Question 7.22: A flitched beam consists of a wooden joist 10 cm wide and 20...

Given :
Let suffix 1 represent steel and suffix 2 represent wooden joist.
Width of wooden joist,   b_2 = 10  cm
Depth of wooden joist,   d_2 = 20  cm
Width of one steel plate, b_1 = 1  cm
Depth of one steel plate, d_1 = 20  cm
Number of steel plates     = 2
Max. stress in wood,        σ_2= 7  N/mm^2
E for steel,                          E_1 = 2 × 10^5  N/mm^2
E for wood,                        E_2 = 1 × 10^4  N/mm^2
Now M.O.I. of wooden joist about N.A.,

I_2=\frac{b_2d_2^3}{12}=\frac{10\times 20^3}{12} \\ = 6666.66  cm^4 \\ = 6666.66 × 10^4  mm^4

M.O.I. of two steel plates about N.A.,

I_1=\frac{2\times b_1d_1^3}{12}=\frac{2\times 1 \times 20^3}{12} \\ = 1333.33  cm^4 = 1333.33 × 10^4  mm^4.

Now modular ratio between steel and wood is given by,

m=\frac{E_1}{E_2}=\frac{2\times 10^5}{1\times 10^4}=20

The equivalent moment of inertia (I) is given by equation (7.13).

∴       I = mI_1 + I_2 \\ \quad \quad = 20 × 1333.33 × 10^4 + 6666.66 × 10^4 \\ \quad \quad  = 10^4(26666.6 + 6666.66) = 10^4 × 33333.2

Moment of resistance of the composite section is given by equation (7.14).

∴      M=\frac{\sigma_2}{y}\times I \\ \quad \quad =\frac{7 \times 10^4 \times 33333.2}{10 \times 10} \quad (∵  y = 10  cm = 10 × 10  mm) \\ \quad \quad = 233332.4 × 10^2  N mm = \pmb{23333.24  Nm.}

Maximum Stress in Steel
Let σ_1 = Max. stress in steel.
Now using equation, we get

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

∴            \sigma_1=\frac{E_1}{E_2}\times \sigma_2 \\ \quad \quad = 20 \times 7 \quad (∵  \frac{E_1}{E_2}=m=20 \text{ and } \sigma_2=7  N/mm^2) \\ \quad \quad =\pmb{140  N/mm^2.}

2nd Method

Total moment of resistance is equal to the sum of moment of resistance of individual member.

∴         M = M_1 + M_2           …(i)

where     M_1=\frac{\sigma_1}{y}\times I_1 \quad (∵  \frac{M}{I}=\frac{\sigma}{y}) \\ \quad \quad =\frac{140}{100} \times 1333.33 \times 10^4  \quad (∵  y = 10 × 10 = 100  mm) \\ \quad \quad  = 18666620  Nmm = 18666.620  Nm

and       M_2=\frac{\sigma_2}{y}\times I_2  \\ \quad \quad =\frac{7}{100} \times 6666.66 \times 10^4   Nmm  \\ \quad \quad  = 46666.62  Nmm = 4666.662  Nm

∴            M = M_1 + M_2 = 18666.620 + 4666.662 \\ \quad \quad = \pmb{23333.282  Nm.}

3rd Method

The equivalent moment of inertia (I) is obtained by producing equivalent section.
(a) The equivalent wooden section is obtained by multiplying the dimension of steel plate in the direction parallel to the N.A. by the modular ratio between steel and wood (i.e., by multiplying by \frac{E_s}{E_w}=\frac{2\times 10^5}{1\times 10^4}=20). But the width of one steel plate parallel to N.A. is 1 cm.
Hence equivalent wooden width for this steel plate will be 20 × 1 = 20 cm. This is shown in Fig. 7.29.

∴ Equivalent M.O.I. is given by,

∴          I =\frac{bd^3}{12}\\ \quad \quad =\frac{(20+10+20)\times 20^3}{12} \\ \quad \quad = 33333.33  cm^4 = 33333.33 × 10^4  mm^4

Total moment of resistance
= Moment of resistance of the equivalent wooden section
=\frac{\sigma}{y}\times I \\ =\frac{\text{Stress in wood}}{y}\times I\\ =\frac{7}{100}\times 33333.33 × 10^4 = 23333333.33  Nmm \\ = \pmb{23333.333  Nm.}

(b) The equivalent steel section is obtained by multiplying the dimensions of wooden joist in the direction parallel to N.A. by the modular ratio between wood and steel (i.e., by multiplying by \frac{E_w}{E_s}=\frac{1\times 10^4}{2\times 10^5}=\frac{1}{20}).

But the width of wooden joist parallel to N.A. is 10 cm. Hence equivalent steel width will be 10 \times\frac{1}{20}=0.5  cm. This is shown in Fig. 7.30.

Hence equivalent M.O.I. is given by

I=\frac{bd^3}{12} \\ = \frac{(1+0.5+1)\times 20^3}{12} \\ = 1666.66  cm^4 \\ = 1666.66 × 10^4  mm^4

∴       M=\frac{\sigma}{y}\times I \\ \quad \quad =\frac{140}{100} \times 1666.76 \times 10^4 \\ \quad \quad  (\text{Here σ is the stress in steel and} = 140  N/mm^2) \\ \quad \quad  = 23333240  Nmm \\ \quad \quad = \pmb{23333.240  Nm.}

Note. The width of the single wooden beam for the total moment of resistance of 23333.33 Nm should be 20 + 10 + 20 = 50 cm as shown in Fig. 7.29. But the width of flitched beam for the same moment of resistance is only 1 + 10 + 1 = 12 cm as shown in Fig. 7.28. Hence flitched beams require less space.