Problem 4a titanium [e = 16,500 ksi;  = 5.3×10–6/°f] bar (1) and a bronze [e = 15,200 ksi;  = 12.2×10–6/°f] bar (2), each restrained at one end, are fastened at their free ends by a pin of diameter d = 0.4375 in. the length of bar (1) is l1 = 19 in. and its cross-sectional area is a1 = 0.50 in.2. the length of bar (2) is l2 = 27 in. and its cross-sectional area is a2 = 0.65 in.2. what is the average shear stress in the pin at b if the temperature changes by 40°f?

Explanation:

Given data:

degree F

Diameter is d = 0.4375

force in titanium and bronze will be equal

from equilibrium condition we have

F_T = F_B = p

From the information given as bar is tightened from ends thus net deformation is assummed to be zero

so we have

solving for P we get

P = 341 kips

average shear at B

x+2 because it just doubles hon so the pattern can actually tell you the expression.

so for example,

if x=2 then 2+2 is 4

if x=4 then 4+4 is 8

and so on and so forth.

i really hope that this makes sense and you out in a way hon: )