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

Beam deflection at C: The beam deflection at C for this case is found from the elastic
curve equation [Equation (d)] for a simply supported beam with a concentrated moment
applied at one end. With the variables and values
and
M =−160 kN⋅m
x = 3m
L = 6m(i.e., thelength ofthe centerspan)
3 2
EI = 43.2 × 10 kN⋅m
the beam deflection at C is calculated from Equation (d):
v
C
Mx
=− − +
6LEI x 2 Lx L 2
( 3 2 )
( −160 kN⋅m)(3 m)
2 2
=−
[(3 m) − 3(6 m)(3 m) + 2(6 m)]
3 2
6(6 m)(43.2 × 10 kN⋅m) −3
= 8.333 × 10 m = 8.333 mm
Beam deflection at A: Use Equation (c) to compute the rotation angle at B:
ML (
θ =− =− − 160 kN ⋅ m)(6 m)
−3
B = 3.704 × 10 rad
3 2
6EI 6(43.2 × 10 kN⋅m)
The beam deflection at A is computed from the rotation angle θ B and the length of the
overhang:
= θ = × − 3 −3
v x (3.704 10 rad)( − 3 m) =− 11.112 × 10 m = −11.112 mm
A B AB
The following table summarizes the results of the method of superposition in this example:
Superposition Case
Case 1—Concentrated load on left overhang
v A
(mm)
−14.583
−29.167
v C
(mm)
v E
(mm)
10.938 −9.722
Case 2—Uniformly distributed load on center span 50.001 −31.250 33.334
Case 3—Uniformly distributed load on right overhang −11.112 8.333
−3.704
−14.814
total Beam Deflection −4.86 −11.98 5.09
Ans.
mecmovies
ExAmpLES
m10.11 Determine an expression for the deflection of the
beam at the midpoint of span BD. Assume that EI for the
beam is constant throughout all spans.
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