Question

g 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 = 570 mm, and the elastic modulus of the material is 77 GPa. If the average normal stress in the central portion of the bar is 200 MPa, calculate the overall elongation δ of the bar.

Answer 1

the overall elongation δ of the bar is 1.2337 mm

Step-by-step explanation:

From the information given :

According to the principle of superposition being applied to the axial load P of the system; we have:

`\delta = \delta_(AB) +\delta_(BC) + \delta_(CD)`

where;

δ = overall elongation

`\delta _(AB)` = elongation of bar AB

`\delta _(BC)` = elongation of bar BC

`\delta _(CD) =` elongation of bar CD]

If we replace;
`(PL)/(AE)` for δ and bt for area;

we have:

`\delta = (P_(AB)L_(AB))/((b_(AB)t)E) +(P_(BC)L_(BC))/((b_(BC)t)E)+(P_(CD)L_(CD))/((b_(CD)t)E)`

where ;

P = load

L = length of the bar

A = area of the cross-section

E = young modulus of elasticity

Let once again replace:

P for
`P_(AB), P_(BC) , P_(CD)` (since load in all member of AB, BC and CD will remain the same )

`(L)/(4)` for
`L_(AB)`,

`(L)/(2)` for
`L_(BC)` and

`(L)/(4)` for
`L_(CD)`

`2(b)/(3)` for
`b_(BC)`

b for
`b_(CD)`

`\delta = (P ((L)/(4)))/(btE)+ (P ((L)/(2)))/(2 (b)/(3)tE)+(P ((L)/(4)))/(btE)`

`\delta = (PL)/(btE)[(1)/(4)+ (1)/(2)*(3)/(2)+ (1)/(4)]`

`\delta = (5)/(4)(PL)/(btE) --- \ (1)`

The stress in the central portion can be calculated as:

`\sigma = (P)/(A)`

`\sigma = (P)/((2)/(3)bt)`

`\sigma = (3P)/(2bt)`

So; Now:

`\delta = (5)/(4)* (2 * \sigma)/(3)*(L)/(E)`

`\delta= (5)/(4)* (2 * 200)/(3)*(570)/(77*10^3 \ MPa)`

δ = 1.2337 mm

Therefore, the overall elongation δ of the bar is 1.2337 mm