Concept :

E = Young's Modulus of Rigidity = Stress / strain

G = Shear Modulus or Modulus of rigidity = Shear stress / Shear strain

ν = Poisson’s ratio = - lateral strain / longitudinal strain

K = Bulk Modulus of elasticity = Direct stress / Volumetric strain

Relation between E, K and ν

E = 2G (1 + ν)

E = 3K (1 - 2ν)

Calculation :

G = 0.4 × 10⁵ MPa, ν = 0.34

E = 2G (1 + ν) = 2 × 0.4 × 10⁵ × (1 + 0.34) ≈ 1.1 × 10⁵ MPa

Concept:

Using Maxwell reciprocal theorem,

$\frac{{\left(deflection\right)}_A}{{\left(Load\right)}_B}=\frac{{\left(deflection\right)}_B}{{\left(Load\right)}_A}$

Calculation:

$\therefore \frac{Y}{W}=\frac{{\left(deflection\right)}_B}{\frac{W}{2}}$

$\therefore {\left(deflection\right)}_B=\frac{Y}{2}$

Concept:

Euler’s critical buckling load $\left(P_e\right)=\frac{{\pi }^2\cdot E\cdot I_{min}}{L_e^2}$

Where, Imin = minimum area moment of Inertia, E = young’s modulus of elasticity, Le = Equivalent length = L ⋅ α

α for different loading conditions are different

Calculation:

Given: 

Two columns,

both ends are fixed, L_e = L × 0.5 

both ends are hinged L_e = L × 1

$P_e\propto \frac{1}{{\left(L_e\right)}^2}$

$\frac{{\left(P_e\right)}_{Bothhinged}}{{\left(P_e\right)}_{Bothclamped}}=\frac{1}{{\left(L\times 1\right)}^2}\times \frac{{\left(L\times 0.5\right)}^2}{1}=0.25$

Concept:

Deflection of the cantilever to paint load P at the free end is

${\left(\delta \right)}_{load}=\frac{P{L}^3}{3EI}$ (downward)

Deflection of the cantilever for concentrated moment M at a free end

${\left(\delta \right)}_{moment}=\frac{M{L}^2}{2EI}$ (upward)

∴ Total deflection at free end  = ($\delta$)_load + ($\delta$)_moment

$=\frac{P{L}^3}{3EI}+\left(-\frac{M{l}^2}{2EI}\right)$

$\therefore \delta =\frac{P{L}^3}{3EI}-\frac{M{L}^2}{2EI}$

Concept:

$\mathrm{s}\mathrm{h}\mathrm{e}\mathrm{a}\mathrm{r}\mathrm{f}\mathrm{o}\mathrm{r}\mathrm{c}\mathrm{e},\mathrm{F}=\frac{\mathrm{d}\mathrm{M}}{\mathrm{d}\mathrm{x}}$

Calculation:

Bending moment (M) = (5x² + 20x - 7)

Now,

$\mathrm{s}\mathrm{h}\mathrm{e}\mathrm{a}\mathrm{r}\mathrm{f}\mathrm{o}\mathrm{r}\mathrm{c}\mathrm{e},\mathrm{F}=\frac{\mathrm{d}\mathrm{M}}{\mathrm{d}\mathrm{x}}$

∴ F = 10x + 20

Now,

(F)_{(x = 2 m)} = 10(2) + 20

∴ F = 40 N

Now,

shear stress $(\tau )=\frac{\mathrm{F}}{\mathrm{A}}=\frac{40}{2}$

∴ τ = 20 MPa

Explanation:

The rod is stretched so there will be initial tensile stress.

Initial stress $=\frac{\sigma }{A}=\frac{12000}{250}=48N/m{m}^2$ 

Final stress = 100 N/mm²

Additional stress required due to temperature change:

σ_th = σ_fin = σ_ini = 100 – 48 = 52 N/mm²

Thermal stress:

σ_th = αΔTE

52 = 11.7 × 10^-6 × ΔT × 200 × 10³

ΔT = T₂ – T_i = 22.22 °C 

Note: - (Temperature will be reduced as tensile stress needs to be generated)

40°– T2   = 22.22 ⇒ T₂ = 17.8°C

Concept:

Change in length due to axial loading is,

$\mathrm{\Delta }l=\frac{PL}{AE}$

Where, P = axial load, L =initial length, A = cross-sectional area of bar, E = young’s modulus of elasticity

Calculation:

Total change in length ∆l = ∆l_PQ + ∆l_QR + ∆l_RS

For the section PQ: l = 500 mm, P = 200 N

For the section QR: l = 1000 mm, P = -200 N

For the section RS: l = 500 mm, P = 100 N 

$\mathrm{\Delta }l=\frac{200\times 500}{10\times 2\times {10}^5}+\frac{-200\times 1000}{10\times 2\times {10}^5}+\frac{100\times 500}{10\times 2\times {10}^5}=-25\times {10}^{-3}mm$

Concept:

For a thin cylinder:

Longitudinal stress: ${\sigma }_L=\frac{pd}{4t}$

Hoop stress: ${\sigma }_h=\frac{pd}{2t}=2{\sigma }_L$

Longitudinal Strain:

${ϵ}_L=\frac{1}{E}\left({\sigma }_L-\mu {\sigma }_H\right)=\frac{{\sigma }_L}{E}\left(1-2\mu \right)=\frac{pd}{4tE}\left(1-2\mu \right)$

Circumferential or hoop strain:

${ϵ}_H=\frac{1}{E}\left({\sigma }_H-\mu {\sigma }_L\right)=\frac{{\sigma }_L}{E}\left(2-\mu \right)=\frac{pd}{4tE}\left(2-\mu \right)$

Calculation:

Given, l = 2.5 m = 2500 mm, d = 1.25 m = 1250 mm, t = 20 mm

P = 1.5 MPa, E = 200 GPa = 200 × 10³ MPa, μ = 0.3

In cylindrical shell Hoop strain ${ϵ}_H=\frac{\delta d}{d}=\frac{Pd}{4tE}(2-\mu )$

$\Rightarrow \delta D=\frac{P{d}^2}{4tE}(2-\mu )$

$\delta D=\frac{1.5\times {1250}^2}{4\times 20\times 200\times {10}^3}(2-0.3)$

δD = 0.249 mm

Concept:

$$\begin{gathered} {\sigma }_{1,2}=\frac{1}{2}\left[\left({\sigma }_x+{\sigma }_y\right)\pm \sqrt{{\left({\sigma }_x-{\sigma }_y\right)}^2+4{\tau }_{xy}^2}\right] \\ {\tau }_{max}=\frac{{\sigma }_1-{\sigma }_2}{2} \end{gathered}$$

Calculation:

Given: σ_x = 120 N/mm₂, σ_y = -90 N/mm², σ₁ = 150 N/mm²

${\sigma }_{1,2}=\frac{1}{2}\left[\left({\sigma }_x+{\sigma }_y\right)\pm \sqrt{{\left({\sigma }_x-{\sigma }_y\right)}^2+4{\tau }_{xy}^2}\right]$

${\sigma }_1=\frac{1}{2}\left[\left(120-90\right)+\sqrt{{\left(120+90\right)}^2+4{\left(\tau \right)}^2}\right]=150$

$\Rightarrow 30+\sqrt{{\left(210\right)}^2+4{\left(\tau \right)}^2}=300$  

⇒ (210)² + 4(τ)² = (270)²

τ = 84.85 N/mm²

Shear stress on these plane = 84 MPa

Confusion Point:

Sum of normal stress = sum of principal stress

σ_x + σ_y = σ₁ + σ₂

120 – 90 = 150 + σ₂

σ₂ = -120 N/mm²

${\tau }_{max}=\frac{{\sigma }_1-{\sigma }_2}{2}=\frac{150-\left(-120\right)}{2}=135N/m{m}^2$

This is maximum shear stress, not shear stress on the oblique plane.

Concept:

The neutral axis is an axis in the cross-section of a beam (a member resisting bending) or shaft along which there are no longitudinal stresses or strains. If the section is symmetric, isotropic and is not curved before a bend occurs, then the neutral axis is at the geometric centroid. All fibers on one side of the neutral axis are in a state of tension, while those on the opposite side are in compression.

Bending Stress distribution for symmetric I section

Calculation:

Using similar triangle property,

$\frac{1200}{x}=\frac{300}{80-x}$

1200 (80 - x) = 300x

96000 – 1200x = 300x

96000 = 1500x

x = 64 mm