Timothy A. Philpot - Mechanics of materials _ an integrated learning system-John Wiley (2017)

ExAmpLE 10.8
For the beam shown, use discontinuity functions to compute
the deflection at D. Assume a constant value of EI =
192,000 kip ⋅ ft 2 for the beam.
Plan the Solution
Determine the reactions at the fixed support A. Using
Table 7.2, write w(x) expressions for the linearly distributed
load as well as the two support reactions. Integrate w(x)
four times to determine equations for the beam slope and
deflection. Use the boundary conditions known at the fixed
support to evaluate the constants of integration.
SolutioN
Support Reactions
An FBD of the beam is shown. On the basis of this FBD, the
beam reaction forces can be computed as follows:
1
Σ Fy
= Ay
− (6 kips/ft)(8 ft) = 0
2
∴ A = 24 kips
y
1
Σ M = −M
−
⎡
+
⎤
2 (6 kips/ft)(8 ft) ⎣⎢
4 ft 2(8 ft)
A A
⎦⎥ = 0
3
∴ M = −224 kip⋅ft
Discontinuity Expressions
A
yv
M A
6 kips/ft
A B C D
v
A y
4 ft 8 ft 4 ft
Distributed load between B and C: Use case 6 of Table 7.2 to write the following expression
for the distributed load:
6kips/ft
1 6kips/ft
1 0
wx ( ) =− x − 4ft + x − 12 ft + 6kips/ft x − 12 ft
8ft
8ft
Reaction forces A y and M A : The reaction forces at A are expressed with the use of cases 1
and 2 of Table 7.2:
−2 −1
wx ( ) =−224 kip⋅ft x − 0 ft + 24 kips x − 0ft
Integrate the beam load expression: The load expression w(x) for the beam is thus
−2 −1
wx ( ) =−224 kip⋅ft x − 0 ft + 24 kips x − 0ft
6kips/ft
6kips/ft
− x − 4ft + x − 12 ft + 6kips/ft x −12 ft
8ft
8ft
Integrate w(x) to obtain the shear-force function V(x):
∫
1 1 0
−1 0
V( x) = wx ( ) dx = −224 kip⋅ft x − 0 ft + 24 kips x − 0ft
6kips/ft
6kips/ft
− x − 4ft + x − 12 ft + 6kips/ft x −12 ft
2(8 ft)
2(8 ft)
2 2 1
Then integrate again to obtain the bending-moment function M(x):
∫
0 1
Mx ( ) = V( x) dx = −224 kip⋅ft x − 0 ft + 24 kips x − 0ft
6kips/ft
6kips/ft
6kips/ft
− x − 4ft + x − 12 ft + x −12 ft
6(8 ft)
6(8 ft)
2
3 3 2
6 kips/ft
A B C D
4 ft 8 ft 4 ft
x
x
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