Question

For a bronze alloy, the stress at which plastic deformation begins is 275 MPa, and the modulus of elasticity is 115 GPa. (a) What is the maximum load that may be applied to a specimen with a cross-sectional area of 325 mm2 without plastic deformation

Answer 1

Answer:

89375 N

Explanation:

Rearrange the formula for normal stress for F:

$\sigma=\frac{F}{A}$

$F=\sigma*A$

Convert given values to base units:

275 MPa = $275*10^{6}$ Pa

325 $mm^{2}$ = 0.000325 $m^{2}$

Substituting in given values:

F = $(275*10^{6})*(0.000325)=89375$ N