Mechanical Properties of Matter Concept Page - 5

Example

Force analysis for a beam supported at both ends

Example: An isolated part of a bridge of mass m$m$ is supported by two identical columns each of length l$l$, cross - sectional radius r$r$ and young's modulus Y$Y$. What should be the minimum cross - section radius r$r$, so that the beam bearing more load, can escape from buckling?

Solution:
N1+N2=mg$N_1+N_2=mg$ .................(i)
(N1)(l2)=(N2)l4$\left(N_1\right)\left({\displaystyle \frac{l}{2}}\right)=\left(N_2\right){\displaystyle \frac{l}{4}}$ ............(ii)
Solving we get,
N1=mg3$N_1={\displaystyle \frac{mg}{3}}$, N2=2mg3$N_2={\displaystyle \frac{2mg}{3}}$
To avoid buckling
N2=2mg3<4π3Yr4l24$N_2={\displaystyle \frac{2mg}{3}}<{\displaystyle \frac{4{\pi }^{3}Y{r}^4}{l^{2}4}}$
Hence ,rmin=(2mgl23π3Y)1/4$r_{min}={\left({\displaystyle \frac{2mg{l}^2}{3{\pi }^{3}Y}}\right)}^{1/4}$

Example

Problem on beams loaded at two ends with weight at the centre

Example:  A steel wire of diameter d$d$, area of cross-section A$A$ and length  2l$2l$ is clamped firmly at two points A and B, which are 2lm$2lm$ apart and in the same plane. A body of mass m$m$ is hung from the middle point of the wire such that the middle point sags by x$x$ lower from original position. If Young's modulus is Y$Y$, then find m$m$?

Solution: We have mg=2Tsinθ$mg=2Tsin\theta$
Also for small values of θ$\theta$ we have sinθ=tanθ=xl$sin\theta =tan\theta ={\displaystyle \frac{x}{l}}$
T=YAlΔl$T={\displaystyle \frac{YA}{l}}\mathrm{\Delta }l$
or
T=YAl((x2+l2)1/2−l)$T={\displaystyle \frac{YA}{l}}((x^2+l^2{)}^{1/2}-l)$
As x<<l$x<<l$ we have
T=YAl(l((1+x2l2)1/2−1))$T={\displaystyle \frac{YA}{l}}(l((1+{\displaystyle \frac{x^2}{l^2}}{)}^{1/2}-1))$
or
T=YA((1+x2l2)1/2−1)$T=YA((1+{\displaystyle \frac{x^2}{l^2}}{)}^{1/2}-1)$
or
T=YA((1+12x2l2)−1)$T=YA((1+{\displaystyle \frac{1}{2}}{\displaystyle \frac{x^2}{l^2}})-1)$
=YAx22l2$={\displaystyle \frac{YA{x}^2}{2l^2}}$
Thus we get
mg=2YAx22l2xl$mg=2{\displaystyle \frac{YA{x}^2}{2l^2}}{\displaystyle \frac{x}{l}}$
or
m=YAx3gl3$m={\displaystyle \frac{YA{x}^3}{g{l}^3}}$

Formula

Sag

Sag for a beam supported at both ends is given by:
δ=Wl34bd3Y$\delta ={\displaystyle \frac{W{l}^3}{4b{d}^{3}Y}}$
where
l:$l:$ length of the bar
b:$b:$ breadth of the bar
d:$d:$ depth of the bar
W:$W:$ load
Y:$Y:$ Young's modulus

Example

Buckling and Parameters related to it

Definition: Buckling is characterized by a sudden sideways failure of a structural member subjected to high compressive stress, where the compressive stress at the point of failure is less than the ultimate compressive stress that the material is capable of withstanding. Mathematical analysis of buckling often makes use of an "artificial" axial load eccentricity that introduces a secondary bending moment that is not a part of the primary applied forces being studied. As an applied load is increased on a member, such as a column, it will ultimately become large enough to cause the member to become unstable and is said to have buckled. Further load will cause significant and somewhat unpredictable deformations, possibly leading to complete loss of the member's load-carrying capacity. If the deformations that follow buckling are not catastrophic the member will continue to carry the load that caused it to buckle. If the buckled member is part of a larger assemblage of components such as a building, any load applied to the structure beyond that which caused the member to buckle will be redistributed within the structure.

Maximum vertical load on the column.
Formula: F=π2El(KL)2$F={\displaystyle \frac{{\pi }^{2}El}{(KL{)}^2}}$

E is modulus of elasticity.
I$I$ is moment of inertia.
L is free buckling length.
K is a dimensionless factor. (Material Property)

Definition

Factor of safety

Factor of safety is a term describing the capacity of a system beyond the expected loads or actual loads. Essentially, the factor of safety is how much stronger the system is than it usually needs to be for an intended load. Safety factors are often calculated using detailed analysis because comprehensive testing is impractical on many projects, such as bridges and buildings, but the structure's ability to carry load must be determined to a reasonable accuracy.

Formula

Relation between stress, strain and young modulus

Metal rod either expands or contract on heating or cooling. A stress is produced in rod if it is restricted from expanding or contracting.
This stress is directly proportional to the thermal strain.
Thermal strain =$=$ Change in length / original length

Thermal strain =Δ$=\mathrm{\Delta }$ L / L

Thermal strain =$=$ L T / L Linear thermal expansion

Thermal strain =$=$ T

Youngs modulus of material =$=$ Stress / Strain

Y =$=$ Stress / T
So Stress =$=$ Y T

Therefore, Force =$=$ Y  A T

Y =$=$ Youngs modulus of elasticity

= coefficient of linear expansion.
T = increase in temperature
A = Area

Definition

Spring-Ball model of Intermolecular forces

The property of the body because of which they are able to come back to their original status after removing the external force applied is called elasticity.
In a solid, each atom or molecules surrounded by neighboring atoms or molecules. These are bonded together by inter atomic or intermolecular forces and stay in a stable equilibrium position. When a solid is deformed, the atoms or molecules are displaced from their equilibrium positions causing a change in the inter atomic (or intermolecular) distances. When the deforming force is removed, the inter atomic forces tend to drive them back to their original positions. Thus the body regains its original shape and size.

Definition

Elastic Force

Elasticity is the ability of a material to return to its original shape after being stretched or compressed. When an elastic material is stretched or compressed, it exerts force which is called Elastic Force.

Definition

Bulk Modulus

Definition: The bulk modulus of a substance measures the substance's resistance to uniform compression. It is defined as the ratio of the infinitesimal pressure increase to the resulting relative decrease of the volume. Its SI unit is the pascal.

Bulk Modulus K=−VdPdV$K=-V{\displaystyle \frac{dP}{dV}}$

Example

Problems on finding change in volume using bulk modulus

Example: A copper solid cube of 60$60$ mm side is subjected to a compressible pressure of 2.5×107$2.5\times {10}^7$ Pa. If the bulk modulus of copper is 1.25×1011$1.25\times {10}^{11}$ pascal, find the change in the volume of cube.                

Solution:
B=−PΔv/v$B={\displaystyle \frac{-P}{\mathrm{\Delta }v/v}}$

Δv=−PvB$\mathrm{\Delta }v={\displaystyle \frac{-Pv}{B}}$

=−2.5×107×(60)31.25×1011$={\displaystyle \frac{-2.5\times {10}^7\times (60{)}^3}{1.25\times {10}^{11}}}$

=−43.2mm3$=-43.2m{m}^3$

So, Δv=−43.2mm3$\mathrm{\Delta }v=-43.2m{m}^3$ and it is negative since volume decreases.

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