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

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
you to find out the reaction forces in both simply supported at the end of the
bridge
1
2020-03-26T03:04:35-0400

Here, above we can take the simply supported beam with RA and RB 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"

RA+RB=(5000+3000+1500+1500+1500)g

RA+RB=125000 N ............(i)

and now we take total moment is equal to zero

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

RB="\\frac{5000+6000+5250+8250+10500}{8}"

RB=25812.5 N

RA+25812.5=125000

RA=99187.5 N

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

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