# Solutions (8th ed structural analysis) chapter 7

exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.

7–1. Determine (approximately) the force in each
member of the truss. Assume the diagonals can support
either a tensile or a compressive force.

50 kN

40 kN

20 kN

F

E

D

3m
A

C
B

Support Reactions. Referring to Fig. a,
a + a MA = 0;

Cy(6) - 40(3) - 20(6) = 0

Cy = 40 kN

a + a MC = 0;

40(3) + 50(6) - Ay(6) = 0

Ay = 70 kN

+
: a Fx = 0;

3m

Ax = 0

Method of Sections. It is required that FBF = FAE = F1. Referring to Fig. b,
+ c a Fy = 0;

70 - 50 - 2F1 sin 45° = 0

F1 = 14.14 kN

Therefore,
FBF = 14.1 kN (T)

FAE = 14.1 kN (C)

Ans.

a + a MA = 0;

FEF(3) - 14.14 cos 45°(3) = 0

FEF = 10.0 kN (C)

Ans.

a + a MF = 0;

FAB(3) - 14.14 cos 45°(3) = 0

FAB = 10.0 kN (T)

Ans.

Also, FBD = FCE = F2. Referring to Fig. c,
+ c a Fy = 0;

40 - 20 - 2F2 sin 45° = 0

F2 = 14.14 kN

Therefore,
FBD = 14.1 kN (T)

FCE = 14.1 kN (C)

Ans.

a + a MC = 0;

14.14 cos 45°(3) - FDE(3) = 0

FDE = 10.0 kN (C)

Ans.

a + a MD = 0;

14.14 cos 45°(3) - FBC(3) = 0

FBC = 10.0 kN (T)

Ans.

Method of Joints.
Joint A: Referring to Fig. d,
+ c a Fy = 0;

70 - 14.14 sin 45° - FAF = 0

FAF = 60.0 kN (C)

Ans.

Joint B: Referring to Fig. e,
+ c a Fy = 0; 14.14 sin 45° + 14.14 sin 45° - FBE = 0 FBE = 20.0 kN (C) Ans.
Joint C:
+ c a Fy = 0;

40 - 14.14 sin 45° - FCD = 0

FCD = 30.0 kN (C)

200

Ans.

3m

exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.

7–1.

Contiuned

7–2. Solve Prob. 7–1 assuming that the diagonals cannot
support a compressive force.

50 kN

40 kN

20 kN

F

E

D

3m
A

C
B
3m

3m

Support Reactions. Referring to Fig. a,
a + a MA = 0;

Cy(6) - 40(3) - 20(6) = 0

Cy = 40 kN

a + a MC = 0;

40(3) + 50(6) - Ay(6) = 0

Ay = 70 kN

+
: a Fx = 0;

Ax = 0

Method of Sections. It is required that
FAE = FCE = 0

Ans.

201

exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.

7–2.

Continued

Referring to Fig. b,
+ c a Fy = 0; 70 - 50 - FBF sin 45° = 0 FBF = 28.28 kN (T) = 28.3 kN (T)

Ans.

a + a MA = 0;

Ans.

a + a MF = 0

FEF(3) - 28.28 cos 45°(3) = 0
FAB(3) = 0

FEF = 20.0 kN (C)

FAB = 0

Ans.

Referring to Fig. c,
+ c a Fy = 0; 40 - 20 - FBD sin 45° = 0 FBD = 28.28 kN (T) = 28.3 kN (T)

Ans.

a + a MC = 0;

28.28 cos 45°(3) - FDE(3) = 0

Ans.

a + a MD = 0;

-FBC(3) = 0

FDE = 20.0 kN (C)

FBC = 0

Ans.

FAF = 70.0 kN (C)

Ans.

Method of Joints.
Joint A: Referring to Fig. d,
+ c a Fy = 0;

70 - FAF = 0

Joint B: Referring to Fig. e,
+ c a Fy = 0;

28.28 sin 45° + 28.28 sin 45° - FBE = 0
FBE = 40.0 kN (C)

Ans.

Joint C: Referring to Fig. f,
+ c a Fy = 0;

40 - FCD = 0

FCD = 40.0 kN (C)

Ans.

202

exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.

7–3. Determine (approximately) the force in each member
of the truss. Assume the diagonals can support either a tensile
or a compressive force.

10 k

10 k

H

10 k

G

10 k

F

E

20 ft
A

D
B
20 ft

VPanel = 8.33 k
Assume VPanel is carried equally by FHB and FAG, so
8.33
2
=
= 5.89 k (T)
cos 45°

FHB

Ans.

8.33
2
= 5.89 k (C)
=
cos 45°

FAG

Ans.

Joint A:
+
:
a Fx = 0;

FAB - 5 - 5.89 cos 45° = 0;

+ c a Fy = 0;

-FAH + 18.33 - 5.89 sin 45° = 0;

FAB = 9.17 k (T)

FAH = 14.16 k (C)

Ans.

Ans.

Joint H:
+
:
a Fx = 0;

-FHG + 5.89 cos 45° = 0;

FHG = 4.17 k (C)

Ans.

VPanel = 1.667 k
1.667
2
= 1.18 k (C)
=
cos 45°

FGC

FBF

Ans.

1.667
2
=
= 1.18 k (T)
cos 45°

Ans.

Joint G:
+
:
a Fx = 0;

4.17 + 5.89 cos 45° - 1.18 cos 45° - FGF = 0
FGF = 7.5 k (C)

+ c a Fy = 0;

Ans.

-10 + FGB + 5.89 sin 45° + 1.18 sin 45° = 0
FGB = 5.0 k (C)

Ans.

203

C
20 ft

20 ft

5k

exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.

7–3.

Continued

Joint B:
+
:
a Fx = 0;

FBC + 1.18 cos 45° - 9.17 - 5.89 cos 45° = 0

FBC = 12.5 k (T)

Ans.

VPanel = 21.667 - 10 = 11.667 k

FEC

11.667
2
=
= 8.25 k (T)
cos 45°

FDF

Ans.

11.567
2
=
= 8.25 k (C)
cos 45°

Ans.

Joint D:
+
:
a Fx = 0;

FCD = 8.25 cos 45° = 5.83 k (T)

+ c a Fy = 0;

21.667 - 8.25 sin 45° - FED = 0

Ans.

FED = 15.83 k (C)

Ans.

Joint E:
+
:
a Fx = 0;

5 + FFE - 8.25 cos 45° = 0

FFE = 0.833 k (C)

Ans.

Joint C:
+ c a Fy = 0;

-FFC + 8.25 sin 45° - 1.18 sin 45° = 0

FFC = 5.0 k (C)

Ans.

*7–4. Solve Prob. 7–3 assuming that the diagonals cannot
support a compressive force.

10 k

10 k

H

10 k

G

10 k

F

E

20 ft
A

D
B

VPanel = 8.33 k

20 ft

FAG = 0
FHB

Ans.

8.33
=
= 11.785 = 11.8 k
sin 45°

Ans.

Joint A:
+
:
a Fx = 0; FAB = 5 k (T)

Ans.

204

C
20 ft

20 ft

5k

exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.

7–4.

Continued

+ c a Fy = 0; FAN = 18.3 k (C)

Ans.

Joint H:
+
:
a Fx = 0;

11.785 cos 45° - FHG = 0

FHG = 8.33 k (C)

Ans.

VPanel = 1.667 k
FGC = 0
FBF =

Ans.

1.667
= 2.36 k (T)
sin 45°

Ans.

Joint B:
+
:
a Fx = 0;

FBC + 2.36 cos 45° - 11.785 cos 45° - 5 = 0

FBC = 11.7 k (T)
+ c a Fy = 0;

Ans.

- FGB + 11.785 sin 45° + 2.36 sin 45° = 0

FGB = 10 k (C)

Ans.

Joint G:
+
:
a Fx = 0;

FGF = 8.33 k (C)

Ans.

VPanel = 11.667 k
FDF = 0
FEC =

Ans.

11.667
= 16.5 k (T)
sin 45°

Ans.

FCD = 0

Ans.

Joint D:
+
:
a Fx = 0;
+ c a Fy = 0;

FED = 21.7 k (C)

Ans.

Joint E:
+
:
a Fx = 0;

FEF + 5 - 16.5 cos 45° = 0

FEF = 6.67 k (C)

Ans.

Joint F:
+ c a Fy = 0; FFC - 10 - 2.36 sin 45° = 0
FFC = 11.7 k (C)

Ans.

205

exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.

7–5. Determine (approximately) the force in each member
of the truss. Assume the diagonals can support either a tensile
or a compressive force.

7k

14 k

H

14 k

G

7k

F

E

6 ft
A

D
B
8 ft

8 ft

Support Reactions. Referring to, Fig. a
+
:
a Fx = 0;

Ax - 2 = 0 Ax = 2 k

a + a MA = 0; Dy(24) + 2(6) - 7(24) - 14(16) - 14(8) = 0 Dy = 20.5 k
a + a MD = 0; 14(8) + 14(16) + 7(24) + 2(6) - Ay(24) = 0 Ay = 21.5 k
Method of Sections. It is required that FBH = FAG = F1. Referring to Fig. b,
3
+ c a Fy = 0; 21.5 - 7 - 2F1 a b = 0 F1 = 12.08 k
5
Therefore,
FBH = 12.1 k (T) FAG = 12.1 k (C)

Ans.

4
a + a MH = 0; FAB(6) + 2(6) - 12.08a b (6) = 0 FAB = 7.667 k (T) = 7.67 k (T) Ans.
5
4
a + a MA = 0; FGH(6) - 12.08a b(6) = 0 FGH = 9.667 k (C) = 9.67 k (C)
Ans.
5
It is required that FCG = FBF = F2.

Referring to Fig. c,

3
+ c a Fy = 0; 21.5 - 7 - 14 - 2F2 a b = 0 F2 = 0.4167 k
5

206

C
8 ft

2k

exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.

7–5.

Continued

Therefore,
FCG = 0.417 k (T) FBF = 0.417 k (C)

Ans.

4
a + a MB = 0; FFG(6) - 0.4167 a b (6) + 7(8) - 21.5(8) = 0
5
FFG = 19.67 k (C) = 19.7 k (C)

Ans.

4
a + a MG = 0; FBC(6) + 7(8) + 2(6) - 21.5(8) - 0.4167 a b(6) = 0
5
FBC = 17.67 k (T) = 17.7 k (T)

Ans.

It is required that FCE = FDF = F3. Referring to Fig. d
3
+ c a Fy = 0; 20.5 - 7 - 2F3 a b = 0 F3 = 11.25 k
5
Therefore,
FCE = 11.25 k (T) FDF = 11.25 k (C)

Ans.

4
a + a MD = 0; 2(6) + 11.25 a b(6) - FEF(6) = 0 FEF = 11.0 k (C)
5

Ans.

a + a ME = 0; 11.25(0.8)(6) - FCD(6) = 0 FCD = 9.00 k (T)

Ans.

Method of Joints.
Joint A: Referring to Fig. e,
3
+ c a Fy = 0; 21.5 - 12.08 a b - FAH = 0 FAH = 14.25 k (C)
5
Referring to Fig. f,
3
3
+ c a Fy = 0; 12.08 a b - 0.4167 a b - FBG = 0 FBG = 7.00 k (C)
5
5

Ans.

Joint B:

Joint C:

Referring Fig. g,
3
3
+ c a Fy = 0; 11.25 a b + 0.4167 a b - FCF = 0 FCF = 7.00 k (C)
5
5

207

Ans.

exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.

7–5.

Continued

Joint D:

Referring to Fig. h,

3
+ c a Fy = 0; 20.5 - 11.25a b - FDE = 0
5
FDE = 13.75 k

Ans.

7–6. Solve Prob. 7–5 assuming that the diagonals cannot
support a compressive force.

7k

14 k

H

14 k

G

7k

F

E

6 ft
A

D
B
8 ft

Support Reactions. Referring to Fig. a,
+
:
a Fx = 0;

Ax - 2 = 0

Ax = 2 k

a + a MA = 0;

Dy(24) + 2(6) - 7(24) - 14(16) - 14(8) = 0

a + a MD = 0;

14(8) + 14(16) + 7(24) + 2(6) - Ay(24) = 0

Dy = 20.5 k
Ay = 21.5 k

Method of Sections. It is required that
FAG = FBF = FDF = 0

Ans.

Referring to Fig. b,
+ c a Fy = 0;

3
21.5 - 7 - FBH a b = 0
5

FBH = 24.17 k (T) = 24.2 k (T) Ans.

4
a + a MA = 0; FGH(6) - 24.17a b(6) = 0 FGH = 19.33 k (C) = 19.3 k (C)
5
Ans.
a + a MH = 0; 2(6) - FAB(6) = 0 FAB = 2.00 k (C)

Ans.
208

C
8 ft

8 ft

2k

exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.

7–6.

Continued

Referring to Fig. c
3
+ c a Fy = 0; 21.5 - 7 - 14 - FCG a b = 0 FCG = 0.8333 k (T) = 0.833 k (T)
5

Ans.

4
a + a MB = 0; FFG(6) + 7(8) - 21.5(8) - 0.8333a b(6) = 0
5
FFG = 20.0 k (C)

Ans.

a + a MG = 0; FBC(6) + 7(8) + 2(6) - 21.5(8) = 0 FBC = 17.33 k (T) = 17.3 k (T) Ans.
Referring to Fig. d,
3
+ c a Fy = 0; 20.5 - 7 - FCE a b = 0 FCE = 22.5 k (T)
5

Ans.

4
a + a MD = 0; 2(6) + 22.5 a b(6) - FEF(6) = 0 FEF = 20.0 kN (C)
5

Ans.

a + a ME = 0;

Ans.

-FCD(6) = 0 FCD = 0

Method of Joints.
Joint A: Referring to Fig. e,
+ c a Fy = 0; 21.5 - FAH = 0 FAH = 21.5 k (C)

Ans.

Joint B: Referring to Fig. f,
3
+ c a Fy = 0; 24.17 a b - FBG = 0 FBG = 14.5 k (C)
5

Ans.

Joint C: Referring to Fig. g,
3
3
+ c a Fy = 0; 0.8333 a b + 22.5 a b - FCF = 0 FCF = 14.0 k (C)
5
5
Joint D: Referring to Fig. h,
+ c a Fy = 0; 20.5 - FDE = 0 FDE = 20.5 k (C)

Ans.

Ans.

209

exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.

7–6.

Continued

210

exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.

7–7. Determine (approximately) the force in each member of the truss.
Assume the diagonals can support either a tensile or compressive force.

2m
F

2m
E

D

1.5 m

A

8 kN

Assume FBD = FEC
+ c a Fy = 0; 2FEC a

1.5
b - 4 = 0
2.5

FEC = 3.333 kN = 3.33 kN (T)

Ans.

FBD = 3.333 kN = 3.33 kN (C)

Ans.

a + a MC = 0; FED(1.5) - a

2
b(3.333)(1.5) = 0
2.5

FED = 2.67 kN (T)

Ans.

+
:
a Fx = 0; FBC = 2.67 kN (C)

Ans.

Joint C:
+ c a Fy = 0; FCD + 3.333a

1.5
b - 4 = 0
2.5

FCD = 2.00 kN (T)

Ans.

Assume FFB = FAE
+ c a Fy = 0; 2FFB a

B

1.5
b - 8 - 4 = 0
2.5

FFB = 10.0 kN (T)

Ans.

FAE = 10.0 kN (C)

Ans.

a + a MB = 0; FFE(1.5) - 10.0 a

2
b (1.5) - 4(2) = 0
2.5

FFE = 13.3 kN (T)

Ans.

+
:
a Fx = 0; FAB = 13.3 kN (C)

Ans.

Joint B:
+ c a Fy = 0; FBE + 10.0 a

1.5
1.5
b - 3.333 a
b - 8 = 0
2.5
2.5

FBE = 4.00 kN (T)

Ans.

Joint A:
+ c a Fy = 0; FAF - 10.0 a

1.5
b = 0
2.5

FAF = 6.00 kN (T)

Ans.

211

C
4 kN

exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.

*7–8. Solve Prob. 7–7 assuming that the diagonals cannot
support a compressive force.

2m
F

2m
E

D

1.5 m

A

B
8 kN

FBD = 0

Assume

+ c a Fy = 0; FEC a

Ans.

1.5
b - 4 = 0
2.5

FEC = 6.667 kN = 6.67 kN (T)

Ans.

a + a MC = 0; FED = 0

Ans.

2
+
:
a Fx = 0; FBC - 6.667 a 2.5 b = 0
FBC = 5.33 kN (C)

Ans.

Joint D:
From Inspection:
Assume

FCD = 0

Ans.

FAE = 0

Ans.

+ c a Fy = 0; FFB a

1.5
b - 8 - 4 = 0
2.5

FFB = 20.0 kN (T)

Ans.

a + a MB = 0; FFE(1.5) - 4(2) = 0
FFE = 5.333 kN = 5.33 kN (T)

Ans.

2
+
:
a Fx = 0; FAB - 5.333 - 20.0 a 2.5 b = 0
FAB = 21.3 kN (C)

Ans.

Joint B:
+ c a Fy = 0;

- FBE - 8 + 20.0 a

1.5
b = 0
2.5

FBE = 4.00 kN (T)

Ans.

Joint A:
+ c a Fy = 0; FAF = 0

Ans.

212

C
4 kN

exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.

7–9. Determine (approximately) the force in each member
of the truss. Assume the diagonals can support both tensile
and compressive forces.

E
1.5 k

15 ft
F

2k

D
15 ft

Method of Sections. It is required that FCF = FDG = F1. Referring to Fig. a,
G

2k

C

+
:
a Fx = 0; 2F1 sin 45° - 2 - 1.5 = 0 F1 = 2.475 k

15 ft

Therefore,
FCF = 2.48 k (T) FDG = 2.48 k (C)

Ans.

a + a MD = 0; 1.5(15) + 2.475 cos 45° (15) - FFG(15) = 0
FFG = 3.25 k (C)

15 ft

Ans.

a + a MF = 0; 1.5(15) + 2.475 cos 45° (15) - FCD(15) = 0
FCD = 3.25 k (T)

Ans.

It is required that FBG = FAC = F2. Referring to Fig. b,
+
:
a Fx = 0; 2F2 sin 45° - 2 - 2 - 1.5 = 0 F2 = 3.889 k
Therefore,
FBG = 3.89 k (T) FAC = 3.89 k (C)

Ans.

a + a MG = 0; 1.5(30) + 2(15) + 3.889 cos 45° (15) - FBC(15) = 0
FBC = 7.75 k (T)

Ans.

a + a MC = 0; 1.5(30) + 2(15) + 3.889 cos 45° (15) - FAG(15) = 0
FAG = 7.75 k (C)

Ans.

Method of Joints.
Joint E: Referring to Fig. c,
+
:
a Fx = 0; FEF cos 45° - 1.5 = 0 FEF = 2.121 k (C) = 2.12 k (C)

Ans.

+ c a Fy = 0; 2.121 sin 45° - FDE = 0 FDE = 1.50 k (T)

Ans.

Joint F: Referring to Fig. d,
+
:
a Fx = 0; 2.475 sin 45° - 2.121 cos 45° - FDF = 0
FDF = 0.250 k (C)

Ans.

213

B

A

exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.

7–9.

Continued

Joint G: Referring to Fig. e,
+
:
a Fx = 0; 3.889 sin 45° - 2.475 cos 45° - FCG = 0
FCG = 1.00 k (C)

Ans.

Joint A: Referring to Fig. f,
+
:
a Fx = 0; FAB - 3.889 cos 45° = 0
FAB = 2.75 k

Ans.

214

exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.

7–10. Determine (approximately) the force in each member
of the truss. Assume the diagonals DG and AC cannot
support a compressive force.

E
1.5 k

15 ft
F

2k

D
15 ft

Method of Sections. It is required that

G

FDG = FAC = 0

2k

C

Ans.

15 ft

Referring to Fig. a,
+
:
a Fx = 0; FCF sin 45° - 1.5 - 2 = 0 FCF = 4.950 k (T) = 4.95 k (T) Ans.
a + a MF = 0; 1.5(15) - FCD(15) = 0 FCD = 1.50 k (T)

Ans.

a + a MD = 0; 1.5(15) + 4.950 cos 45°(15) - FFG(15) = 0
FFG = 5.00 k (C)

Ans.

Referring to Fig. b,
+
:
a Fx = 0; FBG sin 45° - 2 - 2 - 1.5 = 0 FBG = 7.778 k (T) = 7.78 k (T)
Ans.
a + a MG = 0; 1.5(30) + 2(15) - FBC(15) = 0 FBC = 5.00 k (T)

Ans.

a + a MC = 0; 1.5(30) + 2(15) + 7.778 cos 45° - FAG(15) = 0
FAG = 10.5 k (C)

Ans.

Method of Joints.
Joint E: Referring to Fig. c,
+
:
a Fx = 0; FEF cos 45° - 1.5 = 0 FEF = 2.121 k (C) = 2.12 k (C)

Ans.

+ c a Fy = 0; 2.121 sin 45° - FDE = 0 FDE = 1.50 k (T)

Ans.

Joint F: Referring to Fig. d,
+
:
a Fx = 0; 4.950 sin 45° - 2.121 cos 45° - FDF = 0 FDF = 2.00 k (C) Ans.
Joint G: Referring to Fig. e,
+
:
a Fx = 0; 7.778 sin 45° - FCG = 0 FCG = 5.50 k (C)

Ans.

Joint A: Referring to Fig. f,
+
:
a Fx = 0; FAB = 0

Ans.

215

B

A
15 ft

exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.

7–10.

Continued

216

exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.

7–11. Determine (approximately) the force in each
member of the truss. Assume the diagonals can support
either a tensile or compressive force.

1.5 m
E

8 kN

D

2m

10 kN

Method of Sections. It is required that FCE = FDF = F1. Referring to Fig. a,

C
F

3
+
:
a Fx = 0; 8 - 2F1 a 5 b = 0 F1 = 6.667 kN
2m

Therefore,
FCE = 6.67 kN (C) FDF = 6.67 kN (T)

Ans.

4
a + a ME = 0; FCD(1.5) - 6.667 a b (1.5) = 0 FCD = 5.333 kN (C) = 5.33 kN (C)
5
Ans.
4
a + a MD = 0; FEF(1.5) - 6.667 a b(1.5) = 0 FEF = 5.333 kN (T) = 5.33 kN (T)
5
Ans.
It is required that FBF = FAC = F2 Referring to Fig. b,
3
+
:
a Fx = 0; 8 + 10 - 2F2 a 5 b = 0 F2 = 15.0 kN
Therefore,
FBF = 15.0 kN (C) FAC = 15.0 kN (T)

Ans.

4
a + a MF = 0; FBC(1.5) - 15.0 a b (1.5) - 8(2) = 0
5
FBC = 22.67 kN (C) = 22.7 kN (C)

Ans.

4
a + a MC = 0; FAF(1.5) - 15.0 a b(1.5) - 8(2) = 0
5
FAF = 22.67 kN (T) = 22.7 kN (T)

Ans.

Method of Joints.
Joint D: Referring to Fig. c,
3
+
:
a Fx = 0; FDE - 6.667 a 5 b = 0 FDE = 4.00 kN (C)

Ans.

Joint C: Referring to Fig. d,
3
3
+
:
a Fx = 0; FCF + 6.667a 5 b - 15.0 a 5 b = 0 FCF = 5.00 kN (C)

Ans.

Joint B: Referring to Fig. e,
3
+
:
a Fx = 0; 15.0 a 5 b - FAB = 9.00 kN (T)

Ans.

217

A

B

exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.

7–11.

Continued

218

exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.

*7–12. Determine (approximately) the force in each
member of the truss. Assume the diagonals cannot support
a compressive force.

1.5 m
E

8 kN

D

2m

10 kN

Method of Sections. It is required that

C
F

FCE = FBF = 0

Ans.

Referring to Fig. a,
2m

3
+
:
a Fx = 0; 8 - FDF a 5 b = 0 FDF = 13.33 kN (T) = 13.3 kN (T)

Ans.

4
a + a ME = 0; FCD(1.5) - 13.33 a b (1.5) = 0 FCD = 10.67 kN (C) = 10.7 kN (C)
5
Ans.
a + a MD = 0; FEF(1.5) = 0 FEF = 0

Ans.

Referring to Fig. b,
3
+
:
a Fx = 0; 8 + 10 - FAC a 5 b = 0 FAC = 30.0 kN (T)

Ans.

a + a MC = 0; FAF(1.5) - 8(2) = 0 FAF = 10.67 kN (T)

Ans.

4
a + a MF = 0; FBC(1.5) - 30.0 a b (1.5) - 8(2) = 0
5
FBC = 34.67 kN (C) = 34.7 kN (C)

Ans.

Method of Joints.
Joint E: Referring to Fig. c,
+
:
a Fx = 0; 8 - FDE = 0 FDE = 8.00 kN (C)

Ans.

Joint C: Referring to Fig. d,
3
+
:
a Fx = 0; FCF - 30.0 a 5 b = 0 FCF = 18.0 kN (C)

Ans.

Joint B: Referring to Fig. e,
+
:
a Fx = 0; FAB = 0

Ans.

219

A

B

exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.

7–12.

Continued

220

exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.

7–13. Determine (approximately) the internal moments
at joints A and B of the frame.

3 kN/m

E

F

G

H

6m
A

The frame can be simplified to that shown in Fig. a, referring to Fig. b,
a + a MA = 0; MA - 7.2(0.6) - 3(0.6)(0.3) = 0 MA = 4.86 kN # m

6m

Ans.

Referring to Fig. c,
a + a MB = 0; MB - 9.6(0.8) - 3(1.4)(0.1) + 7.2(0.6) = 0
MB = 3.78 kN # m

Ans.

221

B

C
8m

D
6m

exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.

7–14. Determine (approximately) the internal moments
at joints F and D of the frame.

400 lb/ft
F

A

a + a MF = 0; MF - 0.6(0.75) - 2.4(1.5) = 0

a + a MD = 0;

B
15 ft

MF = 4.05 k # ft

Ans.
- MD + 0.8(1) + 3.2(2) = 0

MD = 7.20 k # ft

Ans.

222

D

E

C
20 ft

exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.

7–15. Determine (approximately) the internal moment at

5 kN/m
E

D

9 kN/m
C

F

The frame can be simplified to that shown in Fig. a, Referring to Fig. b,
a + a MA = 0; MA - 5(0.8)(0.4) - 16(0.8) - 9(0.8)(0.4) - 28.8(0.8) = 0
MA = 40.32 kN # m = 40.3 kN # m

Ans.
A

B

8m

223

exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.

*7–16. Determine (approximately) the internal moments

3 kN/m
F

G
5 kN/m

L

K

5 kN/m

The frame can be simplified to that shown in Fig. a. The reactions of the
3 kN/m and 5 kN/m uniform distributed loads are shown in Fig. b and c
respectively. Referring to Fig. d,

H

I

J

E

a + a MA = 0; MA - 3(0.8)(0.4) - 9.6(0.8) - 5(0.8)(0.4) - 16(0.8) = 0

A

B

C

D

MA = 23.04 kN # m = 23.0 kN # m

Ans.

Referring to Fig. e,
a + a MB = 0; 9.60(0.8) - 9.60(0.8) + 5(0.8)(0.4) + 16(0.8) - MB = 0
MB = 14.4 kN # m

Ans.

224

8m

8m

8m

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