Strength of Materials Test 1

Strength of Materials primarily focuses on the analysis of stress and strain in materials under various loading conditions.

The tensile stress is the ratio of change in length to the original length. It is the stress induced in a body when subjected to two equal and opposite pulls. The ratio of increase in length to the original length is the tensile strain.

The strain is given by = dL / L = 2/2000 = 0.001.

Hooke’s law is valid under the elastic limit of a body. It states that stress is proportional to the strain within the elastic limit.

The factor of safety is the ratio of ultimate stress to the permissible stress.

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

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

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

Hooke’s law is valid only up to the elastic range of the material, where the material behaves elastically.

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

Creep is the property by which a metal specimen undergoes additional deformation over time under sustained loading within the elastic limit. It is permanent in nature and cannot be recovered after the removal of the load, hence it is plastic in nature.

When there is an eccentric load, it means the load is applied at some distance from the axis. This causes compression on one side and tension on the other, resulting in bending stress.

The bulk modulus is defined as the ratio of direct stress to the corresponding volumetric strain when a body is subjected to mutually perpendicular equal direct stresses within a certain limit of deformation.

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 \({T_{eq}} = \sqrt {{M^2} + {T^2}}\)

If Poisson’s ratio is zero, then the material is rigid.

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:

Since E = 2G(1 + μ) = 3K(1 – 2μ) = 9KG / (3K + G), knowing any two of these four allows for the calculation of the other two using the relations between them.

Concrete has the property of taking a good amount of compressive stress. So, in the reinforcement bar, the compressive stress is taken by the concrete.

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:-

When there is an eccentric load, it means that the load is at some distance from the axis. This causes compression on one side and tension on the other, resulting in bending stress.

Annealing is the process of heating the metal in the furnace to a temperature slightly above the upper critical temperature and cooling slowly in the furnace. It produces an even grain structure, reduces hardness, and increases ductility usually at a reduction of strength.

Concept involved:

The interface will fail only if the applied normal stress on the interface is greater than the permissible normal stress or generated shear stress is greater than the permissible shear stress.

The maximum normal stress at inclined plane (interface) in case of uniaxial loading, σθ = σ cos2 θ

and shear stress over the interface \({\tau _{\rm{\theta }}} = \frac{\sigma }{2}\sin \theta\)

Calculation:

Criterion 1: Maximum normal stress is 2.5 MPa

Now, calculating normal stress over the interface

σ_θ = σ cos² θ

θ = 30°

\({\sigma _x} = 4 \times {\left( {\frac{{\sqrt 3 }}{2}} \right)^2} = 3\;MPa\)

3 > 2.5 hence, the interface will fail

Criterion 2:

Maximum shear stress is 1.5 MPa

Now, calculating shear stress over the interface

\({\tau _{\rm{\theta }}} = \frac{\sigma }{2}\sin \theta\)

θ = 30°

\({\tau _{\rm{\theta }}} = \frac{4}{2} \times \sin 60^\circ = 2 \times \frac{{\sqrt 3 }}{2} = \sqrt 3 = 1.732\)

1.732 > 1.5

Hence, the interface will fail.

Mistakes: The angle ‘θ’ here is always taken from the vertical for applying in the formula

σ_n = σ cos² θ and τ = σ/2 sin 2θ

so, instead of θ = 30 if we take (90 – θ) = 60°

then, \({\sigma _n} = \sigma {\cos ^2}60^\circ = 4 \times \frac{1}{4} = 1\;MPa\) 

then, the answer will change.

Concrete is more likely to fail in shear due to its brittle nature compared to other materials.

Ductile materials typically fail through yielding, where they undergo plastic deformation before fracture.

Generally, the tensile strength of most materials decreases as temperature increases, as the material becomes more ductile and less stiff.

Shear stress is defined as the force per unit area acting parallel to the surface of a material.

Axial loads produce tensile or compressive stress depending on the direction of the load applied to the material.