For Homework 6 Handout
Problem 7.40 - MODIFIED
A 20-mm-diameter shaft transmits a variable torque of 200 ± 100 N-m. The frequency of
the torque variation is 0.1 s−1. The shaft is made of high-carbon steel (AISI 1080,
quenched and tempered at 800°C). Find the endurance life of the shaft. Ans. 62,900
cycles, or 174 hours.
Notes: Hamrock again solves this as a fully-reversing fatigue problem, which it isn’t.
We solve it correctly as a fluctuating fatigue problem, but reduce the mean and
alternating torque values to give it infinite life. The question then becomes finding the
Factor of Safety for infinite life – we’ll find the FOS for if the mean and alternating
stresses increase proportionately.
Solution: For this AISI 1080, the inside front cover gives Sy = 380 MPa and Sut = 615
MPa. This is a torsional shear problem, so we calculate Sus = 0.67 Sut = 412 MPa, and
Ssy = 0.577 Sy = 219.3 MPa. Then, from Eq. (7.6), one obtains:
To keep life simple, we won’t modify the endurance strength.
The shaft sees a mean torque of 200 Nm, and an alternating torque of 100 Nm, with
Tmax = 300 Nm and Tmin = 100 Nm.
The shear stress amplitude on the 20mm shaft is
τ alt =
τ max − τ min
2
=
(Tmax − Tmin )r (300 Nm − 100 Nm)(0.01m )
=
= 63.66MPa
π
2J
4
2 (0.02m)
32
The mean torque is twice the alternating torque, so
τ mean = 2τ alt = 127.33 MPa
Then we use the equation from Lecture 4 to find the Factor of Safety for proportionately
increasing mean and alternating stresses:
1 σ alt σ mean
=
+
n Se
S ut
But we need to change it a bit to account for our having shear stresses:
1 τ alt τ mean 63.67 127.33
=
+
=
+
= 0.358 + 0.309 = 0.667, so
n Se
Sus
178
412
n= 1
0.667
= 1.5
Problem 7.41
For the shaft in Problem 7.40 determine how large the shaft diameter has to be for
infinite life. Ans. D = 17.5 mm.
Notes: The stress amplitude must equal the fatigue endurance limit for this case. This
solution uses the results of Problem 7.40.
Solution: Being just at infinite life means the FOS = 1.0. With a FOS of 1.5 from Prob.
7.40,
SigmaAlt was 63.7MPa. So SigmaAlt would now be 1.5 x 63.7 = 95.5MPa.
The shear stress amplitude is
τ alt =
τ max − τ min
2
Therefore,
d3 =
( )
d
(Tmax − Tmin )r (300Nm − 100 Nm) 2 509.3 Pa
=
=
=
= 95.5MPa
π
2J
d3
4
2 (d )
32
509.3
= 5.33 × 10 −6 m 3
6
95.5 × 10
or d = 0.0175 m = 17.5 mm
Problem 7.63
A toy with a bouncing 50-mm-diameter steel ball has a compression spring with a spring
constant of 100,000 N/m. The ball falls from a 3-m height down onto the spring (which
can be assumed to be weightless) and bounces away and lands in a hole. Calculate the
maximum force on the spring and the maximum deflection during the impact. The steel
ball density is 7840 kg/m3.
Ans. 1743 N.
Notes: The weight of the ball is calculated from its volume and density, and then Eq.
(7.54) solves the problem.
Solution: The weight of the ball is the product of its volume and density, or
We compute the static spring deflection
δ st =
W
5.03
=
= 5.032 × 10 −5 m
k 100,000
and the Impact factor
Im =
δ max Pmax
2h
(2)(3)
=
= 1+ 1+
=1+ 1+
δ static
W
δ st
5.023 × 10 −5
= 1 + 119450.5 = 1 + 345.6 = 346.6
Then
δ max = I mδ static = (346.6)(5.023 × 10 −5 ) = 0.01743m = 17.43mm
And
Pmax = I mWstatic = (346.6)(5.03) = 1743N