Question #106429

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

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

Expert's answer

Here, above we can take the simply supported beam with R_{A} and R_{B} are the reaction forces, and here P_{1= }Load of bus as point load= 5000g N, P_{2}= Load of truck= 3000g N

P_{3=}P_{4}=P_{5}= distributed load of car= 1500g N here

g= acceleration due to gravity =10 m/s^{2}

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

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