Question 7.23: A timber beam 100 mm wide and 200 mm deep is to be reinforce...

Given :
1st Case. Flitches attached symmetrically at the top and bottom.
(See Fig. 7.31).
Let suffix 1 represents steel and suffix 2 represents timber.

Width of steel,   b_1 = 150  mm
Depth of steel,   d_1 = 12.5  mm
Width of timber, b_2 = 150  mm
Depth of timber,  d_2 = 200  mm
Number of steel plates = 2
Max. stress in timber, σ_2 = 6  N/mm^2
E for steel, E_1 = E_s = 2 × 10^5  N/mm^2
E for timber, E_2 = E_t = 1 × 10^4  N/mm^2
Distance of extreme fibre of timber from N.A.,

y_2 = 100  mm

Distance of extreme fibre of steel from N.A.,

y_1 = 100 + 12.5 = 112.5  mm.

Let σ_1^* = Max. stress in steel
\quad \quad σ_1 = Stress in steel at a distance of 100 mm from N.A.

Now we know that strain at the common surface is same. The strain at a common distance of 100 mm from N.A. is steel and wood would be same. Hence using equation (7.11), we get

\frac{\sigma_1}{E_1}=\frac{\sigma_2}{E_2}

∴     \sigma_1=\frac{E_1}{E_2}\times \sigma_2=\frac{2\times 10^5}{1\times 10^4}\times 6=120  N/mm^2.

But σ_1 is the stress in steel at a distance of 100 mm from N.A. Maximum stress in steel would be at a distance of 112.5 mm from N.A. As bending stresses are proportional to the distance from N.A.

Hence       \frac{\sigma_1}{100}=\frac{\sigma_1^*}{112.5}

∴                \sigma_1^*=\frac{112.5}{100}\times \sigma_1= \frac{112.5}{100}\times  120 =\pmb{135  N/mm^2.}

Now moment of resistance of steel is given by

M_1=\frac{\sigma_1^*}{y_1}\times I_1 \quad (\text{where } \sigma_1^* \text{ is the maximum stress in steel})\\ =\frac{135}{112.5} \times I_1

where I_1 = M.O.I. of two steel plates about N.A.
= 2 × [M.O.I. one steel plate about its C.G. + Area of one steel plate × (Distance between its C.G. and N.A.)²]

=2 \times [\frac{b_1d_1^3}{12}+b_1d_1\times (100+\frac{d_1}{2})^2]\\ \space \\ =2\times [\frac{150\times 12.5^3}{12}+150 \times 12.5 \times (100+\frac{12.5}{2})^2] \\ \space \\ = 2 × [24414.06 + 21166992.18] \\ \space \\ = 42382812.48  mm^4

∴       M_1=\frac{135}{112.5}\times 42382812.48 \\ = 50859374.96  Nmm = 50859.375  Nm

Similarly, M_2=\frac{\sigma_2}{y_2}\times I_2 \\  \quad \quad =\frac{6}{100} \times \frac{150\times 200^3}{12} \\ \quad \quad = 6000000  Nmm = 6000  Nm

∴ Total moment of resistance is given by,

M = M_1 + M_2 \\ = 50859.375 + 6000 = \pmb{56859.375  Nm.}

2nd Case. Flitches attached symmetrically at the sides (See Fig. 7.32)
Here distance of the extreme fibre of steel from N.A.

=\frac{150}{2}=75  mm.

In the first case we have seen that stress in steel at a distance of 100 mm from N.A. is 120 N/mm².
Hence the stress in steel at a distance of 75 mm from N.A. is given by,

\sigma_1^{**}=\frac{120}{100}\times 75\\ (∵ \text{ Stress are proportional to the distance from N.A.}) \\ = 90  N/mm^2

∴ Maximum stress in steel

= σ_1^{**} = \pmb{90  N/mm^2.}

Total moment of resistance is given by,

M = M_1 + M_2

where M_1 = Moment of resistance of two steel plates

=\frac{\sigma_1^{**}}{y_{\max}}\times I_1 \\ (\text{Here } σ_1^{**} = \text{ Maximum stress in steel } = 90  N/mm^2)\\ =\frac{90}{75}\times I_1 \quad (y_{\max}=75  mm)\\ I_1=\text{ M.O.I. of two steel plates about N.A.}\\ =2\times \frac{12.5\times 150^3}{12}=7031250  mm^4

∴   M_1 = \frac{90}{75} × 7031250  Nmm = 8437500  Nmm = 8437.5  Nm.

Similarly, M_2 = Moment of resistance of timber section

=\frac{\sigma_2}{y_2}\times I_2 \\ =\frac{6}{100}\times \frac{150\times 200^3}{12} \quad (∵  I_2=\frac{150\times 200^8}{12})\\= 6000000  Nmm = 6000  Nm

∴ Total moment of resistance,

M = M_1 + M_2 \\ = 8437.5 + 6000 = \pmb{14437.5  Nm.}