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100 Mechanical Engineering Design
The magnitude of the idealized transverse shear stress profile through the beam
depth will be as shown in Fig. 3–20d.
(b) The bending stresses at each point of interest are
Answer σ a = My a = (1098)(0) = 0 psi
I 2.50
My b (1098)(1.24)
σ b = σ c =− =− =−545 psi
I 2.50
My d (1098)(1.50)
σ d =− =− =−659 psi
I 2.50
(c) Now at each point of interest, consider a stress element that includes the bend-
ing stress and the transverse shear stress. The maximum shear stress for each stress
element can be determined by Mohr’s circle, or analytically by Eq. (3–14) with
σ y = 0,
σ
2
τ max = + τ 2
2
Thus, at each point
2
τ max,a = 0 + (828) = 828 psi
2
−545
2
τ max,b = + (715) = 765 psi
2
2
−545
2
τ max,c = + (52.2) = 277 psi
2
2
−659
τ max,d = + 0 = 330 psi
2
Answer Interestingly, the critical location is at point a where the maximum shear stress is the
largest, even though the bending stress is zero. The next critical location is at point b in
the web, where the thin web thickness dramatically increases the transverse shear stress
compared to points c or d. These results are counterintuitive, since both points a and b
turn out to be more critical than point d, even though the bending stress is maximum at
point d. The thin web and wide flange increase the impact of the transverse shear stress.
If the beam length to height ratio were increased, the critical point would move from
point a to point b, since the transverse shear stress at point a would remain constant,
but the bending stress at point b would increase. The designer should be particularly
alert to the possibility of the critical stress element not being on the outer surface with
cross sections that get wider farther from the neutral axis, particularly in cases with
thin web sections and wide flanges. For rectangular and circular cross sections, how-
ever, the maximum bending stresses at the outer surfaces will dominate, as was shown
in Fig. 3–19.