Question

The cantilever beam of length 3L is subject to concentrated loads P and 2P and has the thin-walled cross-section illustrated below. The cross-section has uniform thickness t and moment of inertia with respect to the centroidal z-axis given by I = 3140 t4. Use t = 5 mm, L = 135 mm and P= 6 kN. 2P H 10 t 5t A P 10 t 22 L 10 t 1) Determine the numerical value of the shear force and bending moment at a cross-section located at x = 45 mm. Remember to use the sign conventions for shear forces and bending moments. V2) = kN M(x) = N.m 2) For the cross-section located at 2 = 45 mm, determine the state of plane stress at point H (i.e., y=5 t) in the x - y coordinate system. MPa Oy MPa Tay MPa 3) Determine the principal stresses 01 and 02 at point H: 01 = MPa 02 = MPa 4) Determine the maximum in-plane shear stress Tmar at point H: Tmax = MPa 5) What is the value of the corresponding normal stress when the element is oriented so that the shear stress is maximum? = MPa

Answer 1

Solution Criven that I = 3140 ++ ret = 5mm A L= 135 mm, P=6KN 12P RE 41=MC A A IVANT h 2L- *0 L 1 Vm= P = 6 KN Ans Min = - MA EVA Mr = - 4PL + PX2 = -446X10340-135+ 66103 x 0.045 Mz= -2970 Nam Ang 2 at 50 515 = 25 mm Bending Strees at 'n on = (66) = M xin I = 37.83 MPa - 2970 X103 x 25 3:140 x (514 or = 37.83 MPa Ane

Date Page Toy zo mea because No bending moment about X-agis I=VQ for shear shey IZ Irey 6XL03 x [ Set xt6(265.4454) + 104ktx9:54] 3140 x (54)4 x 6242 Tzy = 5:063 MPa ☺ principal stress (61.62) 04/2 = Entry & + له =6) -/Excy)? 2 2 6162 = 37.83 (37383) & (323) (37.83)?+ (5.063)2 (1/2 z 18.915 + 19.58088 hat point 61 = 3805 mpang HO 162 = 0. 66 588 MPa Are Times 6-02 = 38.5+ 0166888 2 At point Tmas = 19.5829 4 MPa 'H Ang