Question

Part A - Support Reactions and Internal Loading Learning Goal: To use the superposition principle to find the state stress on a bear under multiple loadings. The beam shown below is subjected to a horizontal force P via the rope wound around the pulley. The state of stress at point A is to be determined Determine the support reactions, and C, and the intemal normal force, shear force, and moment on the cross-section containing point A. Express your answers, separated by comras, to three significant figures. View Available Hints) VO AEQ vec c ? C-.-.-.-.- N. KN, KN, N, kN. d. 20 mm Submit 100 mm 200 mm 15 mm 20 mm 150 mm 1 Part B - Properties of the Cross-Section The dimensions are di – 2.2 m, da – 0.8 m. ds – 1.25 m, dy 335 mm, and r' - 200 mm. The applied force has magnitude P-7.5 kN. What are the cross-sectional area, moment of inertia with respect to the neutral axis, and first moment of the area with respect to point for the beam? Express your answers, separated by commas, to three significant figures. View Available Hints) VE ΑΣΦ | 11 vec 1-.1-.0- m.

Answer 1

PART A:- E fy=0 Coy +Ccy=o. o Pxda +8) k Ccx -C Ź Fx = 0 d3 di Ccy p - Cex = 0 Coy Cex=p=7.5KN ŞMcco. Coy x(d,+d2 +d3) - Paldæoto. Coy * (2.270.8 +1.25) – 7.5X.10.33570-2)=d Сру = 0.944 KN (upward) ZO. Cox+Ccy Ccy = Cay = - Coy = -0.944 en 0.944 KN Downward)

4.0125 kv.m 3 0.8m A C 1.25m 2.2 m 7.SKNI 0.944 KN 0.944 KN 0.944 KN 0.944 А. C SFD. shear force At A FA= 0.944 KN 2.0768 KN.m A D -1.18 - 2.0482 KN-m BMD. Moment at A, cross section MA --1.18 kn.m D A Internal Normal force, С Axial force diagram N 7.5 IN

PART(): cross section Area of I section. comm A= 15x200 + 2x20x150 Ipomin 2oomm A = 9ooo mm² - 6 A = 9000 x16 ta 15mm 20mm = 9x10-3 = 9X10-3 m² isomm Moment of Inertia, 150 x 2403-2% 67.5 x 2003 IA = I= Xx [tex 675*2007] IA= 828x 105 mmt me -12 828 x lots X10 -5 4 8. 28 x 10 I m first moment of Area about x-x= Moment of Area about mentral Asia Axis Due to symmetry first moment of Area about Neutral axis is ZERO

PART():- + Normal stress at A, Normal stress = Normal stress due to Normal due to Bending force >(ZERO' At Neutral Axig) on tog 2 = 66 69 - ak to. А A - 7.9 x 103 0.8333 MPa gooo Shear stress at A, bz Az TA F . A Ib 9, ý » AI FAX [A, 5 , +4257 bi А I. + 8.x8xT TA 0.944 8163 100x 15×50 150820 X 110 *8.28 X10 15 Iso TA = 0.082 MPa