An R.C.C footing is having dimension as shown in fig. It carries 4 load as indicated. Calculate the resultant stress at corners A,B,C,D.

Note: Myy B.M coming negative as 80kN(D) and 600kN(A) taking as compression side but it results in tension

[1] Area of footing = $(2 + 2) \times (1 + 1) = 8 m^2$

Section modulus about XX – axis = $\frac{\frac{bd^3}{12}}{(\frac{d}{2})} = \frac{bd^2}{6}$

$( Z = \frac{I}{Y})$

$= \frac{4 \times 2^2}{6} = 2.67 \ m^3$

Section modulus about YY – axis = $\frac{db^3}{6} = \frac{2 \times 4^3}{6}$

= $ 21.33 \ m^3$

[2] Direct stress due to direct load

$P_d = \frac{Total \ Load}{Area} = \frac{600 + 500 + 700 + 800}{8}$

$= 325 \ KN/m^2$ - - -[C]

[3] B.M due to load eccentric about the axis XX:

$Mxx = [800 \times 0.6 + 700 \times 0.8] – [600 \times 0.4 + 500 \times 0.5]$

$\therefore$ Max stress due to above B.M = $ \pm \frac{Mxx}{Zxx}$

$= \frac{550}{2.67} = \pm 206 \ KN/m^2$

(Compress at C and D, tensile at A & B).

B.M due to eccentric about the axis YY.

$M_{yy} = (600 \times 0.9 + 800 \times 0.6) – (500 \times 0.8 + 700 \times 1.5)$

= -430 KNm.

$\therefore$ Max stress due to above B.M = $\frac{+M_{yy}}{Z_{yy}} = \pm \frac{430} {21.33}$

$= \pm 20.15 \ KN/m^2$

(Compressive at B, C & tensile at A & D).

Note: Myy B.M coming negative as 80kN(D) and 600kN(A) taking as compression side but it results in tension

Hence resultant stress at corners of footing:

Stress at A = 325 – 206 – 20.15 = 98.85 [C]

Stress at B = 325 – 206 + 20.15 = 139.15 [C]

Stress at C = 325 = 206 + 20.15 = 551.15 [C]

Stress at D = 325 + 206 – 20.15 = 510.85 [C]