Solid Mechanics. Reactions and internal force diagrams. Step by step example. part 1

Solid Mechanics – Academic Year 2014/2015
Prof: Maribel Castilla Heredia @maribelcastilla v.1.0 October 2014
Bloque A. Sistemas isostáticos reticulados planos.
Solid Mechanics Block A. Internal forces. Diagrams.
Degree in Architecture. San Pablo CEU University – Institute of Technology.

1 – General layout.

Note that the 50 kN load that rests on the hinge can be located on any of the three systems that converge at the hinge.

As the hinge bonds both horizontal and vertical movement of the three solids that meet on it, two unknowns appear at point “C” on each of the three subsystems.

Vertical movement is the only one linked by the roller at E, so just one reaction will appear on each of the solids that meet at that point.

Hint:
If unknowns are positive you’ll always obtain the correct sign in the result.
Vertical movement is the only one linked by the roller at E, so just one reaction will appear on each of the solids that meet at that point.

If any of the subsystems is a 3-unknown one, this will mean that it’s a statically determiate one and it’s reactions don’t depend on the behaviour of the rest of the solids.

If any of the subsystems is a 3-unknown one, this will mean that it’s a statically determiate one and it’s reactions don’t depend on the behaviour of the rest of the solids.
System CDE meets this condition, so we’ll solve it in first place.

Exercise 1 - Reactions on each subsystem    33333310003010031003

Exercise 1 - Reactions on each subsystem  

 
3 3
3
3 3
3 100
0
3
0
1003
100
3

Solid Mechanics – Academic Year 2014/2015 Prof: Maribel Castilla Heredia @maribelcastilla v.1.0 October 2014
Exercise 1 - Reactions on each subsystem
   

    
3 3
3
3 3
3 100
0
3
0
100 3
100
3
     3 4 3 4
0

   
        
   
50 0 50
100
100 3 50 2 0 100
3
100 100
0
3 3
   

    
3 3
3
3 3
3 100
0
3
0
100 3
100
3

Subsistema 4
   50050100100350201003100100033    33333310003010031003

Exercise 1 - Reactions on each subsystem   1122436035341000232500231212

  
    
 
       
    
 
2 2
2 2
2
2
4 3 340
4 3 0
590
0 73 75 8
90
15
250 6
     
     
  1 1
2 2
4 3 60 3
5
3 4 100 0
2
3 250
0
2 3
1 2 1 2    

Exercise 1 - Reactions on each subsystem     8606806035810002332500223325060323

    860680603581000233250500625223832505906037375238

Bloque A. Sistemas isostáticos reticulados planos.
If you want to learn how to draw internal force diagrams for this example go for it on Part 2.
Solid Mechanics Block A. Internal forces. Diagrams.
Degree in Architecture. San Pablo CEU University – Institute of Technology.

Solid Mechanics. Reactions and internal force diagrams. Step by step example. part 1

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Solid Mechanics. Reactions and internal force diagrams. Step by step example. part 1