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Influence Lines
Calculation of Maximum and Minimum Shear Force and Moments on a Statically Determinate Continuous Beam

problem statement

Determine the resulting forces for RA, RB, RC, Ms1, Vs1, Ms1, Vs2, VBL, and VBR under a uniform live load of 2 k/ft and a uniform dead load of 3 k/ft for the beam below.
note: influence lines for this beam are developed in the Statically Determinate Continuous Beam example.

Beam structure to analyze
Figure 1 - Beam structure to analyze

From the Continuous Beam with a Hinge example, the required influence lines for the structure are:

In order to calculate the forces due to uniform dead and live loads on a structure, a relationship between the influence line and the uniform load is required. Referring to Figure 2, each segment dx, of a uniform load w, creates an equivalent concentrated load, dF = w dx, acting a distance x from an origin.

From the general properties for influence lines, given in the introduction, it is known that the resulting value of the function for a force acting at a point is equivalent to the magnitude of the force, dF, multiplied by the ordinate value, y, of the influence line at the point of application.

Equivalent concentrated load
Figure 2 - Equivalent concentrated load

In order to determine the effect of the uniform load, the effect of all series loads, dF, must be determined for the beam. This is accomplished by integrating y dF over the length of the beam, i.e., w y dx = wy dx. The integration of y dx equal to the area under the influence line. Thus, the value of the function caused by a uniform load is equal to the magnitude of the uniform load multiplied by the area under the influence line diagram.

In order to find the resulting minimum and maximum values for the reactions, shears, and moments required, create a table which contains the resulting positive and negative values for the areas enclosed by the influence lines for each function. The effect of the dead load is determined by multiplying the net area under the influence line by the dead load. For the live load, multiply the respective positive and negative areas by the live load, yields to the positive and negative forces, respectively. The resulting maximum and minimum forces for dead load plus the effects of positive and negative live loads are then found by adding the respective values.

The resulting forces due to a uniformly distributed dead load = 3 k/ft and a live load = 2 k/ft applied to the beam above, are as follows:

Positive area under the influence line
Negative area under the influence line
Net area
Force due to DL
Positive force due to LL
Negative force due to LL
Maximum force (DL+LL)
Minimum force
RA 4 -1 3 9 8 -2 17 7
RB 10 - 10 30 20 - 50 30
RC 3 - 3 9 6 - 15 9
MS1 8 -4 4 12 16 8 28 4
VS1 1 -2 -1 -3 2 4 -1 -7
MS2 4.5 - 4.5 13.5 9 - 22.5 13.5
VS2 0.75 -0.75 - - 1.5 -1.5 1.5 -1.5
VB-R 5 - 5 15 10 - 25 15
VB-L - 5 -5 -15 - 10 -15 -25

Column IV = Column II + Column III
Column V = Dead Load * Column IV
Column VI = Live Load * Column II
Column VII = Live Load * Column III
Column VIII = Column V + Column VI
Column IX = Column V + Column VII

Contact Dr. Fouad Fanous for more information.