Beam span and loading.
Static Equilibrium.
The beam illustrated above is supported at A by a pin and at B by a roller. These reactions have to following type of unknown force components:
- @ A - Vertical reaction, Ry, and @ B - Horizontal reaction, Ry
- @ B - Vertical reaction, Ry
The first step is to draw a free body diagram (see Figure T103.4) that graphically illustrates the beam and the external forces acting on it. This problem is unlike the previous ones in that the beam has an internal pin. From the previous examples we know that the sum of the moments at a pinned support equate to zero. This is also true for an internal pin: in effect we have a forth equation of equilibrium that states:
Recognising this forth equation enables us to split the beam as shown in Figures T107.0 and T107.2. It can be seen that each part is in static equilibrium, and for each part there are three equations of equilibrium, therefore each beam part statically determinate.
Figure T105.1 T5 - Free Body Diagram
From Figure T105.1, it can be seen that there unknown forces, for which there are three equations of equilibrium:
Therefore: The beam can be said to be statically determinate.
Horizontal Reactions
The beam, under the loading indicated above, is not subjected to a lateral load, i.e. there is no load acting horizontally either to the left or the right. Therefore to maintain static equilibrium, where the sum of the horizontal forces equates to zero there cannot be any horizontal action at either A or B. Therefore:
Vertical Reactions
For a simply supported beam, subjected to any combination of loads, the vertical reactions at the supports can be calculated by taking moments about one support to determine the other. It should be noted that moments can be taken about either support to determine the vertical reaction at the other. There is no hard and fast rule as to which support to consider first. The decision is usually made by considering the beam geometry and the arrangement of loads and calculating the support reaction that is "easiest" first.
When taking moments about any point you should remember that a moment is simply a force multiplied be a lever arm (the perpendicular distance between the point of interest and the centroid of the load).
Figure _T107.2 Lever Arm - Moments about A
Figure _T107.3 Lever Arm - Moments about B
Figures T105.2 & T105.3 illustrate the lever arms associated with differing load types. In each case the lever arm is taken from the centroid of the load. Figure T103.2 shows the lever arm associated with the three loads when moments are taken about the support at A, whereas Figure T103.3 considers the moments being taken about the support at B.
Consider the reaction at "B". This can be calculated by taking moments about "A". For this, it is usual to assume that any loads or forces acting in a clockwise direction about the point of interest are positive, i.e. any loads acting clockwise about "A" will be treated as positive".
Using the third equation of equilibrium, we can say that the sum of all the moments at A equate to zero.
Considering the loads individually:
- For the partially distributed load (PDL), this acts clockwise about A and is therefore POSITIVE.
- For the point loads (PL1 & PL2) you should note the sign i.e. if they are positive or negative. When the point load is positive it acts in the direction of the arrow - it acts clockwise about A and is treated as POSITIVE. If the point load is negative, it acts anti-clockwise about A and is therefore treated as being NEGATIVE.
- For the applied moment (M1), this when acting clockwise as indicated by the arrow acts vertically downwards at B. This generates a POSITIVE vertical reaction Vb - see Figure T105.5. It should be noted that when calculating the reaction associated with a moment there is no lever arm to consider - reference should be made to your formal lecture notes regarding Couples.
- For the vertical reaction at B (Vb), this acts anti-clockwise about A and is therefore NEGATIVE.
Figure _T107.5 Couple - Reaction at support
The first step in calculating the vertical reaction at B is to formulate an expression using the third equation of equilibrium: ∑ MB = 0.
Summing the moments at A gives:
It's now a simple case of substituting into the above expression the relevant values for the beam spans and loads. This can then be solved to give the reaction at B.
Simplifying gives:
The previous equation only considers the vertical load. We have not considered the reaction at B caused by the rotational moment (M1). Remember this has no lever arm. Also remember to take note of the direction in which the moment is action. A clockwise moment causes a POSITIVE reaction at B, and an anti-clockwise moment causes a NEGATIVE reaction at B.
The reactions at A can now be calculated from simple statics, recognising that the sum of the vertical forces equates to zero i.e. ΣV = 0.
In the calculation above for the reaction at A, the moment is not included. It has already been accounted for in the calculation of the reaction at B so implicitly included in the above.
To demonstrate the point made earlier, that you can take moments about either support to determine the vertical reaction of the other support, the following will take moments about B to determine the vertical reaction at A.
Considering the loads individually:
- For the partially distributed load (PDL), this acts anti-clockwise about B and is therefore NEGATIVE.
- For the point loads (PL1 & PL2) you should note the sign i.e. if they are positive or negative (not to be confused with which direction they rotate about B). PL1 acts NEGATIVELY about B, and PL2 acts POSITIVELY about B.
- For the reaction at A (Va), this acts clockwise about B and is therefore POSITIVE.
Putting the above three equations together gives a expression for the sum of the moments about B. This can then be solved to give the vertical reaction at A. Note: This again excludes the applied moment - this will be added later.
Adding in the moment (look carefully at the direction of rotation!) and simplifying gives:
