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

Beam deflection at C: The beam deflection at C will be equal to the beam deflection at B
plus an additional deflection caused by the slope of the beam between B and C. The rotation
angle of the beam at B is given by Equation (b), using the same variables as before:
3
θ =− =− − 3
wL ( 80kN/m)(3 m)
B
3 2
6EI 6(130 × 10 kN⋅m)
−3
= 2.769 × 10 rad
The deflection at C is computed from v B , θ B , and the length of the beam between
B and C:
−3 −3
v = v + θ (2 m) = (6.231 × 10 m) + (2.769 × 10 rad)(2 m)
C B B
−3
= 11.769 × 10 m = 11.769 mm
The positive value indicates an upward deflection.
v
v
A
3 m
B
2 m
C
x
150 kN.m
Case 2—Cantilever with Concentrated Moment
From the beam table in Appendix C, the elastic curve equation for
a cantilever beam subjected to a concentrated moment applied at
its free end is given as
2
Mx
v =− (c)
2EI
A
3 m
B
v B
2 m
C
v C
x
150 kN.m
Beam deflection at B: The elastic curve equation will be used to
compute the beam deflections at both B and C for this case. For
this beam,
M =−150 kN⋅m
3 2
EI = 130 × 10 kN⋅m
Note: The concentrated moment M is negative because it acts in
the direction opposite that shown in the beam table.
Substitute the foregoing values into Equation (c), using x = 3 m
to compute the beam deflection at B:
v
B
2
=− =− − ⋅
2
Mx ( 150 kN m)(3 m)
3 2
2EI 2(130 × 10 kN⋅m)
−3
= 5.192 × 10 m = 5.192 m
Beam deflection at C: Substitute the same values into Equation (c),
using x = 5 m to compute the beam deflection at C:
v
C
2
=− =− − ⋅
2
Mx ( 150 kN m)(5 m)
3 2
2EI 2(130 × 10 kN⋅m)
−3
= 14.423 × 10 m = 14.423 mm
v
A
80 kN/m
3 m
B
v B
2 m
C
v C
x
150 kN.m
Combine the two Cases
The deflections at B and C are found from the sum of the separate
deflections in cases 1 and 2:
vB = 6.231 mm + 5.192 mm = 11.42 mm Ans.
vC = 11.769 mm + 14.423 mm = 26.2 mm Ans.
428