Question

For a bronze alloy, the stress at which plastic deformation begins is 267 MPa and the modulus of elasticity is 115 GPa. (a) What is the maximum load (in N) that may be applied to a specimen having a cross-sectional area of 206 mm2 without plastic deformation?

Answer 1

Answer:

a) $F_y$ = 55,200N

b) 115.27mm

Explanation:

a) The portion of the problem calls for a determination of the maximum load that can be applied without plastic deformation $(F_{y} )$ Taking the yield strength to be 265MPa

$F_y$ = σ$_y$A$_0$ = (267 x $10^{6}$ N/$m^{2}$)(206 x  $10^{-6}$

$F_y$ = 55,200N

b) The maximum length to which the sample may be deformed without plastic deformation is determined as:

$l_{i} =l_{0}$(1 + σ/E)

= $115mm[1+\frac{265MPa}{115 X 10^{3} MPa } ] =$ 115.27mm

Answer 2

Answer:

The maximum load that may be applied without plastic deformation should be less than 55,002 N.

Explanation:

The load that may be applied at which plastic deformation begins is obtained by multiplying the stress of the bronze alloy by its cross-sectional area.

Stress = 267 MPa = 267×10^6 = 2.67×10^8 Pa = 2.67×10^8 N/m^2

Cross-sectional area = 206 mm^2 = 206 mm^2 × (1 m/1000 mm)^2 = 2.06×10^-4 m^2

Load at which plastic deformation begins = 2.67×10^8 × 2.06×10^-4 = 55,002 N

Maximum load that may be applied without plastic deformation should be less than 55,002 N.