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

A B C D E F
500 mm
B y
400 mm 600 mm 600 mm 400 mm
200 N 350 N 400 N 200 N
E y
Construct the Shear-Force and
Bending-Moment Diagrams
The shear-force and bending-moment diagrams can be
constructed in accordance with the six rules outlined in
Section 7.3.
The maximum internal bending moment occurs at D
and has a magnitude of M = 115 N · m.
Moment of inertia
The moment of inertia for the 40 mm diameter solid steel
shaft is
A B C D E F
I
z
π π
= d = (40mm) = 125,664 mm
64 64
4 4 4
V
M
500 mm
400 mm 600 mm 600 mm 400 mm
200 N 350 N 400 N 200 N
–200 N
625 N
–100 N·m
425 N
70 N·m
75 N
115 N·m
–325 N
525 N
–80 N·m
200 N
Flexure Formula
The maximum bending stress in the shaft occurs at D.
Since the circular cross section is symmetric about the axis
of bending, both the tensile and compressive bending
stresses have the same magnitude. In this situation, the
flexure formula in the form of Equation (8.10) is convenient
for calculating bending stresses. The distance c used
in Equation (8.10) is simply the shaft radius. From this
form of the flexure formula, the magnitude of the maximum
bending stress in the shaft is
Mc (115 Nm)(20 ⋅ mm)(1, 000 mm/m)
σ max = =
Iz
125,664 mm
4
= 18.30 MPa
Ans.
Section Modulus for a Solid Circular Section
Alternatively, the magnitude of the maximum bending
stress in the shaft can be computed from the section modulus.
For a solid circular section, the following formula
can be derived for the section modulus:
4
Iz
( π/64)
d π
S = = = d
c d/2 32
3
For the 40 mm diameter solid steel shaft considered here, the section modulus is,
therefore,
S
π π
= d = (40mm) = 6,283 mm
32 32
3 3 3
and the magnitude of the maximum bending stress in the shaft can be computed as
M (115 Nm)(1, ⋅ 000 mm/m)
σ max = =
3
S 6,283 mm
= 18.30 MPa
Ans.
260