Question 5.21: A mass of 50 kg drops through 25 mm at the centre of a 250 m...

\text { Given } m=50 kg \quad h=25 mm \quad l=250 mm .

S_{y t}=400 N / mm ^{2} \quad(f s)=2 \quad E=207000 N / mm ^{2} .

Step I Impact stress \left(\sigma_{i}\right) .
From Eq. (5.52),

\sigma_{i}=\frac{W}{A}\left[1+\sqrt{1+\frac{2 h A E}{W l}}\right]           (5.52).

\sigma_{i}=\frac{W}{A}\left[1+\sqrt{1+\frac{2 h A E}{W l}}\right] .

In the above equation,

\frac{W}{A}=\text { static stress }=\sigma_{s}             (a)

\frac{W l}{A E}=\text { static deflection }=\delta_{s}         (b).

Substituting (a) and (b) in Eq. (5.52),

\sigma_{i}=\frac{W}{A}\left[1+\sqrt{1+\frac{2 h A E}{W l}}\right]           (5.52).

\sigma_{i}=\sigma_{s}\left[1+\sqrt{1+\frac{2 h}{\delta_{s}}}\right]             (c).

Step II Static stress \left(\sigma_{s}\right) .
For a simply supported beam,

W=m g=50(9.81)=490.5 N .

M_{b}=\frac{W l}{4}=\frac{490.5(250)}{4}=30656.25 N – mm .

I=\frac{b d^{3}}{12}=\frac{a(a)^{3}}{12}=\frac{a^{4}}{12} mm ^{4} \quad y=\frac{a}{2} .

where a is the side of the square cross-section.

Therefore,

\sigma_{s}=\sigma_{b}=\frac{M_{b} y}{I}=\frac{(30656.25)\left(\frac{a}{2}\right)}{\left(\frac{a^{4}}{12}\right)} .

=\frac{183973.5}{a^{3}} N / mm ^{2}               (d)

Step III Static deflection.

\delta_{s}=\frac{W l^{3}}{48 E I}=\frac{(490.5)(250)^{3}(12)}{48(207000) a^{4}}=\frac{9256.11}{a^{4}} mm         (e).

Step IV Cross-section of beam
Equating impact stress to permissible stress,

\sigma_{i}=\frac{S_{y t}}{(f s)}=\frac{400}{2}=200 N / mm ^{2}         (f).

Substituting (d), (e) and (f) in Eq. (c),

200=\frac{183973.5}{a^{3}}\left[1+\sqrt{1+\frac{2(25) a^{4}}{9256.11}}\right] .

or      \frac{a^{3}}{919.87}=\left[1+\sqrt{1+\frac{2(25) a^{4}}{9256.11}}\right] .

\left(\frac{a^{3}}{919.87}-1\right)^{2}=1+\frac{a^{4}}{185.12}.

Simplifying,

\frac{a^{3}}{846160.82}-\frac{1}{459.94}=\frac{a}{185.12} .

The term (1/459.94) is very small and neglected.
Therefore, a = 67.6 or 70 mm.
The cross-section of the beam is 70 × 70 mm.
Step V Check for impact stresses

\sigma_{s}=\frac{183973.5}{a^{3}}=\frac{183973.5}{(70)^{3}}=0.5363 N / mm ^{2} .

\delta_{s}=\frac{9256.11}{a^{4}}=\frac{9256.11}{(70)^{4}}=3.855 \times 10^{-4} mm .

\sigma_{i}=\sigma_{s}\left[1+\sqrt{1+\frac{2 h}{\delta_{s}}}\right].

=0.5363\left[1+\sqrt{1+\frac{2(25)}{\left(3.855 \times 10^{-4}\right)}}\right] .

=193.68 N / mm ^{2} .

\therefore \quad \sigma_{i}<200 N / mm ^{2} .