Question 7.20: Prove that the moment of a resistance of a beam of square se...

Given :
Fig. 7.26 (a) shows a square section with diagonal AC vertical. Let the portions AEF and CGH be cut off.
Let I_1 = M.O.I. of the square ABCD about diagonal B.D.
Z_1 = Section modulus of square ABCD
M_1 = Moment of resistance of the square ABCD
I_2 = M.O.I. of the new section with cut off portion (i.e., M.O.I. of DEFBHG about diagonal BD)
Z_2 = Section modulus of new section
M_2 = Moment of resistance of the new section.
In Fig. 7.26 (a), diagonal AC = 2a

∴       Diagonal DB = AC = 2a.
Now moment of inertia of the square ABCD about N.A. (i.e., diagonal BD) is given by
∴      I_1 = M.O.I. of two triangles ABD and BCD about their base BD

=2 \times \frac{bh^3}{12} = 2\times \frac{2a \times a^3}{12} \quad (\text{ Here } b=2a \text{ and } h=a) \\ =\frac{a^4}{3}

Section modulus, Z_1=\frac{I_1}{y_{\max}}=\frac{(\frac{a^3}{3})}{a} \quad (\text{Here } y_{\max} = a) \\ \quad \quad \quad =\frac{a^4}{3}\times \frac{1}{a}=\frac{1}{3}a^3

Moment of resistance is given by,

M= σ × Z

∴        M_1= σ × Z_1=\sigma \times \frac{1}{3}a^3=\sigma \times 0.3333a^3        …(i)

Now the M.O.I. of the new section with cut off portion (i.e., M.O.I. of DEFBHG) about the diagonal BD is given by [Refer to Fig. 7.26 (b)].
I_2 = M.O.I. of four triangles (i.e., triangles DEK, FLB, DGH and HLB) plus M.O.I. of rectangle EFHG about N.A. (i.e., diagonal BD)

=\frac{4 \times bh^3}{12}+\frac{EF\times EG^3}{12}=\frac{4\times y \times y^3}{12}+\frac{2(a-y)\times(2y)^3}{12} \\ \space \\ (∵ \text{ Here }b=y, h = y, EF = 2(a – y) \text{ and } EG = 2y) \\ \space \\ =\frac{y^4}{3}+\frac{4}{3}(a-y)\times y^3=\frac{y^4}{3}+\frac{4ay^3}{3}-\frac{4y^4}{3}=\frac{4}{3}ay^3-y^4

and section modulus of new section is given by,

Z_2=\frac{I_2}{y_{\max}}=\frac{\frac{4}{3}ay^3-y^4}{y} \quad (∵  \text{Here } y_{\max} = y)  \\ \space \\ =\frac{4}{3}ay^2-y^3

Now moment of resistance of the new section is given by,

M_2=\sigma \times Z_2=\sigma \times [\frac{4}{3}ay^2-y^3]               …(ii)

The moment of the resistance of the new section will be maximum, if
\frac{dM_2}{dy}= 0. Hence differentiating equation (ii) w.r.t. y and equating it to zero, we get

\frac{d}{dy}[\sigma(\frac{4}{3}ay^2-y^3)]=0

or       \sigma(\frac{4}{3}a \times 2y -3y^2)=0      (∵ σ and a are constants)

or       \frac{4}{3}a \times 2y-3y^2=0        (∵ σ cannot be zero)

∴         3y^2=\frac{8}{3}a\times y

or        y=\frac{8}{3}\frac{a\times y}{3\times y}=\frac{8a}{9}            …(iii)

Substituting this value of y in equation (ii), we get

\left(M_2\right)_{\max }=\sigma \times\left[\frac{4}{3} \times a \times\left(\frac{8 a}{9}\right)^2-\left(\frac{8 a}{9}\right)^3\right]=\sigma \times\left[\frac{4 \times 64}{3 \times 81} a^3-\frac{512}{729} a^3\right] \\ \space \\ =\sigma × [1.0535a^3 – 0.7023a^3] = σ × 0.3512 a^3     …(iv)

But from equation (i), M_1 = σ × 0.3333 a^3

∴ M_2 is more than M_1. And from equation (iii), it is clear that M_2 is maximum when

y=\pmb{\frac{8a}{9}.}

Now increase in moment of resistance

= (M_2)_{\max} – M_1 = σ × 0.3512 a^3 – σ × 0.3333 a^2 \\ = σ × 0.0179 a^3

Percentage increase in moment of resistance

=\frac{\text{Increase in moment of resistance}}{\text{Original moment of resistance}}\times 100 \\ \space \\ =\frac{\sigma \times 0.0719 \times a^3}{\sigma \times 0.3333\times a^3}\times 100 = \pmb{5.37\% .}