Question

M4.3 Load capacity of beam-strut structure scenes The structure supports a distributed load of w. The limiting stress in rod (1) is 340 MPa, and the limiting stress in each pin is 190 MPa. If the minimum factor of safety for the structure is 2.00, determine the maximum distributed load magnitude w that may be applied to the structure plus the stresses in the rod and pins at the maximum w. 3.1 m 24-mm-diam. pin double shear w (kN/m) 26-mm-diam. rod 0 (MPa) 7.5 m TB (MPa) single shear 32-mm-diam. pin 9.2 m tc (MPa)

Answer 1

Ans Calculate (6) I limit = 340 =170 Here, Fors=2 fois 2 in r= 170 mpa] calculate (TB), To = 1 / 2 - 0 so, An= 4(71)2 = 4X24)= 530.92 mm AB Now F = 5XA,= 170x530.92 = 90257.95N Now, shear force aticing @ B, due to double shear F = E= 45128.97N i TB = 45128-97 = 99,769 2 (24) 2 ZB= 99.76 mpa Then taking member CBD., '0' is the vertical angle with CBD. tano = 9.2 ie o = 50.8° 7-5 Now calculate Mc Me = o ; F, sin(50-8)x 7.5- WX (3-1+7-5)x(3)+7+5)=0 i. W = 1245.19 N W = 0.1174 KNIM

Now, EF n = 0; -C ₂ - Wt Fi sin (50.81)=0 -ca - 1245.19 + 90 257.97x sin (5081) = 0 n=68709.68N { Fy=0; Cy- Fi cos (50.81)=0 Cy=57033-47 N Calculate resultant load at c c= √ Cut Cyl By substituting c = 99296.34 N calculate (2) Tc = c = 89296.34 = lll.ovy A 7 (32) 2 7 (32) .: E = 11.04 mpa]