Solution to Problem 208 Axial Deformation

$\delta = \dfrac{PL}{AE}$
 

Where:
δ = π (1500.5 - 1500) = 0.5π mm
P = T
L = 1500π mm
A = 10(80) = 800 mm²
E = 200 000 MPa
 

Thus,
$0.5\pi = \dfrac{T( 1500 \pi )}{800(200\,000)}$

$T = 53\,333.33 \, \text{N}$
 

$F = 2T$

$p(1500)(80) = 2(53\,333.33)$

$p = 0.8889 \, \text{MPa}$       → internal pressure
 

Total normal force, N:
N = p × contact area between tire and wheel
N = 0.8889 × π(1500.5)(80)
N = 335 214.92 N

Friction resistance, f:
f = μN = 0.30(335 214.92)
f = 100 564.48 N = 100.56 kN
Torque = f × ½(diameter of wheel)
Torque = 100.56 × 0.75025
Torque = 75.44 kN · m