Question

Chapter 4, Reserve Problem 026 In the figure, the rigid member ABDE is supported at A by a single shear pin connection and at B by a tie rod (1). The tie rod is attached at B and C with double shear pin connections. The pins at A, B, and C each have an ultimate shear strength of 77 ksi, and tie rod (1) has a yield strength of 61 ksi. A concentrated load of P = 30 kips is applied perpendicular to DE, as shown. A factor of safety of 1.6 is required for all components. Assume c = 4 ft, a = 8 ft, b = 3 ft, d = 8 ft, and = 25º. Determine (a) the minimum required diameter for the tie rod. (b) the minimum required diameter for the pin at B. (c) the minimum required diameter for the pin at A. P

Answer 1

given - factor of safety = 1:6 C= 4ft, a= 8ft, b= 3ft d=8ft & 0=25° 8ft 1 - 22.03250 7ft aft || 1377 tot:-* 8ft | | Ultimate shear strength of pin 17)-77 ksi Yield strength y tie : F rod (i) on = 61 ksi concentrated load (P) = 30 kips - AXTA l'iy &- angle made by te rode with horizontal az tani' ( 8) d = 63.43° How Forces in a direction {fuzo Pcos 65 - F, Cos 63.4 3 + Ax=0 -li) Now forces iny direction Efyao - Psin 65 - Fi sin 63.43 +Ay = 0 - (ii)

Taking sum EMA=O +{F, COS 6343)x8 - (PCo565)(1 +85 in 250] - (Psin65) (scos 25)=0 - (ii) Solving cancio) P= 30 kips + F, (3.58) - 182:33 -197.134 = 0 F, = 106 kips Now substituting value of fi in cqn (is plii) we get Ax = 34.73 kips Ay = 121.99 kips Now pin force at A is resultant of forces in as y direction A- An) * + (Ay)? s 126.83 kips allowable normal stress in hie rod (1) Now tellt eine se sent hips ksi Call = 38.125 ksi

a) For tie rod (1) minimum cross-sectional area is Amin - Find Fi = 106 38.125 = 2.78 in : as Amin = ad - da l-e8 in so minimum required diameter for tie rod (1) is 188 in b) 1 Allowable shear stress for pins at A, B xC Tallow - I as = = 48.125 ksi For resultant pin force for supporting the total shear area required is Av? Follow 106 4-8.125

Av = 2.2 in² 1 min Au= 2.2 in? The pin at B is double shear pin so area provided by pin is Av = 2x I d pin double shoar equating both areas, to find minimum pin dia 2x hdpin = 2.2 minimum. Ipin = 1:18 in diameter for pin at B is 1 18in The minimum c) Pin at A, shear area Av 2 Tallow 126.83 48.125 as minimum Av = 2.64 in2 Pin at A is single shear Av= 1* d'pin

equating both area A do pin = 2.64 (dpin = 1.83 in The minimum diameter for Pin at A is 1.83 in
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