Strength of Materials Test 5

Concept:

Bending Moment Equation

\(\frac{M}{I} = \frac{\sigma }{y} = \frac{E}{R}\)

\(M = \frac{I}{y}\sigma = Z\sigma \)

For the constant bending moment: σ ∝ 1/Z

So, bending stress is inversely proportional to the Section Modulus

The strength of two beams of the same material can be compared by the section modulus values.

The beam is stronger when section modulus is more, the strength of the beam depends on section modulus.

Concept:

\({\rm{shear\;force}},{\rm{\;F}} = {\rm{}}\frac{{{\rm{dM}}}}{{{\rm{dx}}}}\)

Calculation:

Bending moment (M) = (5x² + 20x - 7)

Now,

\({\rm{shear\;force}},{\rm{\;F}} = {\rm{}}\frac{{{\rm{dM}}}}{{{\rm{dx}}}}\)

∴ F = 10x + 20

Now,

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

∴ F = 40 N

Now,

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

Concept:

Maximum bending across shaft occurs at either top or bottom fiber and it is given by the bending formula:

\({{\rm{\sigma }}_{{\rm{max}}}} = \frac{{\rm{M}}}{{\rm{Z}}} = \frac{{2{\rm{M}}}}{{\frac{{\rm{\pi }}}{{64}}{{\rm{d}}^4}}} \times \frac{{\rm{d}}}{2} = \frac{{64{\rm{M}}}}{{{\rm{\pi }}{{\rm{d}}^3}}}\)

Maximum shear stress is given by the twisting formula:

\({{\rm{\tau }}_{{\rm{max}}}} = \frac{{{\rm{Tr}}}}{{\rm{I}}} = \frac{{\left( {\frac{1}{2}} \right) \times \left( {\frac{{\rm{d}}}{2}} \right)}}{{\frac{{\rm{\pi }}}{{32}}{{\rm{d}}^4}}} = \frac{{8{\rm{T}}}}{{{\rm{\pi }}{{\rm{d}}^3}}}\)

\(\frac{{{{\rm{\sigma }}_{{\rm{\;max}}}}}}{{{{\rm{\tau }}_{{\rm{max}}}}}} = \frac{{\frac{{64{\rm{M}}}}{{{\rm{\pi }}{{\rm{d}}^3}}}}}{{\frac{{8{\rm{T}}}}{{{\rm{\pi }}{{\rm{d}}^3}}}}} = \frac{{8{\rm{M}}}}{{\rm{T}}}\)

Concept:

Section modules of rectangular section, \(Z = \frac{{d{b^2}}}{6}\;\)

Calculation:

Given:

b = 60 mm, d = 150 mm

\(∴ Z = \frac{{b{d^2}}}{6} = \frac{{60 \times {{150}^2}}}{6} = 225 \times {10^2}m{m^3}\;\)

Now,

The maximum bending moment at the center of a simply supported beam subjected to UDL is, \(M = \frac{{w{l^2}}}{8}\)

w = 4.5 kN/m, l = 4 m = 4 × 10³ mm

\(∴ M = \frac{{w{l^2}}}{8} = \frac{{4.5 \times {{\left( {4 \times {{10}^3}} \right)}^2}}}{8} = 9 \times {10^6}N-mm\)

\(∴ {σ _{bmax}} = \frac{M}{Z} = \frac{{9 \times {{10}^6}}}{{225 \times {{10}^3}}} \)

∴ σ_max = 40 MPa

Points to Remember:

i) For rectangular section \(Z = \frac{{b{d^2}}}{6}\)

ii) Maximum Bending moment in a simply supported beam subjected to UDL in at center and \(M = \frac{{w{l^2}}}{8}\)

iii) Maximum Bending Stress \(= \frac{M}{Z}\)

Concept:

Using the flexural formula,

\(\frac{M}{I} = \frac{E}{R} = \frac{{{\sigma _b}}}{y}\)

And,

\({\sigma _b} = \frac{{M.y}}{I}\)

For the circular shaft, \(y = \frac{d}{2}\) and \(I = {\rm{\;}}\frac{\pi }{{64}}{d^4}\)

\({\sigma _b} = \frac{{32M}}{{\pi {d^3}}}\)

Using torsion equation,

\(\frac{T}{J} = \frac{{G\theta }}{L} = \frac{\tau }{r}\)

And,

\(\tau = \frac{{T.r}}{J}\)

For the circular shaft, \(\tau = {\tau _{max}}\) at \(r = \frac{d}{2}\) and \(J = {\rm{}}\frac{\pi }{{32}}{d^4}\)

\(\tau = \frac{{16T}}{{\pi {d^3}}}\)

Calculation:

Using the given condition,

\({\sigma _b} = 1.5 \times \tau\)

\(\frac{{32M}}{{\pi {d^3}}} = 1.5 \times \frac{{16T}}{{\pi {d^3}}}\)

\(M = 0.75{\rm{\;}}T\)

Maximum principle stress is given as,

\({\sigma _1} = \frac{{16}}{{\pi {d^3}}}\left[ {M + \sqrt {{M^2} + {T^2}} } \right]\)

\(100 \times {10^6} = \frac{{16}}{{\pi\times {{0.08}^3}}}\left[ {0.75T + \sqrt {{{\left( {0.75T} \right)}^2} + {T^2}} } \right]\)

Concept:

Using torsion equation,

\(\frac{T}{J} = \frac{{G\theta }}{l} = \frac{\tau }{r}\)

\(Angle{\rm{\;}}of{\rm{\;}}twist,{\rm{\;}}\theta = {\rm{}}\frac{{Tl}}{{GJ}}\)

Calculation:

Given:

\({d_B} = 0.5{\rm{\;}}{d_A}\)

\({l_B} = 1.5{\rm{\;}}{l_A}\)

\({T_A} = T{\rm{\;}}and{\rm{\;}}{T_B} = 0.75{\rm{\;}}T\)

\({\theta _B} = \frac{{{T_B}{l_B}}}{{G \times \frac{\pi }{{32}}{{\left( {\frac{d}{2}} \right)}^4}}}\)

\({\theta _B} = \frac{{0.75T \times 1.5l}}{{G \times \frac{\pi }{{32}}{{\left( {\frac{d}{2}} \right)}^4}}} = \frac{{576{\rm{\;}}Tl}}{{G\pi {d^4}}}\)

\({\theta _A} = \frac{{{T_A}{l_A}}}{{G \times \frac{\pi }{{32}}{d^4}}}\)

\({\theta _A} = \frac{{Tl}}{{G \times \frac{\pi }{{32}}{d^4}}} = \frac{{32{\rm{\;}}Tl}}{{G\pi {d^4}}}\)

\(\frac{{{\theta _B}}}{{{\theta _A}}} = \frac{{\frac{{576{\rm{\;}}Tl}}{{G\pi {d^4}}}}}{{\frac{{32{\rm{\;}}Tl}}{{G\pi {d^4}}}}} = 18\)