HW #5 Solutions
Prob 1
A cylindrical specimen of titanium alloy with elastic modulus E = 107, original
diameter d = 3.8, tensile load P = 2000, and maximum allowable elongation
δ = 0.42 is considered. The maximum length L before deformation is calculated
using Hooke’s law for elastic deformation:
σ = Eϵ,
σ=
P
,
A
ϵ=
δ
L
where A is the cross-sectional area:
A=
πd2
π(3.8 × 10−3 )2
=
= 1.134e − 5
4
4
Solving for L:
L=
EδA
(107 × 109 ) × (0.42 × 10−3 ) × (1.134 × 10−5 )
=
= 0.2548 = 254.8
P
2000
Thus, the maximum length is approximately 255.
Prob 2
For a bronze alloy with yield stress σy = 275 and modulus of elasticity E = 115:
(a) The maximum load without plastic deformation for a cross-sectional area
A = 325 is:
P = σy A = (275 × 106 ) × (325 × 10−6 ) = 89375 ≈ 89400
(b) The original length L0 = 115. The strain at yield is:
ϵy =
σy
275 × 106
=
= 0.002391
E
115 × 109
The elongation is:
δ = ϵy L0 = 0.002391 × 115 = 0.275
The maximum length without plastic deformation is:
L = L0 + δ = 115 + 0.275 = 115.275 ≈ 115.3
1
Prob 3
A cylindrical aluminum specimen with diameter d = 19, length L = 200, and
force P = 48800 is deformed elastically. Using typical values for aluminum:
E = 69, Poisson’s ratio ν = 0.33.
(a) The elongation in the direction of applied stress is calculated as:
A=
δ=
πd2
π(19 × 10−3 )2
=
= 2.835e − 4
4
4
PL
48800 × 0.2
=
= 0.000499 = 0.499
AE
(2.835 × 10−4 ) × (69 × 109 )
(b) The change in diameter is found from the lateral strain:
ϵlong =
0.499
δ
=
= 0.002495
L
200
ϵlat = −νϵlong = −0.33 × 0.002495 = −0.000823
∆d = ϵlat d = −0.000823 × 19 = −0.0156
The diameter decreases by 0.0156.
Prob 4
For the aluminum specimen with diameter d = 0.505 (gauge length L0 = 2.000)
and load-elongation data:
2
• Cross-sectional area A = πd4 = 0.2003
(a) Engineering stress σ and engineering strain ϵ are computed from the data.
The plot of σ versus ϵ is shown below (not included in LaTeX, but described).
(b) The modulus of elasticity is determined from the linear portion of the
stress-strain curve. Using points up to ϵ = 0.004:
E=
σ
34199
=
= 8549750ψ ≈ 8.55e6ψ
ϵ
0.004
(c) The yield strength at a strain offset of 0.002 is found by drawing a line
parallel to the elastic line from ϵ = 0.002. The intersection with the stress-strain
curve gives:
σy = 40200ψ
(d) The tensile strength is the maximum stress:
2
σts =
Pmax
10700
=
= 53420ψ
A
0.2003
(e) The percent elongation at fracture:
Lf = 2.330,
%elongation =
Lf − L0
2.330 − 2.000
×100% =
×100% = 16.5%
L0
2.000
(f) The modulus of resilience:
Ur =
σy2
(40200)2
= 94.5ψ
=
2E
2 × 8.55 × 106
3