Designing a Tensile Test for an Aluminum Rod to Withstand 200kN Force

[Elitisi Piuela]

I'm asked to design a rod, it is aluminium and it is to withstand force of 200kN.
maximum allowable stress is 170MPa with a strain of 0.0025mm.mm^-1
rod must be at least 3.8m long but must deform elastically no more than 6mm when the force is applied.

UTS = 170 MPa
e = 0.0025mm.mm^-1
F = 200kN
Lf = 6mm + 3.8m (?)
Lo = 3796.5087mm (?)
Gauge length = Lo ??
I'm unsure about the gauge length, Lo and Lf
do i need to also consider the force 200kN

 

[haruspex]

it is to withstand force of 200kN.

Applied how? I'll assume this is tension.

maximum allowable stress is 170MPa

I'm not quite sure what that is saying. Is it that at a tension of 200kN the stress must be no more than 170MPa? What does that tell you about the cross-section?

with a strain of 0.0025mm.mm^-1

at least 3.8m long but must deform elastically no more than 6mm when the force is applied

I'm unclear how those two relate. The first seems to say that at 200kN tension the rod must not extend more than 0.25%, while the second says it must not extend more than 6/38 %, or about .15%.

do i need to also consider the force 200kN

Since it must withstand that force, that seems a rather crucial consideration.

 

[Elitisi Piuela]

I've come up with Lo = 3796.51mm and Ao = 0.0012msquared
using the formulas:
e = Δgaugelength/initial length (taking that Δgaugelength = 3.8m + 6mm)
σ = P/Ao (P=0.2MN , σ = 170MN/msquared)

is this right?

 

[haruspex]

I've come up with Lo = 3796.51mm and Ao = 0.0012msquared
using the formulas:
e = Δgaugelength/initial length (taking that Δgaugelength = 3.8m + 6mm)
σ = P/Ao (P=0.2MN , σ = 170MN/msquared)

is this right?

That satisfies the max allowable stress, but what will the extension be if at that stress it extends 2.5 mm/m?