Advanced Mechanics - Solved Assignments and Essays

ADVANCED MECHANICS
ADVANCED MECHANICS
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1ADVANCED MECHANICS
Question 1:
The diagram of the beam is given below.
The beam AB is hinged at the end A and is being vertically supported at the other end B.
The applied force is distributed triangular force f(x) = f0*(x/L) (x is the positive horizontal
co-ordinate from the point A). The point A is taken as origin of the co-ordinate.
a) The reaction forces at the supporting points A and B are equal to vertical forces applied to
those points.
Reaction force at A or at length x=0 (RA) = f0*(0/L) = 0 kN.
Reaction force at B or at length x=L (RB) = f0*(L/L) = f0 kN.
b) and c) Now, the deflection equation of the beam will be
EI∗
( d4 y
d x4 )=−f 0∗( x
L )
The separating the variable and integrating the equation becomes

2ADVANCED MECHANICS
EIy∗
( d2 y
d x2 )=−f 0∗( x3
6 L )+ C 1∗x +C 2........(1)
Now, again separating the variables and then integrating twice we get
EI*y = −f 0∗( x5
120 L )+C 1∗x3
6 +C 2∗x2
2 +C 3∗x +C 4.....(2)
Now, the equation (1) is the bending moment equation and the equation (2) is the deflection
equation.
Now, applying the boundary conditions at x = 0 and at x= L both moment and deflection
vanish.
Hence, the constants of the equation are
C1 = f0*L/6, C2 = 0, C3 = -7*f0(L^3)/360, C4 = 0.
Substituting the values in the equation (2) gives
y= ( f 0∗x
360 LEI ) ( 7 L4 – 10 L2 x2
+3 x4 )
Now, Moment Mx=EI∗
( d2 y
d x2 )
Hence, the expression of moment from equation (1) will be
y∗Mx=−f 0∗( x3
6 L )+(f 0∗L
6 )∗x
 Mx=−f 0∗( x3
6 yL )+(f 0∗L
6 y )∗x......(3)
The equation (3) is the moment deflection equation.
Now, putting the values of constants in the equation (2) we get the equation of deflection.