Page 511 - Shigley's Mechanical Engineering Design

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486 Mechanical Engineering Design
Figure 9–16
F
′′
Free-body diagram of one of D D
the side plates.

A
′ C ′
C D ′′
D A
r C r D
O
r A
′ r B
B
B A
′′ C ′ A

C

B ′′ B
(f ) Find J. Using case 4 of Table 9–1 again, with Eq. (9–6), we get

3 2 3 4
8(56) + 6(56)(190) + (190) (56)
J = 0.707(6) −
12 2(56) + 190
6
= 7.07(10) mm 4
(g) Find M:
M = Fl = 25(100 + 10.4) = 2760 N · m
(h) Estimate the secondary shear stresses τ at each lettered end or corner:

3
Mr 2760(10) (105)
= 41.0MPa
τ = τ = =
A B 6
J 7.07(10)
3
2760(10) (95.6)
= 37.3MPa
τ = τ =
D
C
7.07(10) 6
(i) Draw the τ stress at each corner and end. See Fig. 9–16. Note that this is a free-body

diagram of one of the side plates, and therefore the τ and τ stresses represent what the

channel is doing to the plate (through the welds) to hold the plate in equilibrium.
(j) At each point labeled, combine the two stress components as vectors (since they
apply to the same area). At point A, the angle that τ A makes with the vertical, α, is also

−1
the angle r A makes with the horizontal, which is α = tan (45.6/95) = 25.64 . This
◦
angle also applies to point B. Thus

◦ 2
◦ 2
τ A = τ B = (19.5 − 41.0 sin 25.64 ) + (41.0 cos 25.64 ) = 37.0MPa
−1
Similarly, for C and D, β = tan (10.4/95) = 6.25 . Thus
◦

◦ 2
◦ 2
τ C = τ D = (19.5 + 37.3 sin 6.25 ) + (37.3 cos 6.25 ) = 43.9MPa
(k) Identify the most highly stressed point:
Answer τ max = τ C = τ D = 43.9MPa