Question

***partial for this answer is "member yields no FS, Sigma p1=112.233. MAKE SURE IT MATCHES

4) For the following member a. Sketch the element stress block for the element at a on the cross section located at B on the horizontal member b. Determine the principle stresses What is the FS for yielding if oy = 20 ksi C. 30 k 1" 20 k 6" 3" / 4' B 2' 3' /

Answer 1

30 20k of eter B x X 2' 3' F. B.D e Me= (30x3)=40x-off 20K > 20K the 3' 30K 30k узоқ 20k 90k-ft. MB a' HB = 90+ (>0x4) = 170k-ft- Blt 20K Зок 30K B go 20R. MB = 170k-ft .= 2040 k-in 't A f Garut 21 NY MB Als 31 3 ÎNA- = 54 in 12 2040 xl due to rok force a tensite Stressribe there, every where (including a) 2) in the cross section . 20K E = 28 loll ksi due to bending moment (MB) also a tensile 386 Stress willte at a. MB X 1 = Als = (346)= 18 in2 30k force is a shear force, due to this shear stress at ". Total Normal stressad a. a millbe Ta vx Ashadex y l Ashaded - TN - (37.97 +1.11) ksio Ở MA X B. 1 273)-6ina Tra = 38.88 ksi (Tensile) Za= 30x6x2 wsi Ita INA sy Ita= 37.77 ksi = IB= 3.11 ) - Sч хЗ {a=2.22 usi. :. State of stressata element at á willbe. a ry 20 = TN 03 = 38.88 ksi exy = 2.22ksi. "Gy=d.

B Principle stresses will be lost PR/80 = Tetry .: 0P, = 38.88 +0 + 138 38.88-0) +(2-22) - 2 + xy 2 39.00ksi Tp. OP2 138:88)72-223" DOLORE 38:38 OP 0.1263 ksi These values Principle stresses are up, = 39.00ki (Tensile) Is TP2 = 0.1263 usi (comp) are absolutely c For stafety agains yielding SX OXUYA Fos. 20 39 Foors < 0.512