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

The beam deflections at A, C, and E will be computed for each
case with the use of standard equations from Appendix C for both
the deflection and the slope. Cases 1 and 3 will require equations
for both simply supported and cantilever beams, whereas case 2
will require equations only for simply supported beams. After
completing the calculations for all three cases, the results will be
added to give the final deflections at the three locations.
SolutioN
For this beam, the elastic modulus is E = 200 GPa and the
moment of inertia is I = 216 × 10 6 mm 4 . Therefore,
6 4 12 2 3 2
EI = (200 GPa)(216 × 10 mm ) = 43.2 + 10 Nmm ⋅ = 43.2 × 10 kN⋅m
Case 1—Concentrated load on left overhang
Both simply supported and cantilever beam equations will be required to compute deflections
at A, but only simply supported beam equations will be necessary to compute the
beam deflections at C and E.
Beam deflection at A: Consider the cantilever beam deflection at
A of the 3 m long overhang. From Appendix C, the maximum
deflection of a cantilever beam with a concentrated load applied
at the tip is given as
v
max
3
PL
=−
3EI
(a)
A
v
B
C
D
80 kN/m
3 m 3 m 3 m 2 m
Case 3—uniformly distributed load on right overhang.
70 kN
A
v
B
C
3 m 3 m 3 m 2 m
D
E
E
x
x
v A
435
Equation (a) will be used to compute one portion of the beam deflection at A. We set
and
P = 70 kN
L = 3m
3 2
EI = 43.2 × 10 kN⋅m
The cantilever beam deflection at A will then be
v
A
3 3
PL (70 kN)(3 m)
−3
=− =−
=− 14.583 × 10 m = −14.583 mm
3 2
3EI 3(43.2 × 10 kN⋅m)
This calculation implicitly assumes that the beam is fixed to a rigid support at B. However,
the overhang is attached, not to a rigid support at B, but rather to a flexible beam that
rotates in response to the moment produced by the 70 kN load. The rotation of the overhang
at B must be accounted for in determining the deflection at A.
The moment at B due to the 70 kN load is M = (70 kN)(3 m) = 210 kN · m, and acts
counterclockwise as shown. The rotation angles at the ends of the span of a simply supported
beam subjected to a concentrated moment can be obtained from Appendix C:
ML
θ 1 =− (atthe end where M is applied)
(b)
3EI
ML
θ 2 =+ (opposite theend whereM
isapplied)
(c)
6EI
The rotation angle at B is required in order to obtain the deflection at A. The rotation angle
at D will be used later to calculate the deflection at E.