So 1 2 and 3 BOLTED CONNECTIONSSituation 1:The lap joint shown consists of bolts 22 mm indiameter with 25 mm diameter holes. The platesare A36 Steel with Fy = 250 MPa and Fu = 400 MPa.Use xi = 50 mm, x2 = 160 mm, X3 = 60 mm and platePthickness of f = 12 mm.Allowable stresses are as follows:Shear strength of bolts, Ev = 210 MPaBearing stress on projected area, F = 1.5 FuTensile stress on gross area, Fi = 0.6 FyTensile stress on net area, Fr = 0.5 FuShear on plates, Fv = 0.3 Fu1.Determine the value of P based on bolt shear.478.97 kN950.40 KNa.C.468.00 kNb.d.504.00 kN2.Find the value of P based on bearing of plates.ä.478.97 KN468.00 kNb.950.40 kNd.504.00 kN3.Compute the value of P based on tension on the gross area.a.478.97 kNC.468.00 kNb.950.40 kNd.504.00 kN4.Solve the value of P based on tension on the net area.a.663.66 kNс.823.32 kNb.756.00 kNd.504.00 kN5.Calculate the value of P based on block shear.633.60 kN489.60 kN568.80 kN554.40 kNa.C.b.d.-----O--

Answered: Image /qna-images/answer/f773f8a3-be3d-4b7d-bcf2-3c222bf677be.jpg

BOLTED CONNECTIONS Situation 1: The lap joint shown consists of bolts 22 mm in diameter with 25 mm diameter holes. The plates are A36 Steel with Fy 250 MPa and Fu = 400 MPa. Use xi = 50 mm, x2 = 160 mm, X3 = 60 mm and plate X3 thickness of t = 12 mm. Allowable stresses are as follows: Shear strength of bolts, Fv = 210 MPa Bearing stress on projected area, Fe = 1.5 Fu Tensile stress on gross area, Fr 0.6 Fy Tensile stress on net area, Fr = 0.5 Fu Shear on plates, Ex = 0.3 Fu 1. Determine the value of P based on bolt shear. 478.97 kN 950.40 kN a. с. 468.00 kN b. d. 504.00 kN Find the value of P based on bearing of plates. 468.00 KN 504.00 KN 2. a. 478.97 KN C. b. 950.40 kN d. 3. Compute the value of P based on tension on the gross area. 478.97 kN 468.00 kN 504.00 KN с. b. 950.40 kN d. 4. Solve the value of P based on tension on the net area. 663.66 kN 756.00 kN a. C. 823.32 kN b. d. 504.00 kN 5. Calculate the value of P based on block shear. 633.60 kN 568.80 kN 554.40 kN a. C. b. 489.60 kN d. O---

BOLTED CONNECTIONS Situation 1: The lap joint shown consists of bolts 22 mm in diameter with 25 mm diameter holes. The plates are A36 Steel with Fy 250 MPa and Fu = 400 MPa. Use xi = 50 mm, x2 = 160 mm, X3 = 60 mm and plate X3 thickness of t = 12 mm. Allowable stresses are as follows: Shear strength of bolts, Fv = 210 MPa Bearing stress on projected area, Fe = 1.5 Fu Tensile stress on gross area, Fr 0.6 Fy Tensile stress on net area, Fr = 0.5 Fu Shear on plates, Ex = 0.3 Fu 1. Determine the value of P based on bolt shear. 478.97 kN 950.40 kN a. с. 468.00 kN b. d. 504.00 kN Find the value of P based on bearing of plates. 468.00 KN 504.00 KN 2. a. 478.97 KN C. b. 950.40 kN d. 3. Compute the value of P based on tension on the gross area. 478.97 kN 468.00 kN 504.00 KN с. b. 950.40 kN d. 4. Solve the value of P based on tension on the net area. 663.66 kN 756.00 kN a. C. 823.32 kN b. d. 504.00 kN 5. Calculate the value of P based on block shear. 633.60 kN 568.80 kN 554.40 kN a. C. b. 489.60 kN d. O---

Mechanics of Materials (MindTap Course List)

9th Edition

ISBN:9781337093347

Author:Barry J. Goodno, James M. Gere

Publisher:Barry J. Goodno, James M. Gere

Chapter1: Tension, Compression, And Shear

Section: Chapter Questions

Problem 1.9.11P: A ship's spar is attached at the base of a mast by a pin connection (see figure). The spar is a...

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