## Transcribed Text

1. A steel component (E = 200 GPa, ν = 0.29) is subjected to a state of plane stress (σz
= τxz = τyz = 0) when the design loads are applied. At the critical point in the
member, the stress components are given in Figure 1. The material has a yield stress,
σyield = 450 MPa. For (a) – (d), calculate the factor of safety based on the following
yield criteria:
(i) Maximum principal stress theory (Rankine’s theory);
(ii) Maximum shear stress theory (Tresca’s theory);
(iii) Maximum distortion energy theory (von Mises theory).
(a) (b)
(c) (d)
Figure 1
x
y
130 MPa
20 MPa
100 MPa
x
y
100 MPa
50 MPa
120 MPa
x
y
100 MPa
30 MPa
150 MPa
x
y
100 MPa
40 MPa
150 MPa
2
2. At a point on the surface of a component the stresses are σx = −130 MPa, σy = 50
MPa, and τxy = 25 MPa as shown in Figure 2. Using Mohr’s circle, determine: (a) the
stresses acting on an element inclined at an angle, θ, of 60° clockwise, (b) the
principal stresses, and (c) the maximum in-plane shear stresses.
Figure 2
x
y
130 MPa
25 MPa
50 MPa
3
3. A shaft containing two fillets, as shown in Figure 3, is machined from alloy steel
heat-treated to 300 BHN and rotates at 3000 rpm, whilst the imposed load, F,
remains static. The shaft is designed to have infinite lifetime with respect to fatigue
failure. It is found when the imposed load is increased, failure will occur at the two
fillets at the same time. Calculate:
(a) the maximum allowable load given a factor of safety of 2;
(b) the distance L4;
(c) the fatigue life in the number of cycles if the shaft is subjected to the loading
pattern as shown in Table 1.
Given: D = 40 mm; d = 30 mm; L1 = 50 mm; L2 = 100 mm; L3 + L4 = 200 mm; R1 = 2
mm; R2 = 1 mm.
Figure 3: A shaft containing two fillets
Table 1: Loading sequence of varying amplitudes
Load Number of cycles (n)
3Fallow 30
7Fallow 20
Fallow 50
6Fallow 40
5Fallow 25
where Fallow is the maximum allowable load found in (a).
L1
R1
F
R2
L4 L3 L2
d
D
d
4
4. A shaft containing two fillets, as shown in Figure 4, is machined from alloy steel and
rotates at 3000 rpm, whilst the imposed load, F, remains static. The shaft is designed
to have infinite lifetime with respect to fatigue failure. The ultimate tensile strength
is 1000 MPa.
(a) If the shaft is not transmitting power, given a factor of safety of 2, calculate the
maximum allowable load;
(b) At the maximum allowable load from (a), if the shaft is transmitting power of 40
kW, calculate the factor of safety.
Given: D = 40 mm; d = 30 mm; L1 = 50 mm; L2 = 150 mm; L3 = 90 mm; R1 = R2 = 3 mm.
Figure 4: A shaft containing two fillets

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