Concept:

Flexure formula:- \(\frac{M}{I} = \frac{\sigma }{y} = \frac{E}{R}\)

Where, σ = Bending Stress, y = Distance from the bending axis, R = Radius of Curvature

This formula is theoretically valid for pure bending, but if we apply it in non-uniform bending, the error introduced is negligible. Analysis of pure bending for symmetrical conditions yields flexure formula.

Calculation:

Given: d = 1 mm, R = 1 m, E = 100 GPa

\({\left( {{\sigma _b}} \right)_{max}} = \frac{{E{Y_{max}}}}{R}\)

\({\left( {{\sigma _b}} \right)_{max}} = \frac{{E\left( {\frac{d}{2}} \right)}}{R} = \frac{{100 \times {{10}^9} \times \left( {0.5} \right) \times {{10}^{ - 3}}}}{1} = 50\;MPa\)

Concept:

The shear stress distribution in a circular section is given as

\({\tau _{max}} = \frac{4}{3}{\tau _{avg}}\)

Calculation:

V = 30 π kN = 30 π × 10³ N

∴ Average shear stress \( = \frac{V}{A} = \frac{{30\pi\; \times \;10^3}}{{\left( {\frac{\pi }{4}\; \times \;{{100}^2}} \right)}} = 12\;MPa\)

∴ Maximum shear stress \(= \frac{4}{3}\;{\tau _{avg}} = \frac{4}{3} \times 12 = 16\;MPa\)

Important Point:

• For a Rectangular section, \({\tau _{max}} = \frac{3}{2}{\tau _{avg}}\)
• For a Triangular section, \(\left\{ {\begin{array}{*{20}{c}} {{\tau _{max}} = \frac{3}{2}{\tau _{avg}}}\\ {{\tau _{NA}} = \frac{4}{3}{\tau _{avg}}} \end{array}} \right.\)
• For a Kite section, \({\tau _{max}} = \frac{9}{8}{\tau _{avg}}\)

Permissible bending stress:

\({\sigma _b} = \frac{{{\sigma _{ut}}}}{{FOS}} = \frac{{480}}{4} = 120\;MPa\)

\(\sigma = \frac{M}{I}y \Rightarrow M = \sigma \times \frac{I}{y}\)

\(I = \frac{{b{d^3}}}{{12}}\)

\(\frac{I}{y} = \frac{{b{d^3}}}{{12}} \times \frac{2}{d} = \frac{{b{d^2}}}{6}\)

\(M = \sigma \times \frac{{b{d^2}}}{6} = \frac{{120 \times 60 \times 100 \times 100}}{6}\)

M = 12 × 10⁶ N.mm

Concept:

Flexure formula:- \(\frac{M}{I} = \frac{\sigma }{y} = \frac{E}{R}\)

Where, σ = Bending Stress, y = Distance from the bending axis, R = Radius of Curvature

This formula is theoretically valid for pure bending, but if we apply it in non-uniform bending, the error introduced is negligible. Analysis of pure bending for symmetrical conditions yields flexure formula –

\(\because \frac{M}{I} = \frac{\sigma }{y} \Rightarrow M = \frac{{\sigma I}}{y}\) 

Calculation:

Due to a hogging moment the bottom fibers below the neutral axis are subjected to compressive stress and the top fibers are subjected to tensile stress.

By flexure formula,

\(\sigma = \frac{{My}}{I}\) 

\(\therefore \frac{{{\sigma _t}}}{{{\sigma _c}}} = \frac{{\left( {\frac{{M{y_2}}}{I}} \right)}}{{\left( {\frac{{M{y_1}}}{I}} \right)}} \Rightarrow \frac{{{\sigma _t}}}{{{\sigma _c}}} = \frac{{{y_2}}}{{{y_1}}} = \frac{{200}}{{100}} = 2\) 

Concept:

For combined loading:

Equivalent bending moment:

\({{\rm{M}}_{\rm{E}}} = \frac{{{\rm{M}} + \sqrt {{{\rm{M}}^2} + {{\rm{T}}^2}} }}{2}\)

Equivalent torque:

\({{\rm{T}}_{\rm{E}}} = \sqrt {{{\rm{M}}^2} + {{\rm{T}}^2}} \)

Given:

M = 400 Nm, T = 600 Nm, τ_b(max) = 80 MPa

Calculation:

The shaft is under the action of combined bending and torsion.

\({{\rm{M}}_{\rm{E}}} = \frac{{{\rm{M}} + \sqrt {{{\rm{M}}^2} + {{\rm{T}}^2}} }}{2} = \frac{{400 + \sqrt {{{400}^2} + {{600}^2}} }}{2} = 560.5\;Nm\)

Form bending equation:

\(\therefore \frac{{{{\rm{f}}_{{{\rm{b}}_{{\rm{max}}}}}}}}{{\rm{R}}} = \frac{{{{\rm{M}}_{\rm{E}}}}}{{\rm{I}}}\)

\(\therefore \frac{{\frac{{80}}{{\rm{D}}}}}{2} = \frac{{560.5 \times {{10}^3}}}{{\frac{{{\rm{\pi }}{{\rm{D}}^4}}}{{64}}}}\)

Concept:

The shear stress at any section is given as-

\(\tau = \frac{{VA\bar y}}{{Ib}}\)

Where, V = Shear force at the section.

Ay̅ = Moment of the area about neutral axis, the difference of force on which gives rise to shear stress.

I = Moment of inertia about the neutral axis.

b = Width at the section where shear stress is to be calculated.

Calculation:

The centroidal axis will act as a neutral axis.

V = 15 kN = 15000 N

Width at the section resisting the shear force (b) \(= \frac{{150}}{{180}} \times \left( {180 - 50} \right)\) = 108.33 mm

\({I_{N.A}} = \frac{{B{h^3}}}{{36}} = \frac{{150 \times {{180}^3}}}{{36}} = 24.3 \times {10^6}\;m{m^4}\)

Ay̅ = Moment of the shaded area about neutral axis.

Ay̅ can also be taken as the moment of the area of ΔAB’C’ about C.G.

\(\therefore A\bar y = \left( {\frac{1}{2} \times b \times 130} \right) \times \left( {\frac{{130}}{3} - 10} \right)\)

\( = \left( {\frac{1}{2} \times 108.33 \times 130} \right) \times \left( {\frac{{130}}{3} - 10} \right)\) = 234715 mm³

\(\therefore \tau = \frac{{15000 \times \left( {234715} \right)}}{{24.3 \times {{10}^6} \times 108.33}} = 1.34\;MPa\)

Important Point:

τ _max in Triangular cross-section \(= \frac{3}{2}{\tau _{avg}}\)

τ _NA in Triangular cross-section \(= \frac{4}{3}{\tau _{avg}}\)

Concept:

Beam of constant strength or fully stressed beam:

If the beam is designed in such a way that at every section of the beam the applied bending moment is equal to the Moment of resistance of the section, then it is said to be a beam of constant strength and under this condition, maximum stress at every section will be equal to σ_per .

Calculation:

Bending moment at any section x = Σ Moment due to forces on either right or left of the section.

Considering the moment due to forces on the right of the section,

\({M_x} = \frac{{w {x^2}}}{2} + Px\)

Section Modulus (Z) = \(\frac{I}{y} = \frac{{\left( {\frac{{b{d^3}}}{{12}}} \right)}}{{\left( {\frac{y}{2}} \right)}} = \frac{{b{d^2}}}{6}\) 

\(\because MOR = {\sigma _{per}} \times Z = {\sigma _{per}} \times \frac{{b{d^2}}}{6}\)

For the beam of constant strength,

\( \Rightarrow \frac{{b{d^2}}}{6} \propto \frac{{w {x^2}}}{2} + Px\)

For the constant width (b), uniformly distributed load and point load P, depth can vary as a function of x.

\({\left( {{d_x}} \right)^2} \propto {x^2} + x \Rightarrow {d_x} \propto \sqrt {{x^2} + x} \)

Concept: 

1) Moment of inertia about x-x axis I_xx = \(\mathop \smallint \nolimits_A^{} {y^2} \cdot dA\)

2) Moment of inertia about y-y axis I_yy = \(\mathop \smallint \nolimits_A^{} {x^2} \cdot dA\)

3) Product moment of inertia = \(\mathop \smallint \nolimits_A^{} x \cdot y.dA\) 

Also,

\({I_{x'x'}} = \frac{{{I_{xx}} + {I_{yy}}}}{2} + \frac{{{I_{xx}} - {I_{yy}}}}{2}\cos 2\theta = {I_{xy}}\sin 2\theta \)

\({I_{x'y'}} = \frac{{{I_{xx}} - {I_{yy}}}}{2}\sin 2\theta + {I_{xy}}\cos 2\theta \)

Calculation:

I_xx = 3 × 10⁶ mm⁴, I_YY = 2 × 10⁶ mm⁴, I_xy = 5 × 10⁶ mm⁴, θ = 30°

Product moment of inertia about x’y’ \(= \frac{{{I_{xx}} - {I_{yy}}}}{2}\sin 2\theta + {I_{xy}}\cos 2\theta\) 

\(= \frac{{\left( {3 \times {{10}^6}} \right) - \left( {2 \times {{10}^6}} \right)}}{2}\sin 60^\circ + \left( {5 \times {{10}^6}} \right) \times \cos 60^\circ \)  = 2.93 × 10⁶ mm⁴

Concept:

Maximum bending stress \(\left( {\rm{\sigma }} \right) = \frac{{\rm{M}}}{{\rm{I}}} \times {\rm{y}} = \frac{{\rm{M}}}{{\rm{z}}}\)

Calculation:

For the given cross section

(Moment of Inertia)_I-section = (M.I)_section-1 - (M.I)_section-2

∴ (M.I)_I-section = (M.I)_section-1 - (M.I)_section-2

\(\therefore {\left( {{\rm{M}}.{\rm{I}}} \right)_{{\rm{I}} - {\rm{section}}}} = \frac{{150 \times {{200}^3}}}{{12}} - \frac{{100 \times {{100}^3}}}{{12}} = 91.67 \times {\rm{\;}}{10^6}{\rm{\;m}}{{\rm{m}}^4}\)

\(\therefore {\left( {\rm{Z}} \right)_{{\rm{I}} - {\rm{section}}}} = \frac{{{{\left( {{\rm{M}}.{\rm{I}}} \right)}_{{\rm{I\;section}}}}}}{{{{\rm{y}}_{{\rm{max}}}}}} = \frac{{91.67 \times {{10}^6}}}{{100}} = 91.67 \times {10^{4}}{\rm{m}}{{\rm{m}}^3}\)

⇒ M = σ × Z

Concept:

Flexure formula:- \(\frac{M}{I} = \frac{\sigma }{y} = \frac{E}{R}\)

Where, σ = Bending Stress, y = Distance from the bending axis, R = Radius of Curvature

This formula is theoretically valid for pure bending, but if we apply it in non-uniform bending, the error introduced is negligible. Analysis of pure bending for symmetrical conditions yields flexure formula –

\(\because \frac{M}{I} = \frac{\sigma }{y} \Rightarrow M = \frac{{\sigma I}}{y}\)

Calculations:

Let the maximum stress is induced at a depth of x-m from the top of the section.

The loading distribution at the section x-x’,

As per flexure formula,

\(\sigma = \frac{{My}}{I} \Rightarrow \sigma = \frac{{\left( {50 \times {{10}^6}} \right) \times \left( {\frac{{75 + 7.5x}}{{\sqrt 2 }}} \right)}}{{\frac{{{{\left( {75 + 7.5x} \right)}^4}}}{{12}}}}\)

Assuming all constant terms as K,

\( \Rightarrow \sigma = K \cdot \frac{x}{{{{\left( {75 + 10x} \right)}^2}}}\)

For the stress (σ) to be maximum, \(\frac{{d\sigma }}{{dx}} = 0\)

\(\therefore \frac{{d\left( {\frac{{Kx}}{{{{\left( {75 + 7.5x} \right)}^3}}}} \right)}}{{dx}} = 0\)

⇒ (75 + 7.5x)3 (1) – x × 3 (75 + 7.5x)2 × 7.5 = 0

⇒ (75 + 7.5x) – 22.5x = 0

⇒ x = 5 m

So at a depth of x = 5 m, the bending stress is maximum.