Question 7.17: A simply supported beam of length 3 m carries a point load o...

Given :
Point load, W = 12 kN = 12000 N
First find the B.M. at X-X. And to do this first calculate reactions R_A \text{ and } R_B.

Taking moments about A, we get

R_B × 3 = 12 × 2

∴          R_B=\frac{12 \times 2}{3} = 8 kN \text{ and } R_A = W – R_B = 12 – 8 = 4  kN

B.M. at X-X      = R_A × 1.5 = 4 × 1.5 = 6  kNm \\ \quad \quad \quad \quad = 6 × 1000  Nm = 6000 × 1000  Nmm \\ \quad \quad \quad \quad  = 6000,000  Nmm

∴         M = 6000,000 Nmm
Now find the position of N.A. of the section of the beam. This can be obtained if we know the position of C.G. of the section.

Let        \bar{y} = Distance of the C.G. of the section from the bottom edge

=\frac{A_1y_1-A_2y_2}{A-1-A_2}       (Negative sign is due to cut out part)

=\frac{(150 \times 100)\times 75 -(75\times 50)\times \Big(50+\frac{75}{2}}{150\times 100 – 75 \times 50}\Big) \\ =\frac{1125000-328125}{15000-3750}=\frac{796875}{11250} = 70.83  mm

Hence N.A. will lie at a distance of 70.83 mm from the bottom edge or 150 – 70.83 = 79.17 mm from the top edge as shown in Fig. 7.23.
Now the moment of inertia of the section about N.A. is given by,

I=I_1-I_2

where I_1 = M.O.I. of outer rectangle about N.A.

\text{= M.O.I. of rectangle 100 × 150 about its C.G. + }A_1 \\ \times \text{ (Distance of its C.G. from N.A.)}^2 \\ =\frac{100 \times 150^3}{12}+100 \times 150 \times (75 – 70.83)^2 \\ = 28125000 + 260833.5 = 28385833.5  mm^4 \\ I_2 = \text{ M.O.I. of cut out rectangular part about N.A.}\\ \text{ = M.O.I. of cut out part about its C.G. + }A_2 \\ \times \text{(Distance of its C.G. from N.A.)}^2 \\ =\frac{50\times 75^3}{12}+50 \times 75 \times \Big(50+\frac{75}{2}-70.83\Big)^2\\  = 1757812.5 + 1042083.375 \\ = 2799895.875  mm^4

∴       I=I-1-I-2= 28385833.5 – 2799895.875 = 25585937.63  mm^4

The bottom edge of the section will be subjected to tensile stress whereas the top edge will be subjected to compressive stress. The top edge is at 79.17 mm from N.A. whereas bottom edge is 70.83 mm from N.A.
Now using the relation,

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

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

(i) For maximum tensile stress, y = 70.83 mm
∴ Maximum tensile stress,

\sigma=\frac{6000000}{25585937.63 }\times 70.83 =\pmb{16.60  N/mm^2.}

(ii) For maximum compressive stress,

y = 79.17 mm.

∴ Maximum compressive stress,

\sigma =\frac{M}{I}\times y = \frac{6000000}{25585937.63}\times 79.17=\pmb{18.56  N/mm^2.}