Question

shear
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: Determine Force per Rivet

Total force \( P = 110\,\text{kN} \). From the figure, there are \( n = 4 \) rivets. So force per rivet \( F=\frac{P}{n}=\frac{110}{4} = 27.5\,\text{kN}=27500\,\text{N} \).

Step2: Check Shearing Stress

Shear area for one rivet: each rivet is in double shear? Wait, no, looking at the diagram, maybe single shear? Wait, the rivet diameter \( d = 1.6\,\text{cm}=0.016\,\text{m} \), thickness \( t = 1\,\text{cm}=0.01\,\text{m} \). Wait, shear stress formula: \( \tau=\frac{F}{A_s} \), where \( A_s=\pi d^2/4 \) (for single shear, but if double shear, \( 2\times\pi d^2/4 \)). Wait, the diagram shows two plates, so each rivet is in double shear? Wait, no, the top plate and bottom plate: each rivet connects two plates, so shear area is \( 2\times(\pi d^2/4) \). Wait, let's recalculate: \( d = 1.6\,\text{cm}=16\,\text{mm} \), so \( A_s = 2\times\frac{\pi\times(16)^2}{4}=2\times64\pi = 128\pi\approx402.12\,\text{mm}^2 \). Then shear stress \( \tau=\frac{F}{A_s}=\frac{27500\,\text{N}}{402.12\times10^{-6}\,\text{m}^2}\approx68.4\,\text{MPa} \). Wait, no, maybe I messed up the number of rivets. Wait the figure: top view has 4 rivets? Wait the first diagram: 4 rivets (2x2). So \( n = 4 \), so \( F = 110/4 = 27.5\,\text{kN} \). Now, shear stress: each rivet is in double shear? Wait, the second diagram: two plates, so each rivet is sheared twice? So shear area per rivet is \( 2\times\frac{\pi d^2}{4} \). \( d = 1.6\,\text{cm}=0.016\,\text{m} \), so \( A_s = 2\times\frac{\pi\times(0.016)^2}{4}=2\times\frac{\pi\times0.000256}{4}=0.000128\pi\approx4.0212\times10^{-4}\,\text{m}^2 \). Then \( \tau=\frac{27500}{4.0212\times10^{-4}}\approx68.4\,\text{MPa} \), which is less than \( [\tau]=140\,\text{MPa} \), so shear stress is okay.

Step3: Check Bearing Stress

Bearing area \( A_b = d\times t \). \( d = 1.6\,\text{cm}=0.016\,\text{m} \), \( t = 1\,\text{cm}=0.01\,\text{m} \), so \( A_b = 0.016\times0.01 = 1.6\times10^{-4}\,\text{m}^2 \). Bearing stress \( \sigma_b=\frac{F}{A_b}=\frac{27500}{1.6\times10^{-4}}\approx171.875\,\text{MPa} \). Wait, but \( [\sigma_b]=320\,\text{MPa} \), so \( 171.875 < 320 \), so bearing stress is also okay? Wait, maybe I made a mistake in the number of rivets. Wait, maybe the rivets are in single shear? Let's recheck. If single shear, \( A_s=\frac{\pi d^2}{4}=\frac{\pi\times(0.016)^2}{4}=6.4\times10^{-5}\pi\approx2.0106\times10^{-4}\,\text{m}^2 \). Then \( \tau=\frac{27500}{2.0106\times10^{-4}}\approx136.8\,\text{MPa} \), which is less than 140, so that's close. Wait, maybe the diagram has 4 rivets, but maybe the shear is double? Wait, the problem says "the force acted on each rivet is equal", so we have to check shear and bearing.

Wait, let's redo with \( d = 1.6\,\text{cm}=16\,\text{mm} \), \( t = 1\,\text{cm}=10\,\text{mm} \), \( F = 110/4 = 27.5\,\text{kN}=27500\,\text{N} \).

Shear stress: if each rivet is in double shear (two shear planes), then \( A_s = 2\times\frac{\pi d^2}{4}=\frac{\pi d^2}{2} \). \( d = 16\,\text{mm} \), so \( A_s=\frac{\pi\times16^2}{2}=128\pi\approx402.12\,\text{mm}^2 \). Then \( \tau=\frac{27500\,\text{N}}{402.12\,\text{mm}^2}\approx68.4\,\text{MPa} < 140\,\text{MPa} \), so shear is safe.

Bearing stress: \( A_b = d\times t = 16\,\text{mm}\times10\,\text{mm}=160\,\text{mm}^2 \). \( \sigma_b=\frac{27500\,\text{N}}{160\,\text{mm}^2}\approx171.875\,\text{MPa} < 320\,\text{MPa} \), so bearing is also safe. Therefore, the riveting is strong enough.

Answer:

The riveting is strong enough as both the shearing stress (\(\approx68.4\,\text{MPa}\) or \(\approx136.8\,\text{MPa}\) depending on shear type) is less than the permissible shearing stress (\(140\,\text{MPa}\)) and the bearing stress (\(\approx171.875\,\text{MPa}\)) is less than the permissible bearing stress (\(320\,\text{MPa}\)).