A simple beam ACE is constructed with square cross sections and a double taper (see figure). The depth of the beam at the supports is dAand at the midpoint is dc= 2d 4. Each half of the beam has length L. Thus, the depth and moment of inertia / at distance x from the left-hand end are, respectively, in which IAis the moment of inertia at end A of the beam. (These equations are valid for .x between 0 and L, that is, for the left-hand half of the beam.)
- Obtain equations for the slope and deflection of the left-hand half of the beam due to the uniform load.
a.
The equations for the slope and deflection of the left-hand half of the beam for the given simple beam due to uniform load.
Answer to Problem 9.7.12P
The equation for the slope and deflection of the left-hand half of the beam are:
Explanation of Solution
Given: .
We have,
Length of the beam, AB =2L..
Depth of the beam at support,
Depth of the beam at midpoint,
Moment of inertia, I
Depth at the left-hand end,
Concept Used: .
Annuity problem requires the use of the differential equation as follows:.
Calculation: .
With the use of we can calculate,
The reaction forces at point A and B would be,
The bending moment would be,
Now we are using second degree differential eqaution as follow,
In the above equation taking integration at both sides,
The boundary conditions at x =L..
In the above equation, Once again taking integration, we will get,
Conclusion: .
The balance in the annuity after
b.
The formulas for the angle of rotation at the support and the deflection at the midpoint with the use of equations found in part (a).
Answer to Problem 9.7.12P
The equation for the angle of rotation
Explanation of Solution
Given: .
We have,
Length of the beam, AB =2L..
Depth of the beam at support,
Depth of the beam at midpoint,
Moment of inertia is I..
Depth at the left-hand end,
Concept Used: .
Annuity problem requires the use of the differential equation as follows:.
Calculation: .
We have the equation,
On this above equation, boundary condition at x =0 and v =0..
So, deflection beam would be,
Deflection at,
Conclusion: .
The equation for the angle of rotation
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Chapter 9 Solutions
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