Calculated the magnitude of the instantaneous stress:
While a concentrated force of 1 kN is applied at the midspan point of a simply supported beam a static deflection of 5 mm is generated. The same load generates a maximum stress of 158 MN/m^{2}. Calculated the magnitude of the instantaneous stress generated when a weight of 10 kg is allowed to fall through a height of 12 mm onto the beam at midspan. What shall be the instantaneous deflection?
Solution
Let We be the equivalent slowly applied load which generates the same static deflection δ_{max} as that of the impact load.
Then,
We/δmax = 1 kN /0.005 m
= 200 kN/m
Now, equating the work completed by the impact load to that done by the equivalent static load, we obtain
(We/2) δ _{max} = W ( h + δ _{max} )
i.e.
(We /2) × (We/200) = ((10 × 9.81)/1000) (0.012 + We /200)
Therefore,
= 0.981 + 0.6932 = 0.791 kN
But
δ _{max} = We / 200 = 0.791 /200
= 0.003955 m = 3.955 mm ; 4 mm
This is given that the load of 1 kN generates a maximum stress of 158 MN/m^{2}. Thus, a load of We must produce a maximum stress, σ_{max }equal to (158 We) MN/m^{2}.
∴ σ _{max} = 158 × 0.791 = 124.978 MN/m^{2} ; 125 MN/m^{2}
Therefore, the maximum stress & maximum deflection are 125 MN/m^{2} & 4 mm, respectively.