Question

example 4 a riveted tie - in acted by force p = 110kn is shown in the figure. knowing the thickness is t = 1cm, width of it is b = 8.5cm. diameter of the rivet is d = 1.6cm and the permissible shearing stress is τ=140mpa, the permissible bearing stress is σ_b=320mpa. try to check the strength of the riveting. (assume the force acted on each rivet is equal.)

Explanation:

Step1: Calculate shear - force on each rivet

There are 4 rivets. The total force $P = 110kN$. So the shear - force on each rivet $Q=\frac{P}{4}$.
$Q=\frac{110\times10^{3}}{4}=27500N$

Step2: Calculate shear - area of each rivet

The shear - area of a circular cross - section of the rivet $A_s=\frac{\pi d^{2}}{4}$, where $d = 1.6cm=0.016m$.
$A_s=\frac{\pi\times(0.016)^{2}}{4}\approx2.01\times10^{-4}m^{2}$

Step3: Calculate actual shear stress

The actual shear stress $\tau=\frac{Q}{A_s}$.
$\tau=\frac{27500}{2.01\times10^{-4}}\approx136.82MPa$
Since $\tau = 136.82MPa<[\tau]=140MPa$, the shear - strength is satisfied.

Step4: Calculate bearing area of each rivet

The bearing area of each rivet $A_b = td$, where $t = 1cm = 0.01m$ and $d = 1.6cm=0.016m$.
$A_b=0.01\times0.016 = 1.6\times10^{-4}m^{2}$

Step5: Calculate actual bearing stress

The actual bearing stress $\sigma_b=\frac{Q}{A_b}$.
$\sigma_b=\frac{27500}{1.6\times10^{-4}}=171.875MPa$
Since $\sigma_b = 171.875MPa<[\sigma_b]=320MPa$, the bearing - strength is satisfied.

Answer:

The strength of the riveting is satisfied as both the shear - stress and the bearing - stress are within the permissible values.