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

M = −6,000 lb · ft. For this negative moment, the tensile bending stress produced on the top of
the flanged shape (at y = +7 in.) is calculated as
My (
σ =− =− − 6,000 lb ⋅ ft)(7in.)(12 in./ft)
x
=+ 1,185 psi = 1,185 psi(T)
4
I
425.333 in.
z
and the compressive bending stress produced on bottom of the flanged shape (at
y = −5 in.) is
My (
σ =− =− − 6,000 lb ⋅ ft)( − 5in.)(12 in./ft)
x
=− 846 psi = 846 psi(C)
4
I
425.333 in.
z
(a) Maximum tensile bending stress: For this beam, the maximum tensile bending stress
occurs on top of the beam, at the location of the maximum negative internal bending
moment. The maximum tensile bending stress is σ x = 1,185 psi (T).
Ans.
(b) Maximum compressive bending stress: The maximum compressive bending stress
also occurs on top of the beam; however, it occurs at the location of the maximum
positive internal bending moment. The maximum compressive bending stress is
σ x = 1,111 psi (C).
Ans.
ExAmpLE 8.4
A solid steel shaft 40 mm in diameter supports the loads
shown. Determine the magnitude and location of the maximum
bending stress in the shaft.
Note: For the purposes of this analysis, the bearing at
B can be idealized as a pin support and the bearing at E
can be idealized as a roller support.
A B C D E F
500 mm
400 mm 600 mm 600 mm 400 mm
Plan the Solution
The shear-force and bending-moment diagrams for the
shaft and loading will be constructed by the graphical
method presented in Section 7.3. Since the circular cross section is symmetric about the
axis of bending, the maximum bending stress will occur at the location of the maximum
internal bending moment.
SolutioN
Support Reactions
An FBD of the beam is shown. From this FBD, the equilibrium equations can be written as
follows:
Σ Fy = By + Ey
− 200 N − 350 N − 400 N − 200 N = 0
Σ M B = (200 N)(500 mm) − (350 N)(400 mm) − (400 N)(1,000 mm)
− (200 N)(2,000 mm) + E (1,600 mm) = 0
From these equilibrium equations, the beam reactions at pin support B and roller
support E are, respectively,
y
200 N 350 N 400 N 200 N
B
y
= 625 N and E = 525 N
y
259