Final 2015, questions PDF

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PRIFYSGOL ABERTAWE SWANSEA UNIVERSITY College of Engineering SEMESTER 1 EXAMINATIONS JANUARY 2015

EG-M97

ADVANCED SOLID MECHANICS LEVEL M

University Calculators Only

Translation dictionaries are not permitted, but an English dictionary may be borrowed from the invigilator on request.

Time allowed:

2 hours Answer THREE questions

A datasheet is provided at the back of the paper.

TURN OVER

Page 1 of 10

Question 1 (a) Explain the Tresca and von Mises yield criteria in terms of their equations, how they differ and which one is the more conservative and why. By means of a sketch, show that a material subjected to three equal principal stresses (either all positive or all negative) will never yield. [5 marks] (b) A thick cylinder, inside radius R1 and outside radius R2, is subjected to an internal pressure, p. The hoop and radial principal stresses are defined by Lame’s equations: B B and σr  A  2 where σθ and σr are the hoop and radial principal stresses 2 r r respectively at radius r. A and B are constants which depend on the boundary conditions. Show that yielding (based on the Tresca criterion) first occurs in the thick cylinder when the σ  A 

internal pressure reaches a value:

σyield  R 2  R 1 2 2 R2 2

2

 [10 marks]

(c) Where does yielding first start ? [3 marks] (d) Show that the collapse pressure for the cylinder (based on an elastic-perfectly plastic material assumption) is given by:

σ yield loge

R2 R1 [7 marks]

The following equations can be used without proof: For the Tresca criterion:-

σθ  σr σyield

For radial equilibrium:-

σθ  σr  r

dσ r dr

(TOTAL 25 MARKS)

TURN OVER

Page 2 of 10

Question 2 (a) A circular hollow section propped cantilever beam, 1000 mm long, supports a load, W, at 200 mm from the prop as shown in Figure Q2. If the outside diameter of the tube is 50 mm and the inside diameter is 25 mm, calculate the yield moment and collapse moment in terms of the yield stress, σyield. Hence find the shape factor for the cross-section. [10 marks] (b) For this beam configuration, use the moment-area method for indeterminate beam structures to find the collapse mechanism and hence the value of the load W if the yield stress for the beam material is 250 MPa. [10 marks] (c) A steel plate, 300 mm long by 70 mm wide by 2 mm thick, is subjected to a compressive load along its shorter edges and its long edges are unsupported. Assuming pin-jointed loaded ends and values for Young’s modulus and Poisson’s ratio of 200 GPa and 0.3 respectively, determine the critical load: (i) using the Euler Buckling theory; (ii)

considering the plate as a wide strut;

By what percentage does the Euler theory under/over predict the value compared with the wide strut assumption? [5 marks]

W

L Figure Q2

(TOTAL 25 MARKS)

TURN OVER Page 3 of 10

Question 3 (a) Sketch the stress-total strain and stress-time curves for the tensile test if the test bar is initially loaded, at a constant rate of change of stress, up to a point beyond the elastic limit, after which the strain is held fixed and uniaxial creep occurs. Explain why this might be a problem in a jointed pipe connection. [5 marks] (b) A uniaxial creep test is to be modelled using finite element analysis. The required predictions are for a specimen subjected to a tensile stress of 150 MPa for a period of 100 hours and then this stress is increased to 250 MPa for a further 50 hours. Determine the predicted accumulated creep strain over 150 hours if a time hardening assumption for variable stress creep is assumed. Assume that the temperature remains constant and a NortonBailey creep law for the material at that temperature of:

cr = 3 x 10-13 σ4 t0.8

(time in hours, stress in MPa) [5 marks]

(c) Repeat part (b) using a strain hardening assumption for variable stress creep and determine the difference in accumulated creep strain predictions over the 150 hours for the two hardening assumptions. Show how these two assumptions are represented on a sketch of the creep strain-time curve for each assumption. [5 marks] (d) A solid circular shaft, 2R diameter, is subjected to an elastic torque, T, and is operating in the creep regime. Working from first principles, determine the equation for the redistributed shear stress distribution, due to creep, if a Norton-Bailey power law for secondary creep (m = 1) is assumed. As a starting point, the following can be assumed: cr 

r  A  n t m at any radius r L

A solid circular shaft is 30 mm diameter and is subjected to an initial elastic torque of 1 kNm. By what percentage will the maximum shear stress reduced due to creep, if the creep law is:

γcr = 1.5 x 10-17 τ4 t

(time in hours, stress in MPa) [10 marks] (TOTAL 25 MARKS) TURN OVER

Page 4 of 10

Question 4 Figure Q4 shows a typical bolted connection used to connect a hook in order to suspend an external load. A permanent connection, using two M16 hexagonal nuts and bolts either side of the hook together with thin rigid washers, is used to join the two 25 mm thick plate members. Details of the bolt and nut are given in Table Q4.

Area of unthreaded section Ad (mm2) 200

Tensile stress area At (mm2)

Unthreaded length Ld (mm)

Proof strength Sp (MPa)

Yield strength Sy (MPa)

154

25

650

720

(a) Calculate the preload, Fi (based on the permanent connection assumption of 0.9.Sp). [2 marks] (b) If the bolted joint is subjected to an external load P, working from first principles, show that: (i)

Pb = CP

(ii)

Pm =(1-C)P

where Pb and Pm are the bolt and members load respectively and C is the stiffness parameter, Kb . Kb  Km [5 marks] (c) Calculate the values of Kb, Km and C for this bolted joint. [9 marks] (d) If an external load of 10 kN is applied to the joint, calculate the safety factor for the bolts and the safety factor for the plates to avoid separation. [5 marks] Page 5 of 10

Assume E= 200 GPa, dw = 1.5 d and pressure distribution angle, degrees. (e) Sketch a typical strain amplitude v number of cycles to failure curve for steel (where the total strain amplitude is also separated into its elastic and plastic components) and use this sketch to explain the differences between high and low cycle fatigue. [4 marks]

(TOTAL 25 MARKS)

TURN OVER

FIGURE Q4 – NOTE: 2 BOLTS ARE USED

Area of unthreaded section Ad (mm2) 200

Tensile stress area At (mm2)

Unthreaded length Ld (mm)

Proof strength Sp (MPa)

Yield strength Sy (MPa)

154

25

650

720

Table Q4- Dimensions and properties of the M16 bolt

Page 6 of 10

END OF PAPER

Page 7 of 10

EG-M97 DATA SHEET 1. Yield Criteria (yielding occurs when:) σ1  σ3 σ yield

a) Tresca:

b) von Mises: σeq = σyield where:

σ eq 

1

or:

σ eq 

1 2

2

 σ1 

σ

σ2 2   σ 2  σ3 2   σ1  σ3 2



 σyy   σ yy  σ zz    σ xx  σ zz   6 τ xy  τ yz  τ xz 2

xx

2

2

2

2

2



2. Plastic Bending a) Rectangular cross-sections:

M yield

σ yieldbd 2  6

M plastic 



σ yield b 2 d  2ad  2a2 6



σ bd 2 M collapse  yield 4

b) Circular cross-sections: M yield

 yield  D 4  d 4   32 D

c) Shape factor:

M collapse

SF 

 yield  D 3  d 3   6

M collapse Myield

d) Residual stresses: Residual stress distribution = applied stress distribution - equivalent elastic stress distribution e) Residual curvature: 1 R res

Rpl

1   1  R pl 

 0.5 d  a  0.5 d

2

 

= the radius of curvature in the plastic condition (i.e. based on the bending equation for the elastic core only)

3. Plastic Torsion a) Solid shaft:

π R3 τyield b) Hollow Tyield  tube: 2

πτ Tplastic  yield  4 R3  R3p  6 Page 8 of 10

Tcollapse

2π R 3 τ yield  3

T yield 







πτ 2π τ yield 3 4 3 3 π τ yield 4 R o Rp  R4p  3 RTi collapse  Ro  Ri  R o 4  R i 4 T plastic  6 Ryield 3 p 2 Ro

T SF  collapse Tyield

c) Shape factor: d) Residual twist:

Residual angle of twist = actual twist (based on the elastic core) - equivalent elastic twist 4. Buckling of Struts and Columns

 EA π2 E I F F 2 →  K L    2 where λ is the slenderness ratio 2

a) Euler theory:

 K L

2 

2

A

I F

b) Engesser modification:

 2 Et I  K L 2

2EI 1  2  a 2

Fcrit 

c) Plates considered as wide struts:

5. Buckling of Plates: a) Plates with long edges unsupported and one-directional loading: F crit 

π2 a2 D   m    m 2   a 

2



2  n      b  

where:

E t3 D 12 1  ν 2 

b) Plates with bi-directional loading and all edges supported:

σcrit



F D π2  m/a   n/b  crit  bt t m/a 2  α n/b 2



2

2

 

2

6. Creep a) Norton-Bailey power law: b) Sinh law: c) Dorn:

ε cr A σ n t m

 σ ε cr A sinh   t m  n

ε cr f1 σ), f 2 (t e Q/RT 

Page 9 of 10



d) Creep relaxation in a bolt (or rivet):

t

 1 1 1   n  1  n 1  σ0  EA  n  1  σ t

   0n 1   t n 1   n 1    EA n t 1 1   0  



or



e) Creep during pure bending of a rectangular bar: 1

M y  2n 1  2y  n σ(y)  . .  I 3n  d

1

f) Creep during torsion of a hollow circular shaft: T Jcr



where

τ(r) r 1/n

J cr

1 3   3 1 2 π n  R o n  Ri n     1  3n 

7. Low Cycle Fatigue  eq ' f  Nf  b  'f  Nf  c  2 E

a) Coffin-Manson equation:

b) Notch stress-strain conversion rules: K t2 = K σ K ε

i)

Neuber rule:

ii)

Linear rule:

iii)

Intermediate rule: K ε Kt

Kt = Kε K    t   Kσ 

I

8. Bolted Joints Under Tensile Loading (a) Bolt stiffness:

1 1 1   Kb K t K d

or

Kb 

(b) Member stiffness:

Page 10 of 10

Kt Kd Kt  Kd

or

Kb 

A tA dE At L d  Ad L t

(c) Load distribution: For the bolt: For the members:

Fb= Pb + Fi F m= P m - F i

Pb = C P Pm = (1 – C) P

 Kb     K b K m 

C=

(d) Safety factors: Sp

For the bolt:

For the members:

nCP Fi  At At

or

S A  Fi n p t CP

Po = Fi/(1 – C)

n= Po /P

9. Bolted Joints Under Shear Loading (includes rivets) (a) Bolts/rivets: (i) Bending:

M = FL/2

max = Md/2I

(ii) Single shear:

 = F/A

(iii) Double shear:

 = F/2A

(b) Members:  = F/A diameter)

for tension (net area) and crushing (thickness of thinner plate x bolt/rivet

Page 11 of 10...