6 3 5
© 2016 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This material is protected under all 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.
7–25.
SOLUTION
a
Ans.
Ans.
a
Ans.
Ans.
Ans.
Ans.ME = - 24.0 kip # ft
©ME = 0; ME + 6 (4) = 0
VE = -9 kip
+ c ©Fy = 0; -VE - 3 - 6 = 0
:+ ©Fx = 0; NE = 0
MD = 13.5 kip # ft
©MD = 0; MD +
1
2 (0.75) (6) (2) - 3 (6) = 0
VD = 0.75 kip
+ c ©Fy = 0; 3 -
1
2 (0.75)(6) - VD = 0
:+ ©Fx = 0; ND = 0
By = 6 kip
+ c ©Fy = 0; By + 3 -
1
2 (1.5)(12) = 0
:+ ©Fx = 0; Bx = 0
Ay = 3 kip
+ ©MB = 0;
1
2 (1.5)(12)(4) -Ay (12) = 0
Determine the normal force, shear force, and moment in
the beam at sections passing through points D and E. Point
E is just to the right of the 3-kip load.
6 ft 4 ft
A
4 ft
B CD E
6 ft
3 kip
1.5 kip/ft
Ans:
ND = 0
VD = 0.75 kip
MD = 13.5 kip # ft
NE = 0
VE = - 9 kip
ME = - 24.0 kip # ft
6 3 6
© 2016 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This material is protected under all 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.
7–26.
SOLUTION
Free body Diagram: The support reactions at A need not be computed.
Internal Forces: Applying equations of equilibrium to segment BC, we have
Ans.
Ans.
a
Ans.MC = - 302 kN # m
- 24.011.52 - 12.0142 - 40 sin 60°16.32 - MC = 0+ ©MC = 0;
VC = 70.6 kN
VC - 24.0 - 12.0 - 40 sin 60° = 0+ c ©Fy = 0;
- 40 cos 60° NC = 0 NC = -20.0 kN:+ ©Fx = 0;
Determine the internal normal force, shear force, and
bending moment at point C.
A
3 m 3 m
0.3 m
C
B
8 kN/ m
40 kN
3 m
60°
-
Ans:
NC = - 20.0 kN
VC = 70.6 kN
MC = - 302 kN # m
jbar
0
