Respuesta :
Answer:
[tex] \sigma_A = 58.43 N/mm^2[/tex]
Explanation:
Given data:
length of Steel bolt [tex]L_1 = 335 mm[/tex]
Length of aluminium cylinder [tex]L_2 = 275 mm[/tex]
Pitch of bolt p = 1mm
Modulus of elasticity of steel E = 215 GPa
Modulus of elasticity of aluminium = 74 GP
Area of bolt [tex]= \frac{\pi}{4} 14^2 = 153.93 mm^2[/tex]
Area of cylinder = 2300 mm^2
n =1
By equilibrium
[tex]\sum F_y = 0[/tex]
[tex]P_A -2P_S = 0[/tex]
[tex]P_A =2P_S[/tex]
By the compatibility
[tex]\delta _s + \delta_A = nP[/tex]
Displacement in steel is [tex]\delta_s = \frac{P_sL_s}{E_sA_s}[/tex]
Displacement in Aluminium is [tex]\delta_A = \frac{P_AL_A}{E_AA_A}[/tex]
from compatibility equation we have
[tex]\frac{P_sL_s}{E_sA_s} + \frac{P_AL_A}{E_AA_A} = nP[/tex]
[tex]\frac{P_s\times 335}{180\times 10^3\times 153.93} + \frac{P_A\times 275}{73\times 10^3\times 2300} = 1\times 1[/tex]
[tex]1.20\times 10^[-5} P_s + 1.44\times 10^{-6}P_A = 1[/tex]
substitute[tex] P_A =2P_S[/tex]
[tex]1.20\times 10^[-5} P_s + 1.44\times 10^{-6} (2\times P_s) = 1[/tex]
[tex]1.488\times 10^{-5} P_s = 1[/tex]
[tex]P_s = 67204.30 N[/tex]
[tex]P_A = 134,408.60 N[/tex]
Stress in Aluminium [tex]\sigma = \frac{P_A}{A_A}[/tex]
[tex]= \frac{134,408.60}{2300}[/tex]
[tex] \sigma_A = 58.43 N/mm^2[/tex]
