Question

A rectangular bar of length L has a slot in the central half of its length. The bar has width b, thickness t, and elastic modulus E. The slot has width b/3. The overall length of the bar is L = 510 mm, and the elastic modulus of the material is 68 GPa. If the average normal stress in the central portion of the bar is 170 MPa, calculate the overall elongation δ of the bar.

Answer 1

Question:

The question is incomplete. The width and radius was not given. For this answer, take the radius of slot as 8mm and the width as 45mm

Answer:

Total elongation = 4.426 * 10⁻³mm

Explanation:

Given Data:

L = 510 mm

E = 68 Gpa = 68 * 10^3N/mm²

b = 45 mm

t = 8 mm

P = 170 MPa

Calculating the cross sectional area for section A, we have

A₁= b*t

 = 45 * 8

= 360mm²

Calculating the deformation of section A, we have

б₁ = (P (L/4))/(A₁E)

Substituting, we have

б₁ = (170*(510/4))/(360*68*10³)

    = 21675/24480000

   = 8.85*10⁻⁴mm

The cross sectional area of section 2 is given as

A₂ = 2/3 * b* t

Substituting, we have

A = 2/3 * 45 * 8

   =240mm

The deformation of section 2 is given by the formula,

б₂ = P (L/2)/A₂E

Substituting, we have

б₂ =  (170*(510/2))/(240*68*10³)

    = 43350/16320000

    = 2.656 * 10⁻³mm

Note, the cross sectional area of section 3 is the same with that of section 1, therefore, deformation for section 1 and section 3 are the same.

б₃ = 8.85*10⁻⁴mm

Total elongation = б₁ +б₂ + б₃

                           =8.85*10⁻⁴mm +2.656 * 10⁻³mm + 8.85*10⁻⁴mm

                           = 4.426 * 10⁻³mm

Answer 2

Answer:

δ = 1.0625 mm

Explanation:

Divide the rectangular bar into three segments AB, BC and CD as shown in the pic.

The elongation of bar will be:

δ = δAB + δBC + δCD

then

δ = (P*(L/4)/(AAB*E)) + (P*(L/2)/(ABC*E)) + (P*(L/4)/(ACD*E))

δ = (P*(L/4)/(b*t*E)) + (P*(L/2)/((2b/3)*t*E)) + (P*(L/4)/(b*t*E))

δ = (P*L/(E*b*t))*(1/4 + 3/4 + 1/4)  

δ = = (5/4)* (P*L/(b*t*E))      (i)

If

L = 510 mm = 0.51 m

E = 68 GPa = 68*10⁹ Pa

σ = 170 MPa = 170*10⁶ Pa (the average normal stress in the central portion of the bar)

Stress in middle region L is

σ = P/A = P/((2b/3)*t) = (3/2)*(P/(b*t))

then

(P/(b*t)) = (2/3)* σ = (2/3)* 170*10⁶ Pa = 113.33*10⁶ Pa

Plug this value in equation (i)

δ = (5/4)*(113.33*10⁶ Pa)*(0.51 m/68*10⁹ Pa) = 0.001 m = 1.0625 mm