Question

Review Learning Goal: To use the superposition principle to find the state of stress on a beam 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. Part A - Support Reactions and Internal Loading Determine the support reactions Cy and Cz and the internal normal force, shear force, and moment on the cross-section containing point A. Express your answers, separated by commas, to three significant figures. P View Available Hint(s) 10. ut vec ΟΙ ΑΣΦ di ? d dz Cy=,C= , N= , V= , M= N, KN, KN, N, INm 20 mm 100 mm 200 mm Submit 15 mm 20 mm 150 mm 1 Part B - Properties of the Cross-Section The dimensions are d = 2.25 m, d2 = 0.85 m dz = 1.15 m. d4 = 300 mm, and r= 170 mm. The applied force has magnitude P= 4 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 A for the beam? Express your answers, separated by commas, to three significant figures. View Available Hint(s)

Answer 1

Solution Ciren, P A G D d. dz di con cy - l'15m, dy = 300 mm = 0.3 mg Bet d, : 2.25m dy = 0,85m, de 8-170 20:17 P = 4KN 24xWN :: Past Ag By using equilibrium Conditions - EF x 20 G = 4kn Take moment at o EM O PCr+dy) = Cy (d tehat dz) H(0.44) - Gy(4-25) Cy = 0,44 235 KN = 442.35 N & Support teactions - HEN Cy = 442.35 N si 1 Internal normal fore N = P IN= 4KN Shear foue va = Cy VE 442.35 N Moment of Cross section → EM zo at A → H-(C) (03) = 0 M = 442-35 X 15 2,508.7025 N-m M= 01508 Kn-m

Part B 150 s Cross-section Acra Area Izo A = (180 x 20)+(200X1S)+(180X20) 100 1200 A = 9000m² A 15 A = 0.009 m² 20 → Moment of Quertia with respect to neutral ang is 160(quos? 135 (20012 Au dimensions are mm Ta 12 12 = 82.8x100 mm I = 82.88 106 m First moment of Euertia with respect to point a Q = 24 = A9, + A₂% (150 820)( 279) +(18x100) (100) = 405 Xcomm? 4.058159 Q

Past C → Normal Stoeks at A TA + My 근 A Since ., pojat A is on Neutral any y zo A - M A 4X03 01009 TA = 0.444 MPa Shear Stress at A TA VQ If 442-35 X667x4.05x164 82. 8x10 6 x 15X103 G = 0.144 MPa please like the Answer Thank you