A plane frame (see figure) consists of column AB and beam BC that carries a triangular distributed load (see figure part a). Support A is fixed, and there is a roller support at C. Beam BC has a shear release just right of joint B.
- Find the support reactions at A and C then plot axial-force (N), shear-force (V), and bending-moment (M) diagrams for both members. Label all critical N,K and M values and also the distance to points where any critical ordinates are zero.
a.
The support reaction at point A and C and plot the shear, moment, and axial force diagram.
Answer to Problem 4.5.39P
Explanation of Solution
Given: .
The given figure.
AB column and BC beam forms the plane frame that carries a load that is distributed in the triangular shape. At C there is roller support and support A is fixed. Below the B joint there is a moment release at column AB
Concept Used:
Vertical force equilibrium is given as,
Horizontal force equilibrium is given as,
Calculation: .
Vertical force equilibrium is given as,
Horizontal force equilibrium is given as,
At the top of the moment release the moment is given as,
In equation (1),
Below B there is moment release at point A,
At x the shear force is given and equated to 0,
At point x bending moment is maximum,
Axial force critical values
Shear force critical values
Moment critical values
Conclusion: .
Thus, the support reaction at point A and C and plot the shear, moment and axial force diagram.
b.
For the load that is parabolic, lateral acts to the right added to AB column and part (a) is repeated.
Answer to Problem 4.5.39P
Explanation of Solution
Given: .
The given figure:.
AB column and BC beam forms the plane frame that carries a load that is distributed in the triangular shape. At C there is roller support and support A is fixed. Below the B joint, there is a moment release at column AB.
Concept Used:
Vertical force equilibrium is given as,
Horizontal force equilibrium is given as,
Calculation: .
With the equilibrium force being vertical,
Moment at point B above moment release,
In (2) substitute
With the equilibrium force being horizontal,
As left of x axis is negative,
At A moment below the moment release,
At x shear force is calculated and equated to 0,
At x bending moment is calculated
For plane frame and critical values of N, V, M, the axial, shear and moment is given.
Axial force critical values
Shear force critical values in beam
Shear force critical values in column
Moment critical values in beam
Moment critical values in column
Conclusion: .
Thus, for the load that is parabolic lateral acts to the right added to AB column and part (a) is repeated.
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Chapter 4 Solutions
Mechanics of Materials, SI Edition
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