Unishivaji Structural Mechanics - II May 2016 Solved Question Paper Q2 B)

A masonry retaining wall of trapezoidal section is 10 m high and retains earth which is level upto the top. The width at the top is 2m and at the bottom 8 m and the exposed face is vertical. Find the maximum and minimum intensities of normal stress at the base. Take density of earth: 16 KN/m³, Density of masonry = 24 KN/m³, Angle of repose : 30⁰.

12 - Marks

>>Given:
Height of retaining wall and earth = H = 10m;
Top width = a = 2m;
Bottom width = b = 8m;
Maximum stress at base = σ_max = ?;
Minimum stress at base = σ_min = ?;
Density of earth = ρ_earth = 16 kN/m³;
Density of masonry = ρ_masonary = 24 kN/m³;
Angle of repose = θ = 30⁰;
>>Solution:
For 1 m length;
Self-weight of dam = Cross-sectional area X 1 X Density of masonry
$W=\left( \frac{Top width + Bottom width}{2} \times H\right)$ $\: \times 1 \times \rho_{masonary}$
$W=\left( \frac{2 + 8}{2} \times 10\right)$ $\: \times 1 \times 24$
$W=1200 kN$
Base area for 1 m length;
$A=Bottom \: width \times 1$
$A=8 \times 1$
$A=8 m^{2}$
Direct stress;
$\sigma_{0}= \frac{W}{A}= \frac{1200}{8}$
$\sigma_{0}= 150 kN/m^{2}$
Force due to earth pressure;
$F= \frac{\rho_{earth} H^{2}}{2} \left( \frac{1-sin \theta }{1+sin \theta} \right)$
$F= \frac{16 \times 10^{2}}{2} \left( \frac{1-sin 30 }{1+sin 30} \right)$
$F= 266.67 kN$
Center of gravity of section;
$\bar{X}= \frac{a_{1}x_{1} + a_{2}x_{2}}{a_{1} + a_{2}}$
$\bar{X}= \frac{\left( 2 \times 10\right) \times 1 + \left( \frac{1}{2} \times 6 \times 10 \right) \times \left(2+ \frac{6}{3} \right)}{\left( 2 \times 10\right) + \left( \frac{1}{2} \times 6 \times 10 \right)}$
$\bar{X}= 2.8m$ (From vertical face of retaining wall.)

Calculate distance of resultant force acting;
assume resultant force acting at 'x' distance form centroidal distance;
taking summation of moment at point 'M' = 0;
$\left(F \times \frac{H}{3} \right) - \left(W \times x \right) = 0$
$x = \frac{\left(F \times \frac{H}{3} \right)}{W}$
$x = \frac{\left(266.67 \times \frac{10}{3} \right)}{1200}$
$x = 0.74m$
Resultant force acting from vertical face;
$=\bar{X} + x = 2.8 + 0.74 = 3.54m$
Eccentricity
$= 3.54 - b/2 = 3.54 - 8/2 = -0.459$
Negative sign indicates eccentricity toward vertical side ignore sign and take eccentricity = 0.46m
$Moment = M = W \cdot e = 1200 \cdot 0.46 = 552 kNm$
Section modulus
$Z = \frac{I_{yy}}{y}= \frac{\frac{db^{3}}{12}}{\frac{b}{2}}$
$Z = \frac{\frac{1 \cdot 8^{3}}{12}}{\frac{8}{2}}=10.67m^{3}$
Bending stress
$\sigma_{b} = \frac{M}{Z}=\frac{552}{10.67}$
$\sigma_{b} = 51.75 kN/m^{2}$
$\sigma_{max} = \sigma_{0} + \sigma_{b}=150+51.75$
$\sigma_{max} = 201.75 kN/m^{2}(C)$
$\sigma_{min} = \sigma_{0} - \sigma_{b}=150-51.75$
$\sigma_{min} = 98.25 kN/m^{2}(C)$
>>Summery:
$\sigma_{max} = 201.75 kN/m^{2}(C)$
$\sigma_{min} = 98.25 kN/m^{2}(C)$

Vinayak Hulwane

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Unishivaji Structural Mechanics - II May 2016 Solved Question Paper Q2 B)

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