Question 1: Stress & Strain

A steel rod of diameter 12 mm is inserted into a brass tube with the same inner diameter and a wall thickness of 2.5 mm. The ends of the tube and rod are joined so that they cannot slide and must move together under load. The bar is 550 mm long. An axial tensile load of 25 kN  is applied to the composite bar. The Young’s modulus of steel is 215 GPa and that of brass is 105 GPa. Calculate the following: [20 Marks]

a)   The loads carried by the steel and brass.

b)  The stresses in the steel and brass

c)   The extension of the composite bar

a) Loads carried by the steel and brass

dBri  = 12 mm P = 25 kN

Area of steel rod:

ASt  =4(几) × 122 mm2

Area of brass tube:

ABr  =4(几) × (172 − 122) mm2

The load is distributed to the steel rod and the brass tube

PSt + PBr  = P                                                  (1)

E = → e =

P

G = A

Since the steel rod and the brass tube are jointed at the ends:

eSt  = eBr

Gst GBr

=

ESt       EBr

PSt PBr

ASt ABr

=

ESt        EBr

ABr EBr

PBr  = PSt

ASt ESt

(2)-(1)

−PSt  = PSt − P

P                                     25 kN

PSt  = + 1 =4(几) × (几172 − 122) mm2 × 105 ×

4 × 122 mm2 × 215 × 109 Pa

PSt  = 16.759 kN

By substituting into (1)

PBT  = 8.241 kN

b) Stress in steel and brass

PSt 16.759 × 103 N

GSt  =       = 几 = 148.182 MPa

GBT  = =4(几) = 72.364 MPa

c) Extension of the composite bar

Since eSt  = eBT  = e

e = = = 0.000689

e =

ΔL = eLo  = 0.000689 × 550 mm = 0.379 mm

Question 2: Creep

1)  Using the Larson-Miller plot shown, calculate (showing all working) the expected stress required to fail this material by creep processes at [10 Marks]

–   520˚C and 100,000 hours

–   930˚C and 10,000 hours

2)  An engineering component made of the material whose data is plotted as an LMP below is expected to last up to 155,000 hours at an applied stress of 100 MPa. [10 Marks]

–   Calculate the temperature at which this component must operate below to satisfy       these requirements?

–   Is it reasonable to use this    number given the maximum test time was less than         10,000 hours?

1) Expected stress required to fail this material by creep processes

LMP = T(log(tf ) + C)

LMP = T(20 + log(tf ))

a)

T = 520 °C = 520 + 273.15 = 793.15 K                       tf  = 100,000 ℎT

LMP = 793.15(20 + log(100,000)) = 19828.75

From the Larson-Miller plot

G ≈ 370 MPa

b)

T = 930 °C = 930 + 273.15 = 1203.15 K

tf  = 10,000 ℎT

LMP = 1203.15(20 + log(10,000)) = 28875.6

From Larson-Miller plot

G ≈ 23 MPa

2)

Temperature at which this component must operate below to satisfy requirements

tf  = 155,000 ℎT                                          G = 100 MPa

From Larson-Miller plot

LMP = 25000

25000 = T(20 + log(155,000))

Is it reasonable to use this number given the maximum test time was less than 10,000 hours?

Yes, it is reasonable to use this number because the LMP value of 25000 corresponding to    100 MPa was obtained from the Larson-Miller plot which is based on both theory and           experimental results, and is within the operational range 15000 < LMP < 30000. Since the LMP value is only dependent on lifetime and temperature, thus, these two factors are proportional. Therefore, if the test time was less than 10,000 hours, by testing with a higher proportional temperature to failure, it validates the temperature value obtained for 155,000 hours.

Question 3: Fracture

A rock bolt is needed to improve the structural integrity of a rock wall created during an        excavation.  The total load required on the bolt is expected not to exceed 502 kN.  During the bolt’s lifetime of 40 years, it is expected that cracks with a maximum size of 4 mm could       form within the bolt.  The material used is a high strength steel with a modulus of 210 GPa    and a yield strength of 320 MPa and fracture toughness of 20 MPa√m.

a)   Calculate the bolt diameter based on maximum load and the yield stress. [10 Marks]

b)  Calculate the bolt diameter based on fracture toughness assuming the crack geometry has a stress intensity factor of K = a , where a is the crack length. [15 Marks]

c)   In one sentence, why is this important? [5 Marks]

Pmax  = 502 kN

E = 210 GPa

tf  = 40 yeaTs

aY  = 320 MPa

2a = 4 mm

Kc  = 20 MPa√m

a = 2 mm

a) Bolt diameter based on maximum load and the yield stress

d = √ mm → d = 44.69 mm

b) Bolt diameter based on fracture toughness

K = a√几a ≤ Kc

amax  = a = √几(2) = 252.31 MPa

d = √4(几) mm → d = 50.33 mm

c) Importance

It is important because it the diameter size helps to predict if the bolt will fail by plastic        deformation or from crack propagation, in this case, if the material has an internal crack of 4 mm it will suffer catastrophic failure since the minimum bolt diameter required to withstand crack propagation is larger than the minimum diameter required to resist plastic deformation with the maximum load applied.

Question 4: Fatigue lifetime.

A controlled fatigue test was carried out on a specimen of aircraft grade Aluminium alloy. The resulting Stage 2 fatigue crack growth data is displayed in the figure below.  The alloy has a fracture toughness of 28 MPaVm.

The same material is now in service in an aircraft and recent Non Destructive Testing has found a through thickness crack of length 4.1 mm in a large plate section which experiences cyclic stresses of 0 MPa to 150 MPa. The material experiences this cyclic stress load with each take- off and landing, thus on average 6 cycles per day.

Useful equations to know: Paris Law of fatigue crack growth: = A∆Km ; Integral of a power: ∫ xn . dx = + C

(a) How many more cycles can the section be exposed to before catastrophic failure? [24 Marks NOTE: If you can’t solve completely, noting exactly which parameters are required and/or calculating those you can get you some marks]

(b) In less than 60 words, summarise if you think it would be reasonable to use this             component up to the number of cycles you’ve calculated?  (if you don’t have an answer from (a) assume its 1000 cycles to catastrophic failure) [6 Marks]

a) Number of cycles before catastrophic failure

= A∆Km

By selecting two points from the Stage 2 fatigue crack growth data plot

10 −6 = A(8.4)m

10 −8 = A(1.5)m

10 −8 = (1.5)

log(100) = mlog () → m = 2.673

A = = 3.383 × 10−9

To obtain Nf , rearrange the Paris Law of fatigue crack growth equation

NfdN = o(a)f = a)m

1           af m

A(Δo√几a)m   ao

Given:

ao  = 4.1 mm = 0.0041 m

omax  = 150 MPa                omin  = 0                 Δo = 150 MPa

Kc  = 28 MPa√m

Kc  = omax√几af

af  = (       )  ×   = (                )  ×    = 0.01109 m

1                           0.01109 m 2.673

3.383 × 10−9(150 MPa × √几)2.673   0.0041 m

a −0.3365 0.01109 m

0.0041 m

−0.01109−0.3365 + 0.0041−0.3365

0.3365