
Assuming complete freedom for the rod, the deformation due to drop of temperature is...
δT=αLΔT
δT=0.0000065(30)(90)
δT=0.01755 ft=0.2106 in.
subscript b ( b ) = refers to the beam
subscript r ( r ) = refers to the rod
δb+δr=δT
(PL33EI)b+(PLAE)r=0.2106
(PL33I)b+(PLA)r=0.2106E
P(63)(123)3(154)+P(30)(12)0.08=0.2106(29×106)
807.90P+4500P=6107400
5307.9P=6107400
P=1150.62 lb=1.151 kips
Stress increase on the rod
σ=PA=1.1510.08
σ=14.39 ksi answer