CIV100
Solutions
Section B/E
Fall 2011
Assignment 7
Due on Monday, November 7
th
2011
Draw a FBD if necessary; Pay attention to significant figures; A ruler is always required.
1 Given forces at point F, E and D are 45 kN
a) Find the forces in members AB, AC, AD and
DB.
b) What would be the forces in these members if
those given forces were increased to 90 kN?
Solution
:
(a)
First find the reaction components at A and B:
Σ
M
A
=0
Bcos45(2) + 45(2) + 45(4) + 45(6) = 0
Æ
B =381.84 kN
Σ
M
B
=0
Ax(2) + 45(2) + 45(4) + 45(6) = 0
Æ
Ax =270 kN
+
↑
Σ
Fy=0
Bsin45 +Ay 3(45)
= 0
Æ
Ay = 135 kN (Direction is opposite to the one shown)
Second, find the forces in the members either by using the method of section or the method
of joints.
Note the following:
 The figure for joints A and B assumes all forces in members to be tension. If the force turns out
a negative magnitude, this implies that the force is compression.
 In this case the method of joints is easier than the method of sections
 F
AB
and F
BA
has same magnitude
 It is easier to start with joint B as it has only 2 unknowns while joint A has 3 unknowns
Take joint B:
+
→
Σ
Fx=0
Bsin45
º
+F
BD
= 0
Æ
F
BD
=270 kN (compression)
+
↑
Σ
Fy=0
Bcos45
º
+F
BA
= 0
Æ
F
BA
=270 kN (compression)
Take joint A:
+
↑
Σ
Fy=0
Ay  F
AB
 F
AD
cos45
º
= 0
Æ
F
AD
=190.9N (tension)
Note that I have indicated the correct direction of AY and thus have used a + value.
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