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*Influence Lines*

Influence Lines for a Simple Beam by Developing the Equations

problem statement

Draw the influence lines for the reactions Y_{A}, Y_{C}, and the shear and
bending moment at point B, of the simply supported beam
shown by developing the equations for the respective influence lines.

Figure 1 - Beam structure to analyze

- Reaction Y
_{A}

The influence line for a reaction at a support is found by
independently applying a unit load at several points on the structure
and determining, through statics, what the resulting reaction at
the support will be for each case. In this example, one such equation for the
influence line of Y_{A} can be found by summing moments around Support C.

Figure 2 - Application of unit load

M_{C} = 0 *(Assume
counter-clockwise positive moment)*

-Y_{A}(25)+1(25-x) = 0

Y_{A} = (25-x)/25 = 1 - (x/25)

The graph of this equation is the influence line for Y_{A} (See Figure 3).
This figure illustrates that if the unit load is applied at A, the
reaction at A will be equal to unity. Similarly, if the unit load is
applied at B, the reaction at A will be equal to 1-(15/25)=0.4, and
if the unit load is applied at C, the reaction at A will be equal to 0.

Figure 3 - Influence line for Y_{A}, the support reaction at A

The fact that Y_{A}=1 when the unit load is applied at A and zero when the unit
load is applied at C can be used to quickly generate the influence line diagram.
Plotting these two values at A and C, respectively, and connecting them with a straight line
will yield the the influence line for Y_{A}. The structure is statically determinate,
therefore, the resulting function is a straight line.

- Reaction at C

The equation for the influence line of the support reaction at C is found by developing an equation that relates the reaction to the position of a downward acting unit load applied at all locations on the structure. This equation is found by summing the moments around support A.

Figure 4 - Application of unit load

M_{A} = 0 *(Assume
counter-clockwise positive moment)*

Y_{C}(25)-1(x) = 0

Y_{C} = x/25

The graph of this equation is the influence line for Y_{C}.
This shows that if the unit load is applied at C, the
reaction at C will be equal to unity. Similarly, if the unit load is
applied at B, the reaction at C will be equal to 15/25=0.6. And,
if the unit load is applied at A, the reaction at C will equal to 0.

Figure 5 - Influence line for the reaction at support C

The fact that Y_{C}=1 when the unit load is applied at C and zero when the unit
load is applied at A can be used to quickly generate the influence line diagram.
Plotting these two values at A and C, respectively, and connecting them with a straight line
will yield the the influence line for Y_{C}. Notice, since the structure is
statically determinate, the resulting function is a straight line.

- Shear at B

The influence line for the shear at point B can be found by developing equations for the shear at the section using statics. This can be accomplished as follows:

a) if the load moves from B to C, the shear diagram will be as shown in Fig. 6 below, this
demonstrates that the shear at B will equal Y_{A} as long as the load is located
to the right of B, i.e., V_{B} = Y_{A}. One can also calculate the shear at
B from the Free Body Diagram (FBD) shown in Fig. 7.

Figure 6 - Shear diagram for load located between B and C

Figure 7 - Free body diagram for section at B with a load located between B and C

b) if the load moves from A to B, the shear diagram will be as shown in Fig. 8, below, this
demonstrates that the shear at B will equal -Y_{C} as long as the load is located
to the left of B, i.e., V_{B} = - Y_{C}. One can also calculate the shear
at B from the FBD shown in Fig. 9.

Figure 8 - Shear diagram for load located between A and B

Figure 9 - Free body diagram for section at B with a load located between A and B

The influence line for the Shear at point B is then constructed by drawing
the influence line for Y_{A} and negative Y_{C}. Then highlight
the portion that represents the sides over which the load was moving. In this case,
highlight the the part from B to C on Y_{A} and from A to B on -Y_{C}.
Notice that at point B, the summation of the absolute values of the positive and
negative shear is equal to 1.

Figure 10 - Influence line for shear at point B

- Moment at B

The influence line for the moment at point B can be found by using statics to develop equations for the moment at the point of interest, due to a unit load acting at any location on the structure. This can be accomplished as follows.

a) if the load is at a location between B and C, the moment at B can be calculated by using the
FBD shown in Fig. 7 above, e.g., at B, M_{B} = 15 Y_{A} - __notice that this relation is valid if and only if the load is
moving from B to C.__

b) if the load is at a location between A and B, the moment at B can be calculated by using the
FBD shown in Fig. 9 above, e.g., at B, M_{B} = 10 Y_{C} - __notice that this
relation is valid if and only if the load is moving from A to B.__

The influence line for the Moment at point B is then constructed by
magnifying the influence lines for Y_{A} and Y_{C} by 15 and 10, respectively,
as shown below. Having plotted the functions, 15 Y_{A} and 10 Y_{C}, highlight
the portion from B to C of the function 15 Y_{A} and from A to B on the function
10 Y_{C}. These are the two portions what correspond to the correct moment
relations as explained above. The two functions must intersect above point B.
The value of the function at B then equals (1 x 10 x 15)/25 = 6. This represents
the moment at B if the load was positioned at B.

Figure 11 - Influence line for moment at point B

Contact Dr. Fouad Fanous for more information.