Question

9 Knowing that the structural tube shown has a uniform wall thickness of 0.29 in., determine the normal and shearing stresses at the three points indicated. (Round the final answer to three decimal places. Input the answer with the appropriate sign.) 6 in. 10 points 600 lb Skipped 600 lb - 1500 lb 5 in. 1500 11 eBook 2.75 in. 20 in. 0.35 in. Hint Print References The normal stresses are: Oa- ksi Ob= ksi Oc ksi The shearing stresses are: Ta ksi Tb ksi Tc ksi

Answer 1

SOLUTION Bin Sin Goolele isodb using.95 in Soalha 2017 - 3000rb 6 1200X25 Firoolb 3000x20 oll ocesin my - 3000x20 = 60 k obrin 60x10 * 3 21.89 bi 23-2xt =2142 $ 4; 26-2xt = 6–280.29 - 5.42 ~= 8.22 ksi (compremion) Ty : (68x3)– (5.42 * 2:42) 12 221.89 in" M м. = 1200 X 25 = 30 kebin 12 Ca 30X1D X1,5 7.098 = 6.34 kesi Comprenion) Ix = (886)–(2-423s.uz) = 7.098 ina .. a = 8.22+634 = 14.56ksi (comprenson) I due to my a 60610% (3-0-25) 27.53 ks (comprenion) 21.89 a due to mx Box10x1.5 7.098 36.34 ksi (comprenion) و هر روز را = 13.87 ksi Compression) = 7.53+6134

ZON a 1 IS 6 - 6.34 ksi' 30x10x15 7.048. 3000 X0.098) jou due to in azo (point į passes through y-arnis) due to My Ya To eb =) T=0 vo 0.21 Q = 6-Buksi (comprenion) Country Tazo (The point à in at top of the surfaces) Fazo ksi 2:42 1200 due to 0.24 21.84 XD.29x2 Igt due to 3000 lb eblina 01262560.2980,125*2+2.42X0:29 ITb = K0.105) 2lahu 30.091814 in न - T=0 2.uzin due to 1200 lb 3000X 4,613 21.84 x 280.29 A sin z Ixt due 3000 lb to 0.29 To = 1092 eblin? = (3x0.2901582)+(2,42 %0.29X2-855 54613 ins

2 ) ditt Foll loin 2018 CC 3500 lb.in M 26000 2 lb.in 7solo 3500 bin l M = 750 (18) - 750x10 "I thiep = 750x8 = 6000 lb.in ارت 167 Tids 16X3:5 TT (1:47.)3 S-6 ksi my I e 321 Tids 5 3 32X6 = 19.24 ksi compiersion) Tilluar 5.6ksi 14.24 ksi