Consider the shearing force and bending moment diagrams for the system of forces act- ing on the beam in Fig. 2.36. For the moment, only a simple system of three point loads will be considered.
24 kN C
B 3 m
5 m
RB A
RA
Figure 2.36 Forces acting on a beam.
It is first necessary to calculate the reactions at A and B as previously described in Section 2.4.4. The beam is simply supported at A and B. This means that it rests on supports at these points giving vertical reactions. The general conditions for equilibrium require that the resultant moment about any point must be zero, and the sum of the upward forces must equal the sum of the downward forces. Therefore, taking moments about A, the moment for RBmust balance the moment for the load C:
RB8 m24 kN5 m RB(24 kN5 m)/8 m RB15 kN
Similarly:
RA24 kN15 kN9 kN
Immediately to the right of A the shearing force is due to RAand is therefore 9 kN. As this force is to the left of the section considered it is upwards, hence the shearing force is posi- tive. The shearing force will be constant for all points between A and C as no other forces are applied to the beam between these points.
When a point to the right of C is considered, the load at C as well as the reaction force RB must be taken into account. Alternatively RBon its own can be considered. The shearing force will be 15 kN. This is the value previously calculated for RBor it can be calculated from the load CRA24 kN9 kN15 kN. For any point between C and B the force to the right is upwards and the shearing force is therefore negative as was shown earlier in Fig. 2.34. It should be noted that the shearing force changes suddenly at point C. This is shown in Fig. 2.37(a).
The corresponding bending moments are shown in Fig. 2.37(b). The bending moment at A is zero, since there are no forces to the left of point A. At a point 1 m to the right of point A the moment of the only force RAto the left of this point is RA1 m9 kN m. As this moment about A is clockwise the moment is positive (9 kN m). At points 2, 3, 4 and 5 m to the right of A the bending moments are respectively:
RA2 m9 kN2 m18 kN m RA3 m9 kN3 m27 kN m RA4 m9 kN4 m36 kN m RA5 m9 kN5 m45 kN m
Since all these are clockwise moments they are all positive bending moments. For points to the right of C, the load at C as well as RA must be considered or, more simply, as pre- viously demonstrated RB alone can be used. At points 5, 6 and 7 m from A, the bending moments are respectively:
RB3 m15 kN3 m45 kN m RB2 m15 kN2 m30 kN m RB1 m15 kN1 m15 kN m
As these moments to the right of the points considered are anticlockwise, they are all posi- tive bending moments. At B the bending moment is zero as there is no force to its right.
The results are summarized in Table 2.5.
SF diagram
BM diagram
A B
9 kN C
15 kN
45 kN m (a)
(b)
Figure 2.37 Shearing force and bending moment diagrams.
Table 2.5 Shearing force and bending moment values for Figure 2.37.
Distance from A (m) 0 1 2 3 4 5 6 7 8
Shearing force (kN) 9 9 9 9 9 9 15 15 15 15
Bending moment (kN m) 0 9 18 27 36 45 30 15 15
Using the results from Table 2.5 the shear force and bending moment diagrams can be drawn as shown in Fig. 2.38. A stepped shearing force diagram with only horizontal and vertical lines can only exist when the beam only carries concentrated, point loads. A sudden change in shearing force occurs where the concentrated loads, including the reactions at the supports, occur. For this type of simple loading the bending moment diagram also con- sists of straight lines, usually sloping. Sudden changes of bending moment cannot occur except in the unusual circumstances of a moment being applied to a beam as distinct from a load.
a x
x
l
Uniform load (W/l)
RB 2 RA Wl
2 Wl
(a) Loading diagram
2 Wl 2
Wl
A
B
(b) Shearing force diagram
(c) Bending moment diagram
Parabolic curve 8
Wl2
A B
Figure 2.38 Beam with uniformly distributed load (UDL).
In practice a beam loaded with concentrated point loads alone cannot exist. This is because the beam itself has mass and, therefore, weight. In a beam of uniform cross-section this represents a uniform load throughout the length of the beam. The effect of uniform load- ing will now be considered.
Figure 2.38(a) shows a uniformly loaded beam of length l and weight W. The only point loads being the reactions at the supports RAand RB. Since the loading is symmetrical RA and RBwill both equal W/2. The shear force diagram shows maximum values at the points of support and zero shear at the midpoint. This time, however, the line joining the shear- ing force is a sloping line passing through the midpoint.
Take a section XXat a distance afrom the left-hand support RA: The shearing force at XX(Wl/2)(Wa)
Since this is the equation of a straight line the shear force diagramis shown in Fig. 2.38(b).
Unlike the previous example, this time the bending moment diagramis not made up of straight lines but is a continuous curve with a maximum value at the midpoint.
The bending moment Mxat XX[(Wl/2)a][Wa(a/2)]
Therefore:
MxW/2(laa2) (1)
This is the equation of a parabola so, its first derivative will be a maximum when:
dMx/da0
Differentiating equation (1):
dMx/daW/2(l2a)0 When:
al/2
The maximum bending moment occurs at the centre of the span which is only to be expected with symmetrical loading. Hence substituting al/2 in equation (1):
MmaxW/2[(l2/2)(l2/4)]Wl2/8
The bending moment diagram is shown in Fig. 2.38(c). Apart from the beam shown in Fig. 2.38(a) where the uniform load resulted from gravity acting on the mass of the beam itself, the only other occasion when a beam is uniformly loaded is when it is carrying a uniform panel of masonry.
More often, there is a combination of point loads and uniform loading as shown in the loading diagram Fig. 2.39(a).
(a) Calculate reactions RAand RCby taking moments about A:
(452)(129)4.5(2412)9RC Thus: RC96 kN and RA81 kN
(b) Shearing forces (let xbe any distance from A):
For 0x2, shear force(8112x) kN (1)
For 2x9, shear force(8112x45)(3612x) kN For 9x12, shear force(811294596) 24 kN These results are depicted by straight lines as shown in Fig. 2.39(b)
(c) Bending moments (let xbe any distance from A):
For 0x2, M81x12(x2/2)81x6x2kN m Therefore: at A, x0 and MA0
at B, x2 and MB16224138 kN m
For 2x9, M81x(12x2/2)45 (x2)36x6x290 kN m (2) At x2 MB722490138 kN m (as above)
At x9 MC32448690 72 kN m
45 kN 24 kN
D C
B
(a) Loading diagram A
2 m
RA
3 m 9 m
24 kN
72 kN
72 kN m
24 kN
3 m
7.9 m 81 kN
57 kN
B
B A
A
(b) Shearing force diagram
(c) Bending moment diagram
C
C Point of
contraflexure
D
D 12 kN
138 kN m
144 kN m
RB UDL 12 kN/m
Figure 2.39 Beam with both point and uniform loads. UDL: Uniformly distributed load.
The maximum bending moment occurs between the points B and C where dM/dx0.
Note this is the same point that the shearing force is also zero. From equation (2):
dM/dx3612x0 Therefore:
x3 m
Substituting x3 in equation (2):
(363)(632)90144 kN m.
Thus the maximum bending moment is 144 kN m and it occurs at a point 3 m from A.
Contraflexure will occur when the bending moment is zero. That is, the positive root of equation (2) when it is equated to zero.
36x6x2900 Therefore:
x7.9 m(the only positive solution)
Therefore contraflexure occurs at a point 7.9 m from A.