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

Axial Stress
On section a–a, the internal force F (which is equal to the clamping force P) produces a
normal stress
F P
σ = = =
A A
P
96 mm
axial 2
This normal stress is uniformly distributed over the entire cross section. By inspection,
the axial stress is a tensile stress.
Bending Stress
Since the tee shape is not symmetrical about its z axis, the bending stress on section a–a
at the top of the flange (point H) will be different from the bending stress at the bottom of
the stem (point K). At point H, the bending stress can be expressed in terms of the clamping
force P as
My P(46mm)(6 mm) P
σ H = = =
I 2,176 mm 7.88406 mm
bend, 4 2
z
By inspection, the bending stress at point H will be a tensile stress.
The bending stress at point K can be expressed as
My P(46mm)(10mm)
P
σ K = = =
I 2,176 mm 4.73043 mm
bend, 4 2
z
By inspection, the bending stress at point K will be a compressive stress.
H
Axial
stress
Bending
stress
Combined
stress
M
F
+ =
K
Combined Stress at H
The combined stress at point H can be expressed in terms of the unknown clamping
force P as
P
P ⎡ 1
1 ⎤ P
σ H = + = P ⎢ +
⎥ =
96 mm 7.88406 mm ⎣96 mm 7.88406 mm ⎦ 7.28572 mm
comb, 2 2 2 2 2
Note that the axial and bending stress expressions are added, since both are tensile
stresses. The expression just obtained can be set equal to the allowable normal stress to
obtain one possible value for P:
P
7.28572 mm
2
2
≤ 108 MPa = 108 N/mm ∴ P ≤ 787 N
(a)
286