Question

PROBLEM ( 2.4 ) Two gage marks are placed exactly ( 250 mathrm{~mm} ) apart on a 12 -mm-diameter aluminum rod with ( E=73 mathrm{GPa} ) and an ultimate strength of ( 140 mathrm{MPa} ). Knowing that the distance between the gage marks is ( 250.28 mathrm{~mm} ) after a load is applied, determine ( (a) ) the stress in the rod, ( (b) ) the factor of safety. SOLUTION (a) [ egin{aligned} delta & =L-L_{0} \ & =250.28 mathrm{~mm}-250 mathrm{~mm} \ & =0.28 mathrm{~mm} \ varepsilon & =frac{delta}{L_{0}} \ & =frac{0.28 mathrm{~mm}}{250 mathrm{~mm}} \ & =1.11643 imes 10^{-4} \ sigma & =E varepsilon \ & =left(73 imes 10^{9} mathrm{~Pa} ight)left(1.11643 imes 10^{-4} ight) \ & =8.1760 imes 10^{7} mathrm{~Pa} end{aligned} ] [ sigma=81.8 mathrm{MPa} ] (b) [ egin{aligned} ext { F.S. } & =frac{sigma_{u}}{sigma} \ & =frac{140 mathrm{MPa}}{81.760 mathrm{MPa}} \ & =1.71233 end{aligned} ] F.S. ( =1.712 )

          PROBLEM ( 2.4 )
Two gage marks are placed exactly ( 250 mathrm{~mm} ) apart on a 12 -mm-diameter aluminum rod with ( E=73 mathrm{GPa} ) and an ultimate strength of ( 140 mathrm{MPa} ). Knowing that the distance between the gage marks is ( 250.28 mathrm{~mm} ) after a load is applied, determine ( (a) ) the stress in the rod, ( (b) ) the factor of safety.
SOLUTION
(a)
[
egin{aligned}
delta & =L-L_{0} \
& =250.28 mathrm{~mm}-250 mathrm{~mm} \
& =0.28 mathrm{~mm} \
varepsilon & =frac{delta}{L_{0}} \
& =frac{0.28 mathrm{~mm}}{250 mathrm{~mm}} \
& =1.11643 	imes 10^{-4} \
sigma & =E varepsilon \
& =left(73 	imes 10^{9} mathrm{~Pa}
ight)left(1.11643 	imes 10^{-4}
ight) \
& =8.1760 	imes 10^{7} mathrm{~Pa}
end{aligned}
]
[
sigma=81.8 mathrm{MPa}
]
(b)
[
egin{aligned}
	ext { F.S. } & =frac{sigma_{u}}{sigma} \
& =frac{140 mathrm{MPa}}{81.760 mathrm{MPa}} \
& =1.71233
end{aligned}
]
F.S. ( =1.712 )

$PROBLEM ( 2.4 ) Two gage marks are placed exactly ( 250 mathrm mm ) apart on a 12 -mm-diameter aluminum rod with ( E=73 mathrmGPa ) and an ultimate strength of ( 140 mathrmMPa ). Knowing that the distance between the gage marks is ( 250.28 mathrm mm ) after a load is applied, determine ( (a) ) the stress in the rod, ( (b) ) the factor of safety. SOLUTION (a) [ eginaligned delta =L-L0 =250.28 mathrm mm-250 mathrm mm =0.28 mathrm mm varepsilon =fracdeltaL0 =frac0.28 mathrm mm250 mathrm mm =1.11643 imes 10^-4 sigma =E varepsilon =left(73 imes 10^9 mathrm Pa ight)left(1.11643 imes 10^-4 ight) =8.1760 imes 10^7 mathrm Pa endaligned ] [ sigma=81.8 mathrmMPa ] (b) [ eginaligned ext F.S. =fracsigmausigma =frac140 mathrmMPa81.760 mathrmMPa =1.71233 endaligned ] F.S. ( =1.712 )$

Added by Ibrahim A.

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