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.

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

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