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Stress and strain

  • Thread starter electr
  • Start date
  • #1
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Homework Statement


The component shown in Fig 1 is made from a material with the following properties and is subjected to a compressive force of 5kN.
Material Properties :
Young’s Modulus of Elasticity – 200 GNm-2
Modulus of Rigidity – 90 GNm-2
Poisons ratio – 0.32
Calculate :
(a) The stress in :
(i) the circular section
(ii) the square section
(b) The strain in :
(i) The circular section
(ii) The square section
(c) The change in length of the component
(d) The change in diameter of the circular section
(e) The change in the 40mm dimension on the square section
(f) If the same component were subjected to a shear force of 7 kN as shown in
FIG 2, calculate the shear strain in :
(i) The circular section
(ii) The square section
FIG

Homework Equations


Area of circle A=Πr^2
STRESS=FORCE/A
E=STREES/STRAIN


The Attempt at a Solution


a(i) stress is 7.073 Mpa
aii)stress is 3.125 Mpa

bi) 7.073 x10^6 /200 x10^9 =35365
bii)3.125 x10^6 / 200 x10^9 =15625

c) Now for the change in length of component Δl=e x lo
so 35365 +15625 x 120 x10^-3=19.86 mm
or circular section 35365 x 60 x 10^-3 = 2,121 mm and square section 15625 x 60 x10^-3 =0.937 mm and add them together 2,121 + 0.937=3.058 mm
or should i use Δl=Fxl0/AxE so for circular Δl= 5x10^3 x 60 x10^-3 / 0.000707 x 200 x 10^9 =0.00000212164 and sqare 5x10^3 x 60 x10^-3 / 0.0016 x 200 x 10^9= 0.0000009375 so total new length 0.00000305914,so new lenght 120 mm - 0.00000305914 = 119.999996941 ?
any of the above is correct?or if not any idea?

d)change in diameter i will find transverse strain εt=-u x e so εt= - 0.32 x 35365 = -11 136
so diameter Δd=εt x d0 so -11136 x 30 x10^-3 =-336.48 μm?

e) stress =5x10^3/ 0.0024=2 mpa
strain 2x10^6 x 200 x 10^9 =0.00001
Δl =0.00001 x 40 x10^-3=0.0000004 μm?

fi)7000/0.000707 =9.9 mpa
so strain is 9.9x10^6/ 90 x10 ^9 =0.00011
fii) 7000/0.0016 =4.3 mpa
so strain is 4.3 x10^6 /90 x10^9 = 0.0000047

is any of the above correct ?and if not i would like your help
 

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Answers and Replies

  • #2
227
58
Hi electr,

I agree with your solutions for part (a) .
In part b), you strain values are off by a factor of 1000
In part c) use the equation and calculate the change in length for both circle and the square, add them together for the final change in length, I get 0.006188 mm $$\varepsilon =\frac{\Delta L}{L}$$

In part d), use the change in length of the cylinder and Poisson's ratio to determine the change in the diameter and the new diameter.
In part e), use the change in length of the square and Poisson's ratio to determine the change in length of the square sides

To calculate shear stress , you use Force/Area , to get shear strain , divide by the modulus of rigidity (make sure your units are consistent).
 
  • #4
29
0
thank you for your reply,i found a solution when i got the right numbers from b part,and checked the link and i have the same as the other one
 

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