Mman2400-June 2007 Paper

Topics: Elasticity, Shear stress, Shear modulus Pages: 12 (595 words) Published: June 11, 2010
The University of New South Wales School of Mechanical and Manufacturing Engineering

June 2007

MMAN 2400-Mechanics of Solids 1

1. Time allowed -Three ( 3 ) hours. 2. Reading Time - 10 Minutes. 3. This examination paper has 11 pages. 4. Total number of questions - Eight (8) 5. Attempt 6 questions (3 questions from part A and 3 questions from part B) out of 8. 6. Questions are of equal value -Total marks

=60.

7. Answers must be written in ink. Except where they are expressly required , pencils may ONLY be used only for drawing, sketching or graphical work. 8. Candidates may bring drawing instruments to the examination. 9. Candidates may NOT bring their own calculators or computers to the examination. 10. The following material will be provided by the Examination Unit: CASIO fx-911 W Calculator. 11. This paper may be retained by the candidate.

Page 1

PART A
Ql.

=

The boom is supported by the winch cable that has a diameter of 6 mm and an allowable normal stress of O"allow 170 MPa. Determine the greatest load including the cage that can be supported without causing the cable to fail when 0 = 30°, and

...

3
The~ factors are for

~

c:
0

.~

'" ~ $
C
U

axial loading but may be used lor bending when "/2r> 3

u

0

! -2

Page 7

(A)

Transformation of Stresses in 2-Dimension

(I)

----"--y-'-cos 28 - r xy sin 28

(a x -a )
2

(2)

(3)

(4)

(5)

aavg

=a c =

ax +ay
2

R=

ax -ay
(

2

2
:

+ r xy

2

max =r·tn-plane

(6)

al

= aavg +R

a2

= aavg- R
a max and a3

(7)

abs r max
(B)

a max -amin

2

;

0'"min are chosen from aI, (72 and

(8)

Transformation of Strains in 2-Dimension

(9)

(10)

(II)

(c x - CY ) r- . ------=--cos 28 - -xy Sin 28 2 2
Page 8

(12)

Y ---:L =
I I

_ \ x

IG

- G

y sin 28 + xy cos 28

)

Y

222
max Yin-plane

(13)

2

(14)

(15)

abs Ymax
(C)

= Gmax

- Gmin ; Gmax and Gmin are chosen from Gl, G2 and G3

(16)

Stress-strain Relationships - Generalised Hooke's Law

(17)

(18)

(19)

(20)

Inverting the above equations:

(l7a)

(18a)

(l9a)

'xy

= Gyxy

;

'yz

= Gyyz

;

':ox

= Gy;;x

(20a)

For principal values, subscripts x, y and;; may be replaced by subscripts 1,2 and 3 respectively.

Page 9

(D)

Some Definitions

ev

=-- =- =
K E

/). V V

0"

(0" x

+ 0" Y + 0"
3K

J

=

volumetrIc stram

..

(21 )

G=

21+v

(

)

=

shear modulus

(22)

E K = ( ) 31-2v

=

bulk modulus

(23)

(24)

+Ud = - -V ~ 0"1 - 0"2 )2
6E

-

(1

H(

+ (0"2

- 0"3 )2

+ (0"3

].. . - 0"1 ) 2 dIstortIon energy densIty =

(25)

(26)

(27)

where 0"nand

Tt

are the normal and shear stresses respectively on an oblique plane.

octahedral normal stress

(28)

'oct =-

1 3
Principal Stresses in 3-Dimension and Failure Criteria

octahedral shear stress

(29)

(E)

(30)
Three roots of this cubic equation gives the principal stresses in 3-dimension.

(0"max - 0"min)

=

Tresca equivalent uniaxial stress

(31 )

Page 10

1

J2
0" a

[(0"1 - 0"2

f + (0"2 - 0"3 f + (0"3 - 0"1 f ]
1
Gerber criterion

=

Mises equivalent uniaxial stress

(32)

+(

0" m J2 =

(33)

Se
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