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electr

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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 length 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