Problem #1
Homework #9 Solution
Problem #2
NOTE: Moment
shown here is
NOT consistent with
positive sign
convention.
Problem #3
NOTE: Moment
shown here is
NOT consistent with
positive sign
convention.
Influence line for moment at A, as shown here, is NOT c
Problem # 1
Homework # 5 Solution
D = 0
FCB(6) - 5(6)(3) = 0
FCB = 15 kN
Fy = 0
FCB - 5(6) + Dy = 0
Dy = 15 kN
A = 0
FCB(6) + 30 + MA = 0
MA = -120 kN-m
Fy = 0
Ay - FCB = 0
Ay = 15 kN
Internal Force Diagrams for all problems in this assignment
were genera
Problem # 1
Homework #4 Solution
kips
Fx = 0
F-8=0
F
F = 8 kips
kip-ft
NOTE: To correctly interpret the
sign of the shear and moment, the
positive sign convention must be
shown.
Problem # 2
kN
kN-m
NOTE: In this method where
internal forces are written as
Problem # 1
Homework # 3 Solution
b + r = 22
2n = 20
b + r > 2n
D = b + r - 2n = 22 - 20 = 2
17 - 16=1
18 - 18 = 0
17 - 18 = - 1
10 - 10 = 0
12 - 12 = 0
10 - 10 = 0
If load is applied at center pin, reactions are concurrent.
As an example, if a vertical
l
Problem # 1
Homework # 2 Solution
= -1. 75 kN = 1.75 kN
Force on hinge at C
is compressive and
equal to sqrt(8^2
+40^2) = 40.8 kips
which is less than
force capacity in
compression (45
kN). Thus, hinge
has sufficient
capacity.
Problem # 2
Cy = -0.27 kips
Name _
CIVL 2670
Exam No. 2
Fall 2010
Write legibly so that you receive the maximum credit you deserve.
All equilibrium equations must be accompanied by a carefully drawn free-body diagram. Zero
points will be given if carefully drawn free-body diagrams a
Name _
CIVL 2670
Exam No. 1
Fall 2010
Write legibly so that you receive the maximum credit you deserve.
All equilibrium equations must be accompanied by a carefully drawn free-body diagram. Zero
points will be given if carefully drawn free-body diagrams a