
Two strips of metal are riveted together at their ends by four rivets, each of diameter $6mm$. What is the maximum tension that can be exerted by the riveted strip if the shearing stress on the rivet is not to exceed $6.9 \times {10^7}Pa$? Assume that each rivet is to carry one quarter of the load.
Answer
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Hint: Shearing is the process of layers that are parallel to each other and sliding past each other. The force per unit area for deformation is shear stress. Here diameter of each rivet is given using it to calculate its radius. Then using the above data determine the maximum tension that can be exerted by the riveted strip.
Formula used:
${\text{Stress = }}\dfrac{{{\text{Shearing force}}}}{{{\text{Area}}}}$
Complete step by step solution:
Shear stress is a force that tends to cause the deformation of a material by slippage along planes. The shear plays an important role in nature, it is closely related to the downslope movement of earth materials. Shear stress can also occur in solids or liquids.
Shear stress is formed at a boundary when any fluid is moving along the vicinity of a solid boundary. The speed of the fluid at the boundary is zero when the no-slip condition dictates. The region between these two points is called the boundary layer.
Suppose the area of each rivet on which the shearing force acts
Shearing stress is given by
${\text{Stress = }}\dfrac{{{\text{Shearing force}}}}{{{\text{Area}}}}$
Stress $ = \dfrac{T}{{4A}}$
$T$ is the maximum tension exerted by the riveted strip,
$\dfrac{T}{{4A}} = 6.9 \times {10^7}$
Simplifying we get,
$ \Rightarrow T = 4A \times 6.9 \times {10^7} - - - - - \left( 1 \right)$
Area of the rivets,
$\Rightarrow$ $A = 3.144 \times {\left( {3 \times {{10}^{ - 3}}} \right)^2} - - - - \left( 2 \right)$
From equation 1 and 2 we get
$\Rightarrow$ $T = 4 \times 3.14 \times {\left( {3 \times {{10}^{ - 3}}} \right)^2} \times 6.9 \times {10^7} = 7.8 \times {10^3}N$.
Note: The strain rate in the fluid is proportional to the shear stress. A shear stress is formed at a boundary when any fluid is moving along the vicinity of a solid boundary. The viscosity is not constant for non-Newtonian fluid. Shear stress can also occur in solids or liquid. Shearing is the process of layers that are parallel to each other and sliding past each other.
Formula used:
${\text{Stress = }}\dfrac{{{\text{Shearing force}}}}{{{\text{Area}}}}$
Complete step by step solution:
Shear stress is a force that tends to cause the deformation of a material by slippage along planes. The shear plays an important role in nature, it is closely related to the downslope movement of earth materials. Shear stress can also occur in solids or liquids.
Shear stress is formed at a boundary when any fluid is moving along the vicinity of a solid boundary. The speed of the fluid at the boundary is zero when the no-slip condition dictates. The region between these two points is called the boundary layer.
Suppose the area of each rivet on which the shearing force acts
Shearing stress is given by
${\text{Stress = }}\dfrac{{{\text{Shearing force}}}}{{{\text{Area}}}}$
Stress $ = \dfrac{T}{{4A}}$
$T$ is the maximum tension exerted by the riveted strip,
$\dfrac{T}{{4A}} = 6.9 \times {10^7}$
Simplifying we get,
$ \Rightarrow T = 4A \times 6.9 \times {10^7} - - - - - \left( 1 \right)$
Area of the rivets,
$\Rightarrow$ $A = 3.144 \times {\left( {3 \times {{10}^{ - 3}}} \right)^2} - - - - \left( 2 \right)$
From equation 1 and 2 we get
$\Rightarrow$ $T = 4 \times 3.14 \times {\left( {3 \times {{10}^{ - 3}}} \right)^2} \times 6.9 \times {10^7} = 7.8 \times {10^3}N$.
Note: The strain rate in the fluid is proportional to the shear stress. A shear stress is formed at a boundary when any fluid is moving along the vicinity of a solid boundary. The viscosity is not constant for non-Newtonian fluid. Shear stress can also occur in solids or liquid. Shearing is the process of layers that are parallel to each other and sliding past each other.
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