Solving AB Beam Reactions with Hyperstatic System

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The discussion focuses on solving for the reactions of an AB beam subjected to a distributed load. The participant outlines their approach, identifying six reactions at the supports and applying equilibrium equations to find vertical and horizontal forces as well as moments. They calculate the total vertical reaction forces as 7.25 KN and determine the center of gravity of the load to be 1.28 m from one end. The participant seeks assistance in deriving the moment equation, noting the complexity due to the beam's potential statically indeterminate nature. The response highlights the need for additional methods, such as deflection compatibility equations, to solve the problem effectively.
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hello everyone
1. Homework Statement

Problem:
An AB beam, loaded with a distributed load q (KN/m), the length of the beam is 2 m.
EI=constant
p_109797omn1.png

find the reactions in the beam ,we use y''=-(1/EI) *M(x)

Homework Equations


y''=-(1/EI) *M(x)[/B]

The Attempt at a Solution


first we have 6 reactions (Va,Ha,Ma,Vb,Hb,Mb) in the supports.
∑F/x=0→ Ha+Hb=0 , Ha=-Hb=0 -------------1
∑F/y=0→ Va+Vb-Q=0 → Va+Vb=Q→ Va+Vb=∫ ((e^x)+(e^-x))dx 0≤ x ≤2
Va+Vb= 7,25 KN -------------2
∑M/b=0 → 2*Va- Q*(center of gravity of the load q)-Ma+Mb=0
center of gravity of the load q = (∫∫x*ds)/(∫∫ds)
∫∫x*ds= ∫∫x*dx*dy 0≤y≤(e^x)+(e^-x) , 0≤x≤2
∫∫x*ds= ∫∫x*dx*dy = 9,25 KN*m
∫∫ds= 7,25 KN
center of gravity of the load q = (∫∫x*ds)/(∫∫ds)=1,28 m (2-1,28)= 0,72m
2*Va- Q*0,72-Ma+Mb=0
2*Va-Ma+Mb=5,22 KN m----------3
Differential equation:
y''=-(1/EI) *M(x)
M(x) = ?? 0≤x<2
my problem is how to find the moment equation?
thanx
 

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Zouatine said:
hello everyone
1. Homework Statement

Problem:
An AB beam, loaded with a distributed load q (KN/m), the length of the beam is 2 m.
EI=constant
View attachment 236680
find the reactions in the beam ,we use y''=-(1/EI) *M(x)
2. Homework Equations
y''=-(1/EI) *M(x)
3. The Attempt at a Solution

first we have 6 reactions (Va,Ha,Ma,Vb,Hb,Mb) in the supports.
∑F/x=0→ Ha+Hb=0 , Ha=-Hb=0 -------------1
∑F/y=0→ Va+Vb-Q=0 → Va+Vb=Q→ Va+Vb=∫ ((e^x)+(e^-x))dx 0≤ x ≤2
Va+Vb= 7,25 KN -------------2
∑M/b=0 → 2*Va- Q*(center of gravity of the load q)-Ma+Mb=0
center of gravity of the load q = (∫∫x*ds)/(∫∫ds)
∫∫x*ds= ∫∫x*dx*dy 0≤y≤(e^x)+(e^-x) , 0≤x≤2
∫∫x*ds= ∫∫x*dx*dy = 9,25 KN*m
∫∫ds= 7,25 KN
center of gravity of the load q = (∫∫x*ds)/(∫∫ds)=1,28 m (2-1,28)= 0,72m
2*Va- Q*0,72-Ma+Mb=0
2*Va-Ma+Mb=5,22 KN m----------3
Differential equation:
y''=-(1/EI) *M(x)
M(x) = ?? 0≤x<2
my problem is how to find the moment equation?
thanx
i haven’t checked your calculus, but you seem to be on the right track. However, are you sure that both ends are fixed? If so, the problem is statically indeterminate to the second degree, and you have to resort to other equations besides the equilibrium equations, like deflection compatibility equations, virtual work, or other methods, which is a bit tedious.
 

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