Question

Diameter of pin 2 mm. All pins are in single shear.

Question 1: A light plane structure is supported by a roller at A and a pin at D and is subjected to a load of Pat F as shown in the figure. All the members have rectangular cross section with height 5 mm and thickness 1 mm. Tensile/Compressive strength of the members = 80 MPa, Shear strength of the pin = 50 MPa. Consider an appropriate factor of safety. Determine the force P such that: 1. The bars ABCD, BEF and CE do not fail under normal stress. 2. The pins at E and C do not fail under shear stress. 3. What is the maximum allowable load P? All the pins have smooth contact. P 50 mm + 50mm * 50 mm * 50 mm*50 mm

Answer 1

P E В Ay 1 EMA = 0 (24(200) = (P)(150), SE Fx = 0 Dx = 0 Dr=o Dy = 3/4P => EFy = 0 Ay + Dy = P, Ay = P-by = P-3/4P = // Ay = P/4. boor bar "ABCD' 14 Cy A G. 1348 » Ety = 0 By tly + P=0 Byt Cy=-p - (1) a plu EMA=0. (34P) (200) + (cy)(150) + (BY) (50) = 0 3 Cy & By = - 31/16 - (2)

- 2 y = – Pt 31 Cy = 6.5p Byr By + Cy=-P By --D- (+6.59) By = 8.50 -7.50 B01 BX+C =0 By = -(x Bx By bor bar "BEF" P Mo=0 Ey (Pfioo= (ty) (50) fy-2P - Ity=0 F+ > MB-0 (P) (100) +()(50) – (Eyl(50) = 0 Ey - Ex = 20 (4). -> I Fy=0 By tP = Ey - 7.5P+P = Ey, ty 6.5p boom above equation, Ey - Ex=2P, 6.59-Ex = 21 E = -8.50 --

Normal borce an EC, EF=0 Bx = t = -8.5p, brom eq. (3). Bx = -6 8.5P for bar "EC calculations JEG. S8+ (-8.58 FEC. = 10.7P Normal force in Way BEFI FBEE BX + By - (8.5) + (7.599 FBEF 11.3358 p Normal borce an AB CD, FAB = 8.58. shear borce an at pin, 'D', to = 3/44 Pin A, A=1/46 al shear borce

- Max. Normal booce, Frer = 11.8858 p Area, 5x1= 5 mm. FREE Abar 80 = 11.33582 5. P = 35.286 N. e shear stras , N= F = Gui (85,286) a (25 y = 8.43 Mpa < T mar. (50m pa). Pe 35,286 N