Influence Line Diagram(ILD)
- Influence lines are important in the design of structures that resist large live loads.
- If a structure is subjected to a live or moving load, the variation in shear and moment is best described using influence lines.
- Constructing an influence line is completely different from constructing a shear or moment diagram.
Definition- An influence line for a member is a graph representing the variation of internal force( reaction, shear, moment, or deflection) in a fixed section of the member, due to a unit load traversing a structure.
Müller Breslau Principle for Qualitative Influence Lines
The Müller Breslau Principle states that the ordinate value of an influence line for any function on any structure is proportional to the ordinates of the deflected shape that is obtained by removing the restraint corresponding to the function from the structure and introducing a force that causes a unit displacement in the positive direction.
Example on ILD for Reaction
To obtain the influence line for the support reaction at A for the beam shown.
- First of all remove the support corresponding to the reaction and apply a force (Figure 2) in the positive direction that will cause a unit displacement in the direction of RA.
- The resulting deflected shape will be proportional to the true influence line (Figure 3) for the support reaction at A.
The deflected shape due to a unit displacement at A is shown in Figure 2 and matches with the actual influence line shape as shown in Figure 3. Note that the deflected shape is linear, i.e., the beam rotates as a rigid body without any curvature. This is true only for statically determinate systems.
Example on ILD for Shear
To draw the qualitative influence line for shear at section C of overhang beam.
- Introduce a roller at section C so that it gives freedom to the beam in the vertical direction as shown in Figure 5.
- Now apply a force in the positive direction that will cause a unit displacement in the direction of VC.
- The resultant deflected shape is shown in Figure 5. Again, note that the deflected shape is linear.
- Figure 6 shows the actual influence, which matches with the qualitative influence.
Example on ILD for Bending Moment
To draw a qualitative influence line for moment at C
- Introduce hinge at C and that will only permit rotation at C.
- Now apply moment in the positive direction that will cause a unit rotation in the direction of MC.
- The deflected shape due to a unit rotation at C is shown in Figure 8 and matches with the actual shape of the influence line as shown in Figure 9.
Maximum Shear in Beam supporting UDLs
If UDL is rolling on the beam from one end to other end then there are two possibilities. Either Uniformly distributed load is longer than the span or uniformly distributed load is shorter than the span. Depending upon the length of the load and span, the maximum shear in beam supporting UDL will change.
Note: For maximum values of shear, maximum areas should be loaded.
Case I - UDL longer than the Span
Given simply supported beam as shown in Figure 10 is loaded with UDL of w moving from left to right where the length of the load is
longer than the span. The influence lines for reactions RA, RB and shear at section C located at x from support A will be as shown in Figure 11, 12 and 13 respectively. UDL of intensity w per unit for the shear at supports A and B will be given by
To know shear at given section at C. As shown in Figure 13, maximum negative shear can be achieved when the head of the load is at the section C. And maximum positive shear can be obtained when the tail of the load is at the section C. As discussed earlier the shear force is computed by intensity of the load multiplied by the area of influence line diagram covered by load. Hence, maximum negative shear is given by
We know that the shear force is computed by the intensity of the load multiplied by the area of influence line diagram covered by load.
Hence, maximum negative shear is given by
and maximum positive shear is given by
Case II - UDL shorter than the Span
In this case also the procedure for calculating maximum(+ve or -ve) shear force at a particular section is same as discussed above.
Only difference is that we have to consider the area under ILD from point under consideration to the point which is x distance away. Where x is the length of udl(x<L)
Maximum Bending Moment at sections in beams supporting UDLs
Similarly here also the maximum moment at sections in beam supporting UDLs can either be due to UDL longer than the span or due to ULD shorter than the span.
Case I - UDL Longer than the Span
Assume the UDL longer than the span is travelling from left end to right hand for the beam as shown in Figure 14. We are interested to know maximum moment at C located at x from the support A.
Maximum bending moment is given by the load intensity multiplied by the area of influence line (Figure15) covered.
In the present case the load will cover the completed span and hence the moment at section C can be given by
Suppose the section C is at mid span, then x = l and maximum moment is given by
Case II - UDL Shorter than the Span
Given UDL of length y is smaller than the span of the beam AB. To find maximum bending moment at section C located at x from support A
Hence, the reaction at B is given by
And moment at C will be
Substituting value of reaction B in above equation, we can obtain
To compute maximum value of moment at C, we need to differentiate above given equation with reference to z and equal to zero.
The expression states that for the UDL shorter than span, the load should be placed in a way so that the section divides it in the same proportion as it divides the span. In that case, the moment calculated at the section will give maximum moment value.
Influence line diagram for Truss members
ILD for Inclined members
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