Strength of Materials Test 1

\(Stiffness\ \alpha \frac{1}{{number\ of\ active\ coils}}\)

So, \(25 \times k = 10 \times k' \Rightarrow k' = \frac{{25}}{{10}}k\)

\(\Rightarrow k' = 2.5\ k\)

We know from the relations of clastic Constant

E = 2G (1 + μ)

E = 3K (1 – 2μ)

On equating 2G (1 + μ) = 3K (1 – 2μ)

\(\frac{{2G}}{{3K}}\left( {1 + \mu } \right) = \left( {1 - 2\mu } \right)\) 

\(\frac{2}{3} \times 0.6\left( {1 + \mu } \right) = 1 - 2\mu \) 

\(0.4\left( {1 + \mu } \right) = 1 - 2\mu \) 

\(\Rightarrow 0.4 + 0.4\mu = 1 - 2\mu \) 

2.4μ = 0.6

Concept:

Equivalent Bending moment:

It is the bending moment which while acting alone produces maximum bending stress equal to the maximum principal stress produced due to combined action of moment and torque. The equivalent bending moment for a shaft which is subjected to the combined twisting moment (T) and bending moment (M) is given by the equation:

\({M_{eq}} = \frac{1}{2}\;\left[ {M + \sqrt {{M^2} + {T^2}} } \right]\)

Calculation:

M_eq = 500 Nm, T = 300 Nm

\({500} = \frac{1}{2}\;\left[ {M + \sqrt {{M^2} + {300^2}} } \right]\)⇒  M =455 Nm

Important Point:

Equivalent torque:

It is the torque which while acting alone produces maximum shear stress equal to maximum shear stress due to combined action of moment and torque.

The magnitude of \({T_{eq}} = \sqrt {{M^2} + {T^2}}\)

Concept:

Maximum strain energy theory (Beltrami-Haigh Theory):-

As per this theory, for no failure maximum strain energy per unit volume should be less than strain energy per unit volume under uniaxial loading when the stress is f_y.

\(\frac{1}{{2E}}\left[ {\sigma _1^2 + \sigma _2^2 + \sigma _3^2 - 2\mu \left( {{\sigma _1}{\sigma _2} + {\sigma _2}{\sigma _3} + {\sigma _3}{\sigma _1}} \right)} \right] < \frac{{f_y^2}}{{2E}}\)

Where σ₁, σ₂, σ₃ are principal stresses.

This theory is applicable for ductile material, not suitable for brittle material, and not suitable for the pure shear case.

Also;  for design purposes:-

\(\left( {\sigma _1^2 + \sigma _2^2 + \sigma _3^2 - 2\mu \left( {{\sigma _1}{\sigma _2} + {\sigma _2}{\sigma _3} + {\sigma _1}{\sigma _3}} \right)} \right) \le {\left( {\frac{{{\sigma _y}}}{{FOS}}} \right)^2}\)

Calculation:

σ₁ = 120 Mpa, σ₂ = 60 MPa, σ₃ = -30 MPa

σ_y = 250 MPa, μ = 0.3

∴ (120² + 60² + 30² – 2 × 0.25 (120 × 60 – 60 × 30 – 120 × 30) \( \le {\left( {\frac{{250}}{{FOS}}} \right)^2}\)

\( \Rightarrow 18000 \le {\left( {\frac{{250}}{{FOS}}} \right)^2} \Rightarrow {\left( {FOS} \right)^2} = \left( {\frac{{62500}}{{18000}}} \right)\) 

⇒ FOS = 1.863

Important Point:

Concept:

Bulk Modulus: The ratio of hydraulic stress to the corresponding hydraulic strain is called bulk modulus.

\(K = - \frac{P}{{{\rm{\Delta }}V/V}}\)

• The negative sign indicates the fact that with an increase in pressure, a decrease in volume occurs. That is, if p is positive, ΔV is negative. Thus, for a system in equilibrium, the value of bulk modulus B is always positive.
• SI unit of bulk modulus is the same as that of pressure i.e., N m^–2 or Pa.

Thin cylindrical shell ⇒ When \(\frac{t}{d} < \frac{1}{{15}}\), it is called a thin shell, otherwise thick shell.

Volumetric strain in a thin cylindrical shell :

\({\varepsilon _v} = \frac{{{\rm{\Delta }}PD}}{{4tE}}\;\left( {5 - 4\mu } \right)\)

Where, P = Internal pressure, D = Diameter of the shell, t = Thickness of shell, E = Modulus of Elasticity, μ = Poisson’s ratio.

Calculation:

Given: D = 50 mm, t = 2 mm, L = 1000 mm, K = 2 × 10⁶ MPa

\(\therefore \frac{t}{D} = \frac{2}{{50}} = \frac{1}{{25}} < \frac{1}{{15}} \Rightarrow \) Thin cylindrical shell.

Let the total fluid that can be pumped ΔV cc.

Initial volume of the fluid \(= V = \frac{\pi }{4} \times {50^2} \times 1000\) = 1963.4495 mm³ = 1963.5 cc.

\(\because K = - \frac{P}{{{\rm{\Delta }}V/V}}\)

Due to a pressure rise of 50 Mpa,

\({\rm{\Delta }}{V_1} = \frac{{{\rm{\Delta }}P}}{K} \times V \Rightarrow {\rm{\Delta }}{{\rm{V}}_1} = \frac{{{\rm{\Delta }}P}}{K} \times 1963.5\)

Volumetric strain in tube, \({\varepsilon _v} = \frac{{{\rm{\Delta }}V}}{V}\)

\(\Rightarrow {\rm{\Delta }}{V_2} = {\varepsilon _v} \times V = \frac{{{\rm{\Delta }}PD}}{{4tE}}\;\left( {5 - 4\mu } \right) \times V\)

\({\bf{\Delta }}{V_1} + {\bf{\Delta }}{V_2}\) = Extra fluid that can be pumped at a pressure rise of 50 MPa.

\(\Rightarrow \left[ {\frac{{{\rm{\Delta }}P}}{K} \times V + \frac{{{\rm{\Delta }}PD}}{{4tE}}\;\left( {5 - 4\mu } \right) \times V} \right] = {\rm{\Delta }}V\)

\(\Rightarrow \left( {\frac{{50}}{{2 \times {{10}^6}}} \times 1963495 + \frac{{50 \times 50}}{{4 \times 2 \times 1 \times {{10}^5}}} \times \left( {5 - 4 \times 0.3} \right) \times 1963495} \right] = {\rm{\Delta }}V\)

⇒ ΔV = 23365.59 mm³ ≈ 23.37 cc

Important Point:

Important Formulae for Thin shells:-