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/m2. 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?
Let We be the equivalent slowly applied load which generates the same static deflection δmax as that of the impact load.
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 )
(We /2) × (We/200) = ((10 × 9.81)/1000) (0.012 + We /200)
= 0.981 + 0.6932 = 0.791 kN
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/m2. Thus, a load of We must produce a maximum stress, σmax equal to (158 We) MN/m2.
∴ σ max = 158 × 0.791 = 124.978 MN/m2 ; 125 MN/m2
Therefore, the maximum stress & maximum deflection are 125 MN/m2 & 4 mm, respectively.