Answer to Question #106429 in Mechanical Engineering for Chan Lap Hang

Question

A simply supported bridge shows below is the primary path for the village people
to travel to the city centre. In the following case, the loading from the bus (5000
kg) and truck (3000 kg) can be treated as concentrated loads and the three private
cars (1500 kg each) can be treated as evenly distributed load. Your supervisor asks
you to find out the reaction forces in both simply supported at the end of the
bridge

Answer 1

Here, above we can take the simply supported beam with R_A and R_B are the reaction forces, and here P₁₌ Load of bus as point load= 5000g N, P₂= Load of truck= 3000g N

P₃₌P₄=P₅= distributed load of car= 1500g N here

g= acceleration due to gravity =10 m/s²

Now for equilibrium condition,

we know that

$\Sigma F= 0$

R_A+R_B=(5000+3000+1500+1500+1500)g

R_A+R_B=125000 N ............(i)

and now we take total moment is equal to zero

R_A$\times0+R_B\times8=5000\times1 +3000\times2+1500\times3.5+1500\times5.5+1500\times7$

R_B=$\frac{5000+6000+5250+8250+10500}{8}$

R_B=25812.5 N

R_A+25812.5=125000

R_A=99187.5 N

So, from here we got the reaction force A and B