Shear Stress Shear Stress develops when the applied loads cause one section of the body to slide past its adjacent section. The force acts parallel to the area. V __ τ= A τ V A P where V || A Shear Stress Resultant Shear Cross – Sectional Area Shearing area parallel to the load P V ∑Fx = 0; V=P TYPES OF SHEARING STRESS SINGLE SHEAR LAP JOINT Rivet under Single shear P B P t P Shearing Area P 130mm DOUBLE SHEAR Rivet under Double shear BUTT JOINT t P/2 splice plate t main plate P P/2 Shearing Areas SINGLE SHEAR with n number of bolt or rivet; ex. n = 2 w/ only 1 bolt or rivet P P Lower Plate Right Side View P P P Lower Plate P P Right Side View P DOUBLE SHEAR with n number of bolt or rivet; ex. n = 4 w/ only 1 bolt or rivet P/2 P P/2 P/2 P P/2 Middle Plate (Top View) Middle Plate (Top View) P P Middle Plate (Bottom View) Middle Plate (Bottom View) P Right Side View P Right Side View P P TYPES OF SHEARING STRESS PUNCHING SHEAR P INDUCED SHEAR P L = width x t D a a Ɵ at Section a-a : x t V L Ashearing = x.L Ashearing = πD.t t πD Pcos(Ɵ) Bearing Stress Contact pressure exerted by one body upon another body. The force acts perpendicular to the area where load is applied. P b __ δb = A δb Pb A where P A ┴ Bearing Stress Perpendicular Force / Bearing Force Projected Contact Area Bearing Stress END STRESS LAP JOINT t P B P P 130mm P 130mm P P Enlargement of Holes t A = Dt D EXAMPLE no. 01 A circular hole is to be punched out of a plate that has a shear strength of 40ksi The working compressive stress in the punch is 50ksi. a.) Compute the maximum thickness of a plate in which a 2.5” diameter hole can be punched. b.) If the plate is 0.25in thick, determine the smallest diameter that can be punched. P answer : a.) tplate = 0.781” b.) Dhole = 0.800 ” EXAMPLE no. 02 The lap joint shown in the figure is fastened by four ¾-in. diameter rivets. Calculate the maximum safe load P that can be applied if the shearing stress in the rivets is limited to 14 ksi and the bearing stress in the plates is limited to 18 ksi. Assume the applied load is uniformly distributed among the four rivets. answer : P = 24,___ N EXAMPLE no. 03 Compute for the shearing stress in the pin at B for the member supported as shown in the figure. The diameter of the pin is 20mm. answer : 94.01 MPa EXAMPLE no. 04 Member B is subjected to a compressive force of 600 lb. If A and B are both made of wood and are 1.5 in. thick, determine to the nearest 1/8” the smallest dimension a of the support so that the average shear stress along the blue line does not exceed 50 psi. answer : a = 6.50 inches Homework no. 02 - b The figure shows a roof truss and the detail of the connection at joint B. Members BC and BE are angle sections with thickness shown in the figure. The working stresses are 70MPa for shear in rivet and 140MPa for bearing stress due to the rivets. How many 19-mm diameter rivets are required to fasten the said members to the gusset plate? DETAIL OF JOINT B 14 mm thick GUSSET PLATE D B G 75X75X13 mm 4m 3m C A 3m E 3m 50KN 3m F 75X75X6 mm P BE 3m 3m 75KN H 50KN answer : P BC
