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A plate is fastened to a fixed member by four 20-mm-diameter rivets arranged as shown in Fig. P-333. Compute the maximum and minimum shearing stress developed

A plate is fastened to a fixed member by four 20-mm-diameter rivets arranged as shown in Fig. P-333. Compute the maximum and minimum shearing stress developed

Elements Of Electromagnetics
Elements Of Electromagnetics
7th Edition
ISBN:9780190698614
Author:Sadiku, Matthew N. O.
Publisher:Sadiku, Matthew N. O.
ChapterMA: Math Assessment
Section: Chapter Questions
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A plate is fastened to a fixed member by four 20-mm-diameter rivets arranged as shown in Fig. P-333. Compute the maximum and minimum shearing stress developed.
radial distance p from the centroid of the rivet group. 2_53032313900897.. Solution 332 The shearing stress on each rivet is P/A T = Tp/J Where: T = PRn p= R J = EAp? = AR'n %3D PRn(R) T = AR?n T= - ok! A This shows that 7 = Tp/J can be used to find the shearing stress at the center of any rivet. Solution to Problem 333 | Flanged bolt couplings A plate is fastened to a fixed member by four 20-mm-diameter rivets arranged as shown in Fig. P-333. Compute the maximum and minimum shearing stress developed. 14 kN mm mm 14 kN Figure P-333 Solution 333 Tp T = 120 40 40. 120 14 kN 14 kN Where: T = 14(1000)(120) = 1 680 000N . mm J = EAp? = {#(20)² [ 2(40²) + 2(120²) ] J = 3 200 000 mm m shearing stress (p = 120 mm): 1 680 000(120) 3 200 000 Tmax = 20.05 MPa answer Minimum shes ri n stress (p = 40 mm): 1 680 000(40) Tuin =
Transcribed Image Text:radial distance p from the centroid of the rivet group. 2_53032313900897.. Solution 332 The shearing stress on each rivet is P/A T = Tp/J Where: T = PRn p= R J = EAp? = AR'n %3D PRn(R) T = AR?n T= - ok! A This shows that 7 = Tp/J can be used to find the shearing stress at the center of any rivet. Solution to Problem 333 | Flanged bolt couplings A plate is fastened to a fixed member by four 20-mm-diameter rivets arranged as shown in Fig. P-333. Compute the maximum and minimum shearing stress developed. 14 kN mm mm 14 kN Figure P-333 Solution 333 Tp T = 120 40 40. 120 14 kN 14 kN Where: T = 14(1000)(120) = 1 680 000N . mm J = EAp? = {#(20)² [ 2(40²) + 2(120²) ] J = 3 200 000 mm m shearing stress (p = 120 mm): 1 680 000(120) 3 200 000 Tmax = 20.05 MPa answer Minimum shes ri n stress (p = 40 mm): 1 680 000(40) Tuin =
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