Question 7.24: Two rectangular plates, one of steel and the other of brass ...

Given :
Width of plates, b = 40 mm
Depth of plates, d = 10 mm
Span,                    L = 1 m
Stress in steel,    σ_s = 112.5  N/mm^2
Stress in brass,   σ_b = 75  N/mm^2
Value of E for steel, E_s = 2 × 10^5  N/mm^2
Value of E for brass, E_b = 8 × 10^4  N/mm^2.

1st Case. The plates are separate and can bend independently.
Since the two materials bend independently, each will have its own neutral axis. It will be assumed that the radius of curvature R is the same for both the plates.

Using the relation      \frac{\sigma}{y}=\frac{E}{R}

or           R=\frac{E\times y}{\sigma}

or            R=\frac{E_s\times y_s}{\sigma_s}=\frac{E_b\times y_b}{\sigma_b}

or             \frac{\sigma_s}{\sigma_b}=\frac{E_s\times y_s}{E_b\times y_b}

But y_s = y_b as the two plates are having their own N.A. The distance of the extreme fibre of brass from its own N.A. is 5 mm. Also the distance of extreme fibre of steel from its N.A. = 5 mm.

∴          \frac{\sigma_s}{\sigma_b}=\frac{E_s}{E_b}=\frac{2\times 10^5}{8\times 10^4}=2.5

Now the allowable stress in steel is 112.5 N/mm²
i.e.,      σ_s = 112.5  N/mm^2.

Then maximum stress in brass will be,

\sigma_b=\frac{\sigma_s}{2.5}=\frac{112.5}{2.5}=45  N/mm^2

This is less than the allowable stress of 75 N/mm².
Note. If maximum stress in brass is taken as 75 N/mm². Then the stress in steel will be σ_s = 2.5 × σ_b = 2.5 × 187.5  N/mm^2. This stress is more than the allowable stress in steel.

The total moment of resistance is given by,

M = M_s + M_b

where M_s = Moment of resistance of steel plate.

and M_b = Moment of resistance of brass plate

=\frac{\sigma_b}{y_b}\times I_b \\ =\frac{45}{5.0}\times \frac{40\times 10^3}{12}= 30000  Nmm = 30  Nm

∴  M = M_s + M_b = 75 + 30 \\ = 105  Nm           …(i)

Let W = Maximum load applied at the centre in N to a simply supported beam. Then maximum bending moment will be at the centre of the beam. And it is equal to,

M=\frac{W\times L}{4}=\frac{W\times 1.0}{4}Nm        …(ii)

Equation (i) and (ii), we get

\frac{W}{4} = 105 \quad \text{ or } \quad W = 4 × 105 = \pmb{420  N.}               …(iii)

2nd Case. The plates are firmly secured throughout their length. In this case, the two plates act as a single unit and thus will have a single N.A. Let us convert the composite section into an equivalent brass section as shown in Fig. 7.34 (b). The equivalent brass section is obtained by multiplying the dimensions of steel plate in the direction parallel to the N.A. by the modular ratio between steel and brass (i.e., by multiplying by \frac{E_s}{E_b}=2.5). But the width of steel plate parallel to N.A. is 40 mm. Hence equivalent brass width for the steel plate will be 40 × 2.5 = 100 mm. This is shown in Fig. 7.34.

Let \bar{y} = Distance between C.G. of the equivalent brass section and bottom face.

= \frac{A_1y_1+A_2y_2}{A_1+A_2} \\ \space \\ =\frac{100 \times 10 \times 5 + 40 \times 10 \times (10+5)}{100\times 10+40\times 10} \\ \space \\ = \frac{5000+6000}{1000+400}=\frac{11000}{1400}= 7.86  mm.

Hence N.A. of the equivalent brass section is at a distance of 7.86 mm from the bottom face.
Now the moment of inertia of the equivalent brass section about N.A. is given as
I = [M.O.I. of rectangle 100 × 10 about its C.G.
+ Area of rectangle 100 × 10 × (Distance between its C.G. and N.A.)²]
+ [M.O.I. of rectangle 40 × 10 about its C.G. + Area of rectangle 40 × 10
× (Distance between its C.G. and N.A.)²]

Distance of upper extreme fibre from N.A.

= 20 – 7.86 = 12.14 mm

Distance of lower extreme fibre from N.A.

= 7.86 mm

Now allowable stress in brass is given 75 N/mm². As the upper plate is of brass. Hence the upper extreme fibre will have a stress of 75 N/mm². Then the lowermost fibre will have the stress = \frac{75}{12.14} × 7.86 = 48.56  N/mm^2. In Fig. 7.34 (b), the lowermost fibre is also of brass. Hence the actual stress in the lowermost fibre of steel will be = 48.56 × 2.5 = 121.4 N/mm².

But the safe stress in steel is given as 112.5 N/mm². Hence the brass cannot be fully stressed.
If we take maximum stress in steel at the bottom to be 112.5 N/mm², then the corresponding stress in brass at the bottom fibre will be
\frac{112.5}{2.5}=45  N/mm^2.

∴     σ_s = 112.5  N/mm^2 \quad \text{ and } \quad σ_b = 45  N/mm^2.

Now using the relation,

\frac{M}{I}=\frac{\sigma}{y}

or      M=\frac{\sigma}{y}\times I \\ \quad \quad \quad =\frac{45}{7.86} \times 40238.1 = 230370.8  Nmm \\ \quad \quad \quad = 230.3708  Nm                      …(iii)

The maximum bending moment at the centre of a simply supported beam, carrying a point load W at the centre is given by,

M=\frac{W\times L}{4}=\frac{W\times 1.0}{4}                …(iv)

Equating (iii) and (iv), we get

\frac{W}{4}=230.3708

∴          W = 4 × 230.3708 = 921.48 N.