ch01-03 stress & strain & properties
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01 Solutions 46060 5/6/10 2:43 PM Page 3<br />
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•1–5. Determine the resultant internal loadings in the<br />
beam at cross sections through points D and E. Point E is<br />
just to the right of the 3-kip load.<br />
1.5 kip/ ft<br />
3 kip<br />
A<br />
D<br />
B<br />
E<br />
C<br />
6 ft 6 ft 4 ft<br />
4 ft<br />
Support Reactions: For member AB<br />
a+ ©M B = 0; 9.00(4) - A y (12) = 0 A y = 3.00 kip<br />
: + ©F x = 0; B x = 0<br />
+ c ©F y = 0; B y + 3.00 - 9.00 = 0 B y = 6.00 kip<br />
Equations of Equilibrium: For point D<br />
: + ©F x = 0; N D = 0<br />
Ans.<br />
+ c ©F y = 0; 3.00 - 2.25 - V D = 0<br />
V D = 0.750 kip<br />
Ans.<br />
a +©M D = 0; M D + 2.25(2) - 3.00(6) = 0<br />
M D = 13.5 kip # ft<br />
Ans.<br />
Equations of Equilibrium: For point E<br />
: + ©F x = 0; N E = 0<br />
Ans.<br />
+ c ©F y = 0; -6.00 - 3 - V E = 0<br />
V E = -9.00 kip<br />
Ans.<br />
a +©M E = 0; M E + 6.00(4) = 0<br />
M E = -24.0 kip # ft<br />
Ans.<br />
Negative signs indicate that M E<br />
and V E<br />
act in the opposite direction to that shown<br />
on FBD.<br />
3
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