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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 GNm2
Modulus of Rigidity – 90 GNm2
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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