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Strength of Materials

Uniaxial Direct Stress

A mild steel ...

Question

A mild steel bar of square cross-section has a sectional area 200 $m m^{2}$ . It is subjected to an axial force of 20 kN as shown in the below figure. The intensity of the normal tensile stress in $N / m m^{2}$ on the oblique plane 1-1 at 45° with the YY-axis which is normal to the longitudinal axis will be

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Solution

The correct option is B 50
$A = 200 m m^{2}$

$σ_{1} = \frac{P}{A} = \frac{20 \times 1000}{200} \frac{N}{m m^{2}} = 100 N / m m^{2}$

$θ = 45^{\circ}$

$σ_{n} = (\frac{σ_{1} + 0}{2}) + (\frac{σ_{1} + 0}{2}) cos 2 θ$

$= \frac{100}{2} + \frac{100}{2} cos 90^{\circ} = 50 N / m m^{2}$

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A mild steel bar of square cross-section has a sectional area 200 mm2. It is subjected to an axial force of 20 kN as shown in the below figure. The intensity of the normal tensile stress in N/mm2 on the oblique plane 1-1 at 45° with the YY-axis which is normal to the longitudinal axis will be

The correct option is B 50A=200mm2 σ1=PA=20×1000200Nmm2=100N/mm2 θ=45∘ σn=(σ1+02)+(σ1+02)cos2θ =1002+1002cos90∘=50N/mm2

The correct option is B 50
$A = 200 m m^{2}$

$σ_{1} = \frac{P}{A} = \frac{20 \times 1000}{200} \frac{N}{m m^{2}} = 100 N / m m^{2}$

$θ = 45^{\circ}$

$σ_{n} = (\frac{σ_{1} + 0}{2}) + (\frac{σ_{1} + 0}{2}) cos 2 θ$

$= \frac{100}{2} + \frac{100}{2} cos 90^{\circ} = 50 N / m m^{2}$