# How Do You Calculate Tractive Force for Accelerating a Car?

• Gixer1127
In summary: Yes, this is correct. The average acceleration is calculated as the net force (tractive force + wind resistance) required to achieve the desired velocity.
(b) Tensile strain (ɛ) = σ/E

Modulus of elasticity (E) of the bolt is 160GPa.

ɛ = 34468656.72Pa / 160x10⁹GPa = 0.0002154 (2.154x10¯⁴) (No units of measurement for strain)

(c) Extension = ɛ x original length

Extension = 0.0002154 x 20mm = 0.004308mm(4.308x10¯³)

I'm thinking the above in bold is not correct? Should I be converting the 20mm to 0.02m and working it that way? In which case the answer should be 4.308x10¯⁶ (0.000004308mm).

Gixer1127 said:
(b) Tensile strain (ɛ) = σ/E

Modulus of elasticity (E) of the bolt is 160GPa.

ɛ = 34468656.72Pa / 160x10⁹GPa = 0.0002154 (2.154x10¯⁴) (No units of measurement for strain)

(c) Extension = ɛ x original length

Extension = 0.0002154 x 20mm = 0.004308mm(4.308x10¯³)

I'm thinking the above in bold is not correct? Should I be converting the 20mm to 0.02m and working it that way? In which case the answer should be 4.308x10¯⁶ (0.000004308mm).
Nope, going with 20mm, not 0.02m.

Gixer1127 said:
Nope, going with 20mm, not 0.02m.
Where those 20 mm come from?
34468656.72Pa may be the wrong amount, based on the erroneously calculated cross-section area.
160x10⁹ GPa is not correct; it is either 160x10⁹ Pa or 160 GPa.

Sorry, yes, I missed out in my original post where the 20mm comes into play.
The 16 mm diameter steel bolt with modulus of elasticity of 160 GPa and modulus of rigidity of 90 GPa shown in the diagram below holds two components together. The thickness of the bolted interface of the components is 20 mm.
And the question reads: (c) Calculate the change in length of the 20mm bolted interface of the bolt.
I'll change, in my calculation summary, to 160x10⁹ Pa, thank you.
I'm going to stick to my original answer, albeit with the correct decimal point placement. (34468656.72Pa) In my original calculation I had not input the data correctly in my calculator which gave me an incorrect decimal point placement.
So here's my final answers below with the 3 questions I got wrong due to my first answer being incorrect.

 Calculate the direct stress in the bolt at the centre of the joined components. Calculate the tensile strain experienced by the bolt. Calculate the change in length of the 20mm bolted interface of the bolt.
Area of cross section of 16mm dia. bolt is.
A = πr² = π x 0.008² = 2.01x10¯⁴mm² (0.000201mm²)

(a)Direct stress (σ) = Force/Area

= 6928.2N / 2.01(10¯⁴)mm² = 34468656.72 Pa (0.3446865672x10⁸)

(b) Tensile strain (ɛ) = σ/E

Modulus of elasticity (E) of the bolt is 160GPa.

ɛ = 34468656.72Pa / 160x10⁹Pa = 0.0002153 (2.153x10¯⁴) (No units of measurement for strain)

(c) Extension = ɛ x original length

Extension = 0.0002153 x 20mm = 0.004306mm(4.306x10¯³mm)

Lnewqban
Gixer1127 said:
Area of cross section of 16mm dia. bolt is.
A = πr² = π x 0.008² = 2.01x10¯⁴mm² (0.000201mm²)
HUH?? How can the area be less than the diameter?

Hi Tom. I thought I'd posted a reply a couple of days ago but obviously didn't push the send button.
You are 100% correct, of course. I used the wrong unit of measurement so thank you for spotting this. I have changed this and submitted for marking.

Tom.G

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