Question

A solid brass rod support a force of 15 kips. The brass rod is 1.25 inches in diameter and has a modulus of elasticity of 14000 ksi. The brass rod is connected to a steel plate which is support by two cast iron pipes. The cast iron pipes have an outside diameter of 4 inches and a modulus of elasticity of 12000 ksi. Determine:

a) the normal stress in the brass rod.

b) the required inside diameter of the cast iron pipe if the pipe has an allowable normal stress of 5 (ksi).

c) The clearance between the bottom of the brass rod and the ground surface after load W is applied.

The normal stress in the brass rod is: ______ (ksi)

The required inside diameter of the cast iron pipe is  (in.)

The deflection of the brass rod is ______(in.)

Cast Iron Pipe E = 12000 ksi Cast Iron Pipe E = 12000 ksi 42 in. Brass Rod (solid) E = 14000 ksi Diam. = 1.25 in. 48 in. W Ground Surface

can someone help me with this strengths problem please?

The deflection of the cast iron pipes are ______(in.)

The clearance between the bottom of the brass rod and the ground surface is ______ (in.)

Answer 1

B cast iron E=12000 på (hollow) cast iron pipe E= 12000 psi Brass (soud) 48 inch 42in. D Ground surface Given - dBE = EBE = 14000 topsi = 1 = 14000 ksi AN= 15 kips 1.25 in. 4in. Couter diameted) EAD = ECF = 12000 kosi 12 ki= 12000 ksi do docka AD di AD dier = ? (mot known => innen diameter) a) To calculate normal stress in the brass rod- Let us make a free body diagram of all the rods -

Fale body diagram B +४/२ I w/2 TW A B tw F otwa pulla The normal stress in the brass rod will be 15 Area Of BE *(1.25)? JBE TBE- 12.29 KST b) According to face body diagrom drawn above- moral The force acting ao cast irom pipes is w/2 Hence, the normal stress acting on the cast iron pipes are o

15 => cast iron a cast 2 (Area) 216(4-dij) Given that the allowable normal stress is 5 ksi - Hence, considering the factor of safety as 1 15 5 1 > (5 (4-di)?) 2 3x3 (4- do) di = 2.62 in Hence, the required inside diameter is 2.62 in.

C Hence, Normal stress in Brass = 12,2 2 Asi (Tensile) Normal stress in Cast iron pipe = 5ksi (compressive Hence, "There will be an elongation in Brass god strain = Shess o Al e shers E => A= Stess x e E Al 36.66 inch x10-3 12. a 2x43 Brass = 14000 Al Brass = 0.037 inch lelongation) For Cast iron pipe - l forma contraction ) > Al cast iron - 5848 0.02 inch 1 2000 (contraction)

Hence, The new length of Brass rod = et al 424 0.037 42.037 inches length The new of cast iron pipes e- Al - 48-0.02 - 47.98 inches. 42.037 inches 47.98 inch f D Hence, the clearance between the bottom of the brass rod (E) and ground swiface 47.98- 42.037 се 5.943 inches

deflection of brass rod = Albrass = 0.037 inches (calcellated in part c) deflection of cast iron pipes 0.02 inches (calculated in pour c)