Question

For the beam cross-section shown below and for an internal shear force Va150 kips and internal bending moment M 1,400 kip-in: opiem 3 3.1) Calculate the moment of inertia of the beam about the x-axis. 3.2) Calculate the maximum bending st 10 Points ress in the section. Indicate on the section where the maximum stress occurs. Points 3.3) Calculate the average shear stress over the cross-section

Answer 1

Given that shear force v = 150 kips Bending moment M = 1,400kip = in 31)for the given isection As the section is symmetric centroid is at centre of the section i.e y^bar = 10^11/2 = 5^11 I times x = I_xx_1 + A_1 d_1^2 + I_x times 2 + I_x times 3 + A_3 d_3^2 = 2 times 8 times 10^3/12 + (8.0 times 1.0) times (5 - 1.0/2) + 0.5 times 8.0^3/12 = 346.67 in^4 therefore moment of inertia about x-axis = 346.67 in^4 I_x = 346.67 in^4 3.2) maximum stress (bending) occurs at the top and bottom of the beam and zero at centre from flexure equation M/I = f/y f = 1400/346.67 times 5 = 32.307 k/n^2 3.3) Average shear stress q_Avar = shear force/Area of cross section Q_Avr = 150/(2 times 8.0 times 1.0) + (0.5 times 8.0) = 150/20 (therefore (2 flnges + web) = Total area q_Aver = 7.5 k/in^2 (therefore total cross sectional area = 20 in^2) therefore Average shearstress = 7.5 kip/in^2