.07 (C)

5 (T)

2.5 (T)

7.5 (C)4 m

2 m

D y = 2.5 kN

2 m4 m

10 kN

2 m

A y = 7.5 kN

F KF = 7.5 + 1( x) = 0; x = – 7.5Thus: F BC = 5 kN (C) Ans

F AB = 0 Ans F FC = 3.54 kN (T) Ans

F AG = 7.5 kN (C) Ans F FH = 10.6 kN (T) Ans

F GB = 3.54 kN (T) Ans F KH = 3.54 kN (T) Ans

F GL = 2.5 kN (C) Ans F KJ = 7.5 kN (C) Ans

F GI = 3.54 kN (C) Ans F JH = 10.6 kN (T) AnsF LI = 3.54 kN (T) Ans F CD = 0 Ans

F LK = 2.5 kN (C) Ans F DE = 2.5 kN (C) Ans

F IK = 3.54 kN (C) Ans F CE = 3.54 kN (C) Ans

F IF = 3.54 kN (T) Ans F HE = 3.54 kN (T) Ans

F BF = 10.6 kN (T) Ans F JE = 7.5 kN (C) Ans

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z

P z = 3 kN

2 m

1 m

x

y

4 m

1.5 m

1 m

A

B

D

C

4 m

1

m

1. 5 m

1 m2 m

4 m

1 m

1. 5 m

0.2 kN

4 m

2 m

1. 5 m

0.05 kN

1 kN

1.8 kN

0.586 kN

4 m

1 m

C z(2.5) – 3(1.5) 0

C z 1.8 kN Ans

3(1) – Az(3) 0

A z 1 kN Ans

B z + 1 + 1.8 – 3 0

B z 0.2 kN Ans

B z 0 Ans

A y – Cy 0

A y(1.5) + C y(1) 0

A y 0 Ans

C y 0 Ans

8

77 F BH + 0.2 0

F BD 0.219 kN 0.22 kN (C)

13

77

3

13

0.219 – F BC 0

F BC 0.075 kN (T)

13

77

2

13

(0.219) – F BA 0

F BA 0.05 kN Ans

8

89F AD + 1 0

F AD 1.179 kN 1.18 kN (C) Ans

5

89

3

5

(1.179) – F AC sin 39.8° 0

F AC 0.586 kN 0.59 kN (T)

0.586 cos 39.81° –5

89

4

5

(1.179) + 0.05 0 Check

8

72F CD + 1.8 = 0

F CD = 1.91 kN (C) Ans

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y

x

1.2 m

1.2 m0.9 m

C

B

A

F

E

D

2.5k kN

z

4k kN

F CE (sin 45°) – 2.5 = 0; F CE = 3.536 kN = 3.54 kN (T) Ans

F CD – 3.536(cos 45°) = 0; F CD = 2.5 kN (C) Ans

F CB = 0 Ans

F BF (sin 45°) – 4 = 0; F BF = 5.657 kN = 5.66 kN (T) Ans

F BA – 5.657 (cos 45°) = 0; F BA = 4 kN (C) Ans

F ED – 3.536(sin 45°) = 0; F ED = 2.5 kN (C)

F EF = 0

3.536(cos 45°) – E y = 0; E y = 2.5 kN

F FA – 5.657(cos 45°) = 0; F FA = 4 kN (C)

5.657(sin 45°) – F y = 0; F y = 4 kN

D y – 2.5 = 0; D y = 2.5 kN

D z – 2.5 = 0; D z = 2.5 kN

2.5 kN

2.5 kN

4 kN

2.5 kN

= 5.657 kN

FCE = 3.536 kN

4 m

1 m

0.75 m

0.75 m

z

y x

AB

D

C

P y = 2 kN

P z = 3 kN

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A

1 m1 m

1 m1 m

1 m

1 m

1 m

10 kN10 kN

10 kN

D

I

G E

F

H

BC

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A

10 kN10 kN

10 kN

D

I

G1 m

1 m

1 m1 m

1 m

1 m

1 m

E

F

H

BC

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x y

z

C

A

B

E

D1.2 m

0.9 m 0.9 m

0.9 m2.4 m

8 kN

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3–1P. The Pratt roof trusses are uniformly spaced every 5 m. Thedeck, roofing material, and the purlins have an average weight of 0.28 kN/m2. The building is located in New York where theanticipated snow load is 1.0 kN/m2 and the anticipated ice load is0.4 kN/m2. These loadings occur over the horizontal projected areaof the roof. Determine the force in each member due to dead load,snow, and ice loads. Neglect the weight of the truss members andassume A is pinned and E is a roller.

1.8 m

1.8 m

2.4 m 2.4 m 2.4 m 2.4 m

21 kN

21 kN

10.5 kN

21 kN

10.5 kN

A y = 42 kN E y = 42 kN

2.4 m2.4 m2.4 m2.4 m

52.5 kN

21 kN

42 kN

10.5 kN

21 kN

35 kN

Loading:

At joints H , G and F :

F = (1.0 + 0.4)(5)(2.4) + 0.28(5)(3) = 21 kN

At joints A and E :

F = (1.0 + 0.4)(5)(1.2) + 0.28(5)(1.5) = 10.5 kNDue to symmetrical loading and geometry

A x = 0, A y = 42 kN, E y = 42 kN

42 – 10.5 – F AH 3

5

= 0

F AH = 52.5 kN (C) Ans

F AB – 52.5 4

5

= 0

F AB = 42 kN (T) Ans

F BC – 42 = 0

F BC = 42 kN (T)

–21 cos 36.87° + F HC cos 16.26° = 0

F HC = 17.5 kN (C) Ans

52.5 – F HG – 21 sin 36.87° – 17.5 sin 16.26° = 0

F HG = 35 kN (C) Ans

4

5(35) –

4

5F GF = 0

F GF = 35 kN (C) Ans

3

5(35) +

3

5(35) – 21 – F GC = 0

F GC = 21 kN (T) Ans

Due to symmetrical loading and geometryF DE = F AB = 42 kN (T) Ans

F DC = F BC = 42 kN (T) Ans

F EF = F AH = 52.5 kN (C) Ans

F BH = F DF = 0 Ans

F HC = F FC = 17.5 kN (C) Ans

2.4 m 2.4 m 2.4 m 2.4 m

1.8 m

1.8 m

AB

H

G

F

C D E

9789810682627_sm_03

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