Question

(7) A moment of M = 4 kip-ft is applied on the cross section shown. Determine (a) the normal stress at point a, (b) the maximum tensile and compressive bending stresses in the beam. Ans: (a) oa = 0.0523 ksi (b) om =-1.779 ksi, om' = 3.72 ksi 10.5 in.al -3 in. +0.5 in. 0.5 in. B 3 in. M 10 in. D +0.5 in.

Answer 1

The solution is provided in the images below. Revert back in case of any query.

solution : o.Sim Lo Sin 3 o 3in. 1 3 in Din tym o. Sin M24_sipoft. we need to first find the neutral avil for section. y. 4,4, +42A +_y Ag + y4 Ay A. + A, A++ 5xl_10X0.5) +28.5x13x0.5) + $ 10:25 x 24x0:5). (10X0.5) + 246.3.2.0.5) + (480.5). 25+ 25.5 + 20.5 = 7.1 in S+ 3+2 So neutral aas is located at base. y = 7.1 in from the Distance of a froma neutral_aris : _Onlin_below the

We know that -my. I Ifaxosxo't_10X0.5X(2017) +4x4xos?t uxo.SX(10.35.7.) + 2x4 4x0.58g+ 380.52(8.5-2.0 I = 63.716 + 19.8866 + 8.13 91.7326 in 4 At 0.1 in a 53:32_psi Ta 4X1000X12_X_0.1 91.2326 0.0523 ksi y = Maximus compressive stress will appear at the_topmost plane le of = 10.5-2.) this section, where =_3.4 in - у - Competeivel mos 4000 X/2 x 3.4 92.7326 = -1.779 ksi. 1779 psi at the bottom- Masumum tensile stress will appear ost the section, i.e. y -7.1 in plane of

T Tensile Imax - 4000 X 12% -7.) 91.7326 at 3715.145 psi 3.72 ksi