Question

A steel bar and an aluminum bar are bonded together as shown to form a composite beam. The vertical shear in the beam is 3.6 kips and the modulus of elasticity is 29 x 106 psi for the steel and 10.6 x 106 psi for the aluminum. 2 in. Aluminum Steel 1 in. 1.5 in. Determine the maximum sharing stress in the beam. (Round the final answer to two decimal places.) (Hint: Both Q and / are computed by using the transformed section of the beam.)

Answer 1

Date. losin soli 2 in Aluminium 4.1037 in Steel * 1.5m Equivalent Aluminium Section 3.6 kips Given, Shear force, sa S 3.68103 eb Elastic Modules of steel, Es = 298106 Psi Elastic Modules of aluminium, Et = 10.6 X106 psi m= 29x106 10.6 X 106 2.7358 EAL ولما width of steel in terms of aluminium mx1.5 = 2.7358X1.5 = 4.1037 in New from figurie 6 Position of Neutral Axis You A, Y + A₂Yz A+Az INA (4.1037x1)(0.5) + (1.5x2) (105) 14.103781 ) + (1.5 X2) You 0.9223 in

Date. moment of Inentia of Area ② about Neutral axis I, (Ia) + where, h, 2 %-TNA |(4.1031) (0)2 + 4.1034 * (-0.4223||n Ah = 0.5-0.9223 -0.4223 in 12 1.6738 iny moment of inentia of Area ③ about Mental Axlf (Fa) + A2 h Where Y - INA (105)(2)3+ (105x2)(0.5720) ho 1.5-0.9222 0.5777 in 12 Iz = 2.001 in4 Aluminium Lo, Moment of Inentia of equivalent Crossection BI こ I It Iz 1.0738+ 2.001 3.0748 in4 I So, Maximum shear stress will be at the Junction steel and Aluminium Tmax z SAY Tmax = Ibi 3.6X103 x (1.5x2)x(1-0.9223) 3.0748 X 1.5 181.94 psi Tmax =