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

The beam deflection at E is computed from the rotation angle θ D and the overhang length:
= θ = − × − 3 −3
v x ( 4.861 10 rad)(2 m) =− 9.722 × 10 m = −9.722 mm
E D DE
Case 2—uniformly Distributed load on Center Span
For the uniformly distributed load acting on the center span, equations for the maximum
deflection acting at midspan and the slopes at the ends of the span will be required.
Beam deflection at A: Since the uniformly distributed load acts only between the supports,
there is no bending moment in the overhang spans. To compute the deflection at A,
begin by computing the slope at the end of the simple span. From Appendix C, the rotation
angles at the ends of the span are given by
To compute the rotation angle at B, let
and
θ
=− θ =− wL 24EI
1 2
w = 80 kN/m
L = 6m
3 2
EI = 43.2 × 10 kN⋅m
3
v A
A
(e)
v
80 kN/m
v C
B C D E
θ B
θD
3 m 3 m 3 m 2 m
v E
x
and compute θ B from Equation (e):
3 3
wL (80 kN/m)(6 m)
−3
θ B =− =−
=− 16.667 × 10 rad
3 2
24EI 24(43.2 × 10 kN⋅m)
The beam deflection at A is computed from the rotation angle θ B and the overhang
length:
= θ = − × − 3 −3
v x ( 16.667 10 rad)( − 3 m) = 50.001 × 10 m = 50.001 mm
A B AB
Beam deflection at C: The equation for the midspan deflection of a simply supported
beam subjected to a uniformly distributed load can be obtained from Appendix C:
v
max
4
5wL
=−
384EI
(f)
From Equation (f), the deflection at C for case 2 is
v
C
4 4
5wL
5(80 kN/m)(6 m)
=− =−
3 2
384EI
384(43.2 × 10 kN⋅m)
−3
=− 31.250 × 10 m = −31.250 mm
Beam deflection at E: The rotation angle at D is calculated from Equation (e):
wL
θ D = =
24EI
(80 kN/m)(6 m)
3 2
24(43.2 × 10 kN⋅m)
3 3
−3
= 16.667 × 10 rad
The beam deflection at E is computed from the rotation angle θ D and the overhang
length:
= θ = × − 3 −3
v x (16.667 10 rad)(2 m) = 33.334 × 10 m = 33.334 mm
E D DE
437