Other cases are: shear force is positive in the interval or 8. shear force is negative in the interval.
Connecting properly earlier points we get the sketch of cross-section forces diagrams, Fig. 5.6, Fig. 5.7 and Fig. 5.8.
Fig. 5.6 Reactions upwards, maximum of bending moment Project “The development of the didactic potential of Cracow University of Technology in the range of modern construction” is co-financed by the European Union within the confines of the European Social Fund and realized under surveillance of Ministry of Science and Higher Education.
Adam Paweł Zaborski
Fig. 5.7 Reactions upward and downward, no moment extremum Fig. 5.8 Reactions upward, no moment extremum Reverse problem
Reconstruct the load applied to the beam from the bending moment diagram in Fig.5. 9.
20
2°
2
2
2
2
20
40
Fig. 5.9 Bending moment diagram
Solution
We begin by determining load type:
1. The first interval (a cantilever) has constant continuous load (the parabola of 2nd order) 2. Remaining intervals have no continuous loading (straight lines of diagram) Project “The development of the didactic potential of Cracow University of Technology in the range of modern construction” is co-financed by the European Union within the confines of the European Social Fund and realized under surveillance of Ministry of Science and Higher Education.
Adam Paweł Zaborski
3. In the next characteristic points there are point loads: upward force of reaction, point moment clockwise (bottom fibers in tension), downward point force and upward point force of reaction, Fig. 5.10. There is no point force applied with point moment because lines are parallel (lack of bend).
q
P
2
2
2
2
RB
RA
Fig. 5.10 Loadings applied to the beam
2
qa
4. From extreme intervals we find: M =
= 2 ,
0
a = 2
→
q = 10 and RB = 20.
2
5. From moment equations for next points (from left and right) we find: RA = 20, P = 30. In this way we reconstructed the entire loading of the beam.
Workshop theme
For the beam in Fig. 5.11 calculate the constraints reactions, write down the cross-section forces functions and draw their diagrams.
q1
q2
M1
P
a
0,8a
1,5a
a
a
Fig. 5.11 Loaded beam
Input data:
a = ……. m (1,2÷4,8)
P = …….. kN (20÷80)
M1 = …….. kNm (10÷50)
q1 = ……. kN/m (12÷45)
q2 = ……. kN/m (14÷40)
Addendum
Tip: Try to guess the cross-section forces diagrams for several beams and next verify your prediction by using a computer program. Carefully analyse your mistakes (if any).
Tip: Remember the diagrams should be drawn in scale.
Tip: Solve the problems with real numbers obtaining numerical results. Don’t use parameters.
Project “The development of the didactic potential of Cracow University of Technology in the range of modern construction” is co-financed by the European Union within the confines of the European Social Fund and realized under surveillance of Ministry of Science and Higher Education.
Adam Paweł Zaborski
Review problems
Fig. 5.11 Review problems – beams
Project “The development of the didactic potential of Cracow University of Technology in the range of modern construction” is co-financed by the European Union within the confines of the European Social Fund and realized under surveillance of Ministry of Science and Higher Education.
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