# SOLVED NUMERICAL PROBLEMS-STRENGTH OF MATERIAL

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On May 10, 2017

**SOLVED NUMERICAL PROBLEMS **

**STRENGTH OF MATERIALS**

**Q. Find the ratio of second moment of area about the centroid axis and about the base of a triangle. Given base=200 mm and altitude = 600 mm.**

#### A.

#### Second moment of area is moment of inertia ‘I’

#### (i) About the centroid axis, I _{centroid axis } = bd^{3}/36, say= I_{1}

#### (ii) About the base, I _{base} = bd^{3}/12, say = I_{2}

#### I_{1}/I_{2} = (bd^{3}/36)/( bd^{3}/36) = 1/3

#### NOTE: Dimensions given are superfluous.

**Q. What is total elongation of a bar of uniform cross-section under its own weight and an external load equal to the weight of the bar**.

#### A.

#### Elongation due to external load ‘W’ =δL_{1}= WL/AE

#### Elongation due to the weight of the bare ‘W’ =δL_{2}= WL/2AE

#### Total elongation =δL_{1} +δL_{2} =(3/2) WL/AE

#### The ratio δL_{2}/ δL_{1} = ½

### Q. **A beam of square section with side ‘a’. Its diagonal is horizontal.** Find the section modulus of the beam.

#### A.

#### There will be one triangle above and another below the horizontal diagonal. The moment of inertia of one triangle about the base is bd^{3}/12= a^{4}/12. Therefore the total moment of inertia will be 2 x a^{4}/12.

#### Length of the diagonal =

#### Height of the triangle = [a^{2} – (a/ )^{2}]^{0.5} =a/

#### Therefore section modulus , Z = I/y_{max} = (2 x a^{4}/12) / a/ = a^{3}/3

**Q. The deflection of free end of a cantilever of rectangular cross-section due to certain loading is 0.8 cm. Find the deflection of the free end if the depth doubles. Keep the width same.**

#### A.

#### Deflection δ at free end = WL^{3}/3EI=0.8

#### I = bd^{3}/12 with depth as ‘d’ and width as ‘b’

#### I=b(2d)^{3}/12 = 4 bd^{3}/12 with depth as 2d and width as b

#### Deflection with depth as 2d will be =0.8/4=0.2 cm

**Q. Find the ratio of torques of solid shaft of diameter ‘D’ and a hollow shaft with diameters ‘D’ and ‘d’. Given allowable shear stress ‘z’.**

#### A.

#### Torsional moment is torque’ T ’.

#### T_{solid} = (πD^{3}/16) z

#### T_{hollow} = π(D^{4}—d^{4})/16 D) z

#### Ratio T_{solid} / T_{hollow} = [(πD^{3}/16) z]/[ π(D^{4}—d^{4})/16 D) z]= D^{4}/(D^{4}—d^{4})

#### T_{solid} / T_{hollow} = D^{4}/(D^{4}—d^{4})

**Q. Bending moment of M and a torque T act on a shaft. What is ratio of maximum bending stress to the maximum shear stress?**

#### A.

#### Equivalent bending moment is

#### M_{eq} = (1/2) [M+(M2 +T2)^{0.5}]

#### Maximum bending stress σ = (M_{eq}/I) (D/2)

#### Equivalent torque is

#### T_{eq} = (M2 +T2)^{0.5}

#### Maximum shear stress z = (T_{eq}/ 2I) (D/2)

#### Therefore ratio σ/z = 2 M_{eq}/T_{eq}

**Q. Write the relationships between uniformly distributed load ‘w’, shear force ‘V’ and bending moment ‘M’.**

#### A.

#### (a) The relation between UDL ‘w’ and shear force ‘V’ is

#### dV/dx = — w

#### (b)The relation between shear force ‘V’ and bending moment ‘M’ is

#### dM/dx = V_{x}

**Q. The Euler load for a column is 1000 kN and the crushing load is 1500 kN. Find the Rankine load.**

#### A.

#### We know 1/P_{Eu} +1/P_{c} = 1/P_{R}

#### 1/1000 + 1/1500 = 1/P_{R}

#### P_{R} = 600 kN

**Q. Explain the difference between Neutral layer and neutral axis.**

**A.**

#### Neutral Layer: It is a plane (area) under zero stress and under zero strain. It passes through the neutral axis of the beam.

#### Neutral axis: It is a line where neutral fiber cuts the cross section of the beam. It does not have any stress or no strain.