Question 8.2: Repeat Example Analysis 1 using the following physical and g......
Repeat Example Analysis 1 using the following physical and geometric data:
σ_{max} = 200 MPa τ_{max} = 70 MPa
r = 6.3 mm h = 13 mm τ = 3.5 mm (a)
For convenience we show the line drawing of the chain link again here:
As before the objective is to find the theoretical maximum force F that the link joint can sustain. We can obtain this objective by following the solution procedure of Example 1.
To this end, in considering pin shear, the cross-section area of the pin, with the new given data is:
A_{pin}=\pi r^{2}=\pi \left(6.3\times 10^{-3} \right) ^{2}=124.7\times 10^{-6} m^{2} (a)
The shear stress τ on the pin is then:
τ=\frac{\left(F/2\right) }{A_{pin} } (b)
where, as before, with ideal geometry, each end of the pin need support only half the applied load F.
From the result of Eq. (a), τ is:
τ= 4.01F kPa (c)
With a maximum shear stress τmax being 70 MPa, the maximum applied force F_{max} before pin shear failure is:
F_{max}=\frac{τ_{max} }{4.01}=\frac{70,000}{4.01}=17.45 kN (d)
Next, for the tension on the link, the cross-section area of the link with the new data is:
A_{link}=\left(3.5\right)\left(13\right)\times 10^{-6}=45.5\times 10^{-6} m^{2} (e)
As before, the tensile stress σ in the links is:
\sigma =\frac{\left(F/2\right) }{A_{link} } (f)
where again each parallel link need support only half the applied load F. From the result of Eq. (e) σ is:
σ= 0.011F kPa
With a maximum tensile strength σ_{max} being 200 MPa the maximum applied force F_{max} before link tension failure is:
F_{max}=\frac{\sigma _{max} }{0.011}=\frac{200}{0.011}=18,182 N=18.182 kN (g)
By comparing the results of Eqs. (d) and (g), we see that the pin is likely to fail in shear before the links fail in tension. Therefore, the theoretical maximum pull load on the chain-link system is: 17.45 kN..