Question

A force of F = 29 KN is applied to the frame as shown. The frame is supported by pins at C and D. A) (20 points) Calculate the internal reactions in member ABC at the midpoint between points B and C (denoted by the green line). B) (10 points) Calculate the normal stress and shear stress in member ABC at the midpoint between B and C (denoted by the green line) if the cross section at that location is 16 x 16 mm. C) (20 points) Calculate the shear stress in pin D if it is in double shear and its radius is 16 mm. 135 mm 240 mm D С 450 mm 510 mm B 120 mm A F

Answer 1

soln: Given F= 29 KN 240 135mm y Sino = Coso 450 510 ck Cx 510 450mm M T Toson Drawing FBD of member ABC. Considering equilibrium of member ABC using equation of equilibrium Taking moment about 'e' EMc = 0 B Tsino A 240mm C D Fx 135 + Tsino x 450 = 0 450 mm 510mm 29x135 + TX 240 x 450=0 0 510 My TMx T = 18.4875 KN (Tension in the B. member BD) (A) Let mid point of BC is M. Drawing FBD -M M 225mm Reactions at point M are Mx, My Taking moment about Tsino M Tsino x 225-FX 150 → M-18.4875 x 240 x 225 - 29x135 = 0 Toso -7-B and 120mm A 510 M 5872.5 KN.mm Now using Efag - My-F + T coso = 0 - My - 29+ 18.4875 x 450 510 My 12.6875 KN

Again Efx=0 - Mx + Tsino =0 - Mx + 18.4875 x 240 510 Mx 8.7 KN Crossection area given A= 16x16 mm² My Sn= Normal stress at point M A 256 mm² 12.6875 X 103 256 mm2 N On 49.56 MPa Mx 8.7x10²N 11 Shear stress at point M 2 = А 256 mm2 て 33.984 MPa (C) R = 16 mm (Double shear) Total Sheared area As 2x I R² = 25 (16)² mm² As 1608.5 mm2 Shear stress in pin D. will be Zp = T As 18.4875x100 1608-5 mm2 Te = 11.493 MPa
i have explained throuly from basics hope you will understand easily please upvote it your positive reviews are motivates us. Thank you