shear\nexample 4 a riveted tie - in acted by force p = 110kn is shown in the figure.\nknowing the thickness…

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})).