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

v A
70 kN
A
v
210 kN.m
B
θ B
C
v C
θ D
3 m 3 m 3 m 2 m
D
E
v E
x
Using the variables and values
and
M =−210 kN⋅m
L = 6m(i.e., thelength ofthe centerspan)
3 2
EI = 43.2 × 10 kN⋅m
we calculate the rotation angle at B from Equation (b):
ML (
θ =− =− − 210 kN ⋅ m)(6 m)
−3
B = 9.722 × 10 rad
3 2
3EI 3(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 (9.722 10 rad)( − 3 m) =− 29.167 × 10 m = −29.167 mm
A B AB
Beam deflection at C: The beam deflection at C for this case is found from the elastic
curve equation for a simply supported beam with a concentrated moment applied at one
end. From Appendix C, the elastic curve equation is
With the variables and values
Mx
v
6LEI x 2
( 3 Lx 2 L 2
=− − + )
(d)
M =−210 kN⋅m
x = 3m
L = 6m(i.e., thelength ofthe centerspan)
and
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 )
( −210 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
= 10.938 × 10 m = 10.938 mm
Beam deflection at E: For this case, the overhang at the right end of the span has no bending
moment; therefore, it does not bend. The rotation angle at D given by Equation (c) and
the overhang length are used to compute the deflection at E. With the variables and values
M =−210 kN⋅m
and
L = 6m(i.e., thelength ofthe simple span)
3 2
EI = 43.2 × 10 kN⋅m
the rotation angle at D is calculated from Equation (c):
ML ( −210 kN⋅m)(6 m)
θ D =+ =
3 2
6EI 6(43.2 × 10 kN⋅m)
−3
=− 4.861 × 10 rad
436