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Hey it's a me again @drifter1! Today we continue with Physics, and more specifically the branch of "Classical Mechanics". Today marks the end of the chapter about Equilibrium and Elasticity. This final article will contain Exercises on Elasticity and Plasticity. So, without further ado, let's get straight into it!
Let's consider a vertical pillar of length L, cross-sectional area A and mass density ρ. On top of it rests a sculpture of weight w.
Find the equation that describes the compressive stress at the cross-section located h (< L) below the top of the pillar, as well as the compressive strain of the top h-segment of the pillar.
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The total force that acts upon the h-segment is equal to the sum of the weight of the sculpture and weight of the pillar segment above the cross-section.
To calculate the weight of the pillar segment, we first have to calculate the volume of the pillar segment and its mass.
Thus, the weight of the pillar segment is:
The total normal force acting on the cross-section located h m below the sculpture is:
The compressive stress is equal to the division of the normal force by the cross-sectional area A:
The compressive strain can be found using Young's Modulus Y for the specific material, and is equal to the division of the stress by Y:
From the results we conclude that the normal force acting on the cross-sectional area is not constant along its length, but increasing as the height of the segment h increases.
Therefore, the stress is larger at the base of the pillar, and smaller at the top.
Consider a suspension cable that is used to carry gondolas at ski resorts of length L. Assume that the diameter of the cable is d = 2r and the maximum tension it can withstand is T.
Calculate the maximum stretch of the cable.
To calculate the maximum stretch we apply a force equal to the maximum tension, F = T.
The cross-sectional area of the cable is:
To calculate the change in length (or stretch in our case) we use Hooke's Law combined with Young's Modulus:
Let's suppose that a hydraulic press is filled with V liters of fluid. The fluid is subjected to Δp pressure increase and its Bulk Modulus is B.
Find the bulk strain and absolute decrease in volume of the fluid when the press is operating (in relation to V, Δp and k).
From the definition of Bulk Modulus, and knowing that bulk stress is basically Δp, solving for the bulk strain we get:
Again from the definition of Bulk Modulus, knowing the equation for bulk strain we can calculate ΔV:
From the results we conclude that the bulk strain can be expressed by only using the pressure/strain and Bulk Modulus of the fluid. The volume change ΔV can be expressed by using only on the bulk strain and initial volume of the fluid.
Let's consider that a massive puzzle of mass 10 kg is hanging from a steel nail with a diameter of 2 mm and from a distance to the wall of L0 = 5 mm.
Calculate the deformation of the nail.
[Custom Figure based on Figure 7 of Reference 2]
The force F on the nail (negleting the nail's own weight) is only the weight of the picture w:
The Shear Modulus of steel is known and equal to: S = 80 × 109 N/m2.
All quantities are known for the Shear Modulus equation and so we can easily calculate the nail deformation/bending Δx:
The bending is clearly undetectable to the eye. Nail strong. Really strong.
Mathematical equations used in this article, where made using quicklatex.
