Influence Line Diagram in Structural Analysis: Definition, Construction, and Applications_ Part - III

 

Influence Lines for Girders with Floor Systems

 

 

Influence Lines for Shears and Bending Moments in Girders

• To understand the procedure for constructing influence lines for shears and bending moments in girders that support bridge or building floor systems, consider a simply supported girder.

• A unit load moves from left to right across the stringers.
• The stringers are assumed to be simply supported on the floor beams.
• The effect of the moving unit load is transmitted to the girder at several specific points, labeled A through F.
• These points (A, B, C, D, E, F) are called panel points.
• The segments of the girder between panel points (for example, AB or BC) are called panels.
• In the illustrative sketch (Fig-a):
• The stringers are shown resting on top of the floor beams.
• The floor beams are shown resting on top of the girder.
• This arrangement is used only for clarity in diagrams to show how loads are transmitted between members.
• In actual floor systems, the members are not stacked vertically as shown in the figure.
• Instead:
• The top edges of the stringers and floor beams are generally aligned (level) with each other.
• They are usually positioned either lower than or at the same level as the top edges of the girders (as shown in Fig. above – plan and section).

 

Influence Lines for Reactions

• The influence lines for the vertical reactions (at support A) and  (at support F) can be determined using the equilibrium equations.
• Considering the simply supported girder in Fig. (a), when a unit load moves across the span, the equilibrium of the structure gives:

• Here:
• = total span length of the girder
•  = distance of the unit load from the left support (A)
• Therefore:
• When the unit load is at A (x = 0) → ,
• When the unit load is at F (x = L) → ,
• The influence line for  is a straight line sloping downward from 1 at A to 0 at F.
• The influence line for  is a straight line sloping upward from 0 at A to 1 at F.
• These influence lines are identical to those for the reactions of a simply supported beam directly subjected to a moving unit load.

 

Influence Line for Shear in Panel BC

• To determine the influence line for shear in panel BC, consider a section just to the right or left of point B (depending on which side the shear is desired).
• The shear at this section changes as the unit load moves across the span.
• The ordinate of the shear influence line at any position of the load is found by:
• Cutting the beam at the desired section.
• Using equilibrium equations for either the left or right portion of the girder.
• When the unit load is between A and B, the shear just to the right of B is equal to the reaction at A.
• When the unit load is between B and F, the shear changes by subtracting the effect of the unit load from the reaction at A.
• The influence line is therefore piecewise linear, with a discontinuity (jump) at B, corresponding to the unit shear passing through that panel.

 

Influence Line for Shear in Panel BC

 

Next, suppose that we wish to construct the influence lines for shears at points G and H, which are located in the panel BC, as shown in Fig. (a).

When the unit load is located to the left of the panel point B, the shear at any point within the panel BC (e.g., the points G and H) can be expressed as

   

Similarly, when the unit load is located to the right of the panel point C, the shear at any point within the panel BC is given by

When the unit load is located within the panel BC, as shown in Fig. (d), the force exerted on the girder by the floor beam at B must be included in the expression for shear in panel BC:

Thus, the equations of the influence line for can be written as

These expressions for shear do not depend on the exact location of a point within the panel; that is, these expressions remain the same for all points located within the panel BC.

The expressions do not change because the loads are transmitted to the girder only at the panel points; therefore, the shear in any panel of the girder remains constant throughout the length of that panel.

Thus, for girders with floor systems, the influence lines for shears are usually constructed for panels rather than for specific points along the girders.

The influence line for the shear in panel BC, obtained by plotting above S_BC Equations, is shown in Fig. (e).

 

Influence Line for Bending Moment at G

 

The influence line for the bending moment at point G, which is located in the panel BC (Fig. (a)), can be constructed by using a similar procedure.

When the unit load is located to the left of the panel point B, the bending moment at G can be expressed as

When the unit load is located to the right of the panel point C, the bending moment at G is given by

When the unit load is located within the panel BC, as shown in Fig. (d), the moment of the force  exerted on the girder by the floor beam at B, about G, must be included in the expression for the bending moment at G:

Thus, the equations of the influence line for  can be written as

Equation above M_G indicates that, unlike shear, which remains constant throughout a panel, the expressions for the bending moment depend on the specific location of the point G within the panel BC.

The influence line for , obtained by plotting Equations, is shown in Fig. (f). It can be seen from this figure that the influence line for , like the influence line for shear constructed previously (Fig. (e)), consists of three straight-line segments, with discontinuities at the ends of the panel containing the response function under consideration.

 

Influence Line for Bending Moment at Panel Point C

 

When the unit load is located to the left of C (Fig. (a)), the bending moment at C is given by

The influence line obtained by plotting these equations is shown in Fig. (g). Note that this influence line is identical to that for the bending moment of a corresponding beam without the floor system.

 

Procedure for Analysis

 

As the foregoing example indicates, the influence lines for girders supporting floor systems with simply supported stringers consist of straight-line segments with discontinuities or slope changes occurring only at the panel points.

In the influence lines for shears and bending moments at points located within panels, the changes in slope occur at the panel points at the ends of the panel containing the response function (see Fig. (e) and (f)).

However, in the influence lines for bending moments at panel points, the change in slope occurs directly at the panel point where the bending moment is evaluated.

Therefore, the influence lines for girders can be conveniently constructed as follows:

 

Step 1: Determine Ordinates at Key Points

Find the influence-line ordinates at:

• the supports, and
• the panel points where slope changes occur,
  by placing a unit load successively at each of these points and applying the equilibrium equations.

 

Step 2: Special Case – Cantilever Girder

For the influence line of bending moment at a panel point of a cantilever girder, the ordinate at that point will be zero.

In this case, determine at least one additional ordinate (usually at the free end of the cantilever) where the value is nonzero, to complete the influence line.

 

Step 3: Internal Hinges

If the girder contains internal hinges, the influence line will be discontinuous at those panel points where hinges are located.

If an internal hinge lies within a panel, discontinuities will appear at the panel points on either side of that panel. Compute the influence-line ordinates at these hinge-related points by:

• placing the unit load at these points, and
• applying the equilibrium equations and/or equation of conditions.

 

Step 4: Connect Ordinates

Finally, complete the influence line by connecting the previously determined ordinates with straight lines, and find any remaining ordinates by applying the geometry of the influence line.

 

Example:

Draw the influence lines for the shear in panel BC and the bending moment at B of the girder with floor system shown in Fig. (a).

1) Influence line for the shear in panel BC,  

Procedure (brief):

• Place a 1-k load successively at the panel points A, B, C, and D (the transferred unit at each panel point is taken as 1 k at that point) and compute the shear in panel BC for each case using equilibrium (reactions by proportion).
• Connect the ordinates by straight lines.

2) Influence line for the bending moment at panel point B,  

Procedure (brief):

• Place a 1-k load successively at panel points A, B and D (these three positions suffice because of the panel layout / symmetry) and compute the bending moment at B each time by equilibrium (use reaction at A times distance AB, etc.).
• Connect the ordinates by straight lines.

·     Compute reactions for a unit load at position  from left:

 

·    

 

 

Example

 

Draw the influence lines for the shear in panel CD and the bending moment at D of the girder with floor system shown in Fig. (a).

 

Influence Line for S_CD : To determine the influence line for the shear in panel CD, we place a 1 kN load successively at the panel points B, C, D and F. For each position of the unit load, the appropriate support reaction is first determined by proportions, and the shear in panel CD is computed.

The ordinates at the ends A and H of the girder are then determined from the geometry of the influence line.

Influence Line for M_D : To determine the influence line for the bending moment at panel point D, we place the1 kN load successively at the panel points B, D and F. For each position of the unit load, the bending moment at D is determined

 

Example:

          Draw the influence lines for the reaction at support A, the shear in panel CD, and the bending moment at D of the girder with floor system shown in Fig. (a).

Solution

Influence Line for A_y : To determine the influence line for the reaction A_y,  place a 1 k load successively at the panel points A, B, and C. For each position of the unit load, the magnitude of A_y is computed by applying the equation of condition ∑ M^AF_F =0.

Influence Line for S_CD : Place the 1 k load successively at each of the five panel points and determine the influence-line ordinates as follows.

Influence Line for M_D :  Place the 1 k load successively at each of the five panel points and determine the influence-line ordinates as follows.

 

Moment Envelope & Absolute Maximum Live Load Moment

 

Case 1 — Single Concentrated Load

• A single moving point load P creates a triangular bending moment diagram.
• Maximum moment always occurs directly under the load.
• As the load moves:
• Moment at that point = 0 at supports.
• Moment increases to 0.25 PL at midspan.
• Moment diagrams for positions at L/6, L/3, L/2 show how the bending moment increases.
• Moment Envelope:
• Formed by connecting maximum moment values at each beam section (top curve of all possible moment diagrams).
• Represents the required flexural capacity for design.
• Absolute maximum moment for a single moving load occurs at midspan.

 

Figure: Moment envelope for a concentrated load on a simply supported beam: (a) four loading positions (A through D) considered for construction of moment envelope; (b) moment curve for load at point B; (c) moment curve for load at point C; (d) moment curve for load at point D (midspan); (e) moment envelope, curve showing maximum value of moment at each section.

 

Case 2 — Series of Wheel Loads

 

• A set of moving loads (e.g., truck wheel loads) produces greater moment near midspan, but:
• Critical section is NOT always midspan.
• Influenced by spacing and magnitudes of wheel loads.

Steps to Find Maximum Moment at a Section

• Draw the influence line for moment at the section being evaluated.
• Use increase–decrease method to position the wheel loads for maximum effect.

Locating the Most Critical Section in the Beam

• Unlike a single load, a series of loads may produce maximum moment under any wheel, not always at midspan.
• Experience shows:
• Maximum moment likely occurs under a wheel near the resultant R of the load system.
• Assume maximum moment occurs under wheel 2, located a distance x from midspan.

Finding the Position for Absolute Maximum Moment

• Express moment under wheel 2 as a function of x.
• Use the resultant R of all wheel loads to compute reactions.
• Differentiate the moment equation w.r.t x:
• The value of x gives the critical location of wheel 2 → which gives the absolute maximum live load moment.

 

Summing moments about support B

To compute the moment M in the beam at wheel 2 by summing moments about a section through the beam at that point

Expression for Moment at Wheel 2

• Take moments about a section directly under wheel 2.
• Reaction at support A = R_A.
• Distance from A to wheel 2 = (L/2 – x).
• Distance between wheel 1 and wheel 2 = a.
• Moment at wheel 2:

 Substitute Reaction Using Resultant R

From earlier:

Substitute this into the moment expression → yields:

Find Position of Wheel 2 That Maximizes Moment

To establish the maximum value M, differentiate the above Equation with respect to x and set the derivative equal to zero

• Differentiate  with respect to :

• Set derivative = 0 for maximum moment:

• Solve for x:

Meaning of the Result

•  means:
• The maximum moment under wheel 2 occurs when the beam’s midspan lies halfway between wheel 2 and the resultant R.
• The centerline (midspan) divides the distance between the resultant and the selected wheel.

 

Example:

Determine the absolute maximum moment produced in a simply supported beam with a span of 30 ft by the set of loads shown in Figure below:

 

Figure: (a) Wheel loads; (b) position of loads to check maximum moment under 30­ kip load; (c) position of loads to check maximum moment under 20­ kip load.

Solution:

1. Resultant magnitude & location

• kips.
• Locate resultant  measured from the 30-kip load by moment equilibrium:

2. Assume maximum under the 30-kip load

• Position loads so beam centerline splits the distance between the 30-kip and the resultant.
• Sum moments about B to get :

• Moment at the 30-kip location:

3. Assume maximum under the 20-kip load

• Position loads so beam centerline is halfway between the 20-kip load and the resultant.
• Sum moments about A to get :

• Moment at the 20-kip location:

4. Result

• Absolute maximum live-load moment = 300.125 kip·ft (≈ 300 kip·ft), occurring under the 30-kip load.

 

Example:

Determine the absolute maximum bending moment in the simply supported beam due to the axle loads of the HL-93 truck shown in Fig. below

Solution

Resultant of Load Series:

The magnitude of the resultant is obtained by summing the magnitudes of the loads of the series.

Location of resultant (measured from the 32-kip rear-axle)

Use the moment balance about that 32-kip axle:

So, the resultant is 9.33 ft to the right of the 32-kip axle

Determine which axle gives absolute maximum moment

• The second load (the 32-kip rear-axle in the series) is closest to the resultant → expected location of absolute maximum moment.
• The resultant is 4.67 ft to the right of the second load (since 9.33 − 4.66? — this is the geometry given in the figure).
• To get the absolute maximum bending moment, place the load series so the midspan of the beam is halfway between that second load and the resultant.
• Hence position the second load 2.33 ft left of midspan (i.e., ft).

Reaction at A when series is positioned for absolute max

Distance from A to the resultant line of action = 22.67 ft.
Sum moments about B (or use equilibrium) → vertical reaction at A:

Absolute maximum bending moment (under the second 32-kip axle)

Compute moment under that second axle

(distance from A to that axle =  ft):

Absolute maximum bending moment = 627.94 kip·ft, occurring under the second (32-kip) axle when the series is positioned so the beam midspan lies halfway between that axle and the resultant.

 

Example:

Given,

Simply supported span = 15 m

UDL = 40 kN/m

Length of UDL = 5 m

Direction of movement of UDL = left to right

Draw ILD at a section 6m from the left and maximum shear and moment at that point.

Solution:

Now IL of 1 unit load at section 6m from left side:

Now, maximum negative SF at C

Hence, Maximum Negative SF at C = Intensity of load x area of ILD under load

                                                        = 40 x [1/2 * (0.4+0.067) * 5]

                                                        = 46 kN

Now, maximum positive SF at C

Hence, Maximum Positive SF at C = Intensity of load x area of ILD under load

                                                        = 40 x [1/2 * (0.267+0.6) * 5]

                                                        = 86.7 kN

 

Now, Maximum Bending Moment at C

Hence, Maximum Bending Moment at point C:

= 40*[1/2 * (2.4+3.6) * 2] + 40*[1/2 * (3.6 + 2.4) * 3]

= 600 kN-m

 

Home Work:

 

1.     Draw the influence lines for the vertical reaction and the reaction moment at support A and the shear and bending moment at point B of the cantilever beam shown in Fig. 1

Fig-1

2.     Draw the influence lines for vertical reactions at supports A and C and the shear and bending moment at point B of the beams shown in Figs. 2 to 4.

Fig-2

Fig-3

Fig-4

3.     Draw the influence lines for the vertical reaction and the reaction moment at support A and the shear and bending moment at point B of the cantilever beam shown in Fig.5

Fig-5

4.     Draw the influence lines for vertical reactions at supports A and C and the shear and bending moment at point B of the beam shown in Fig.6

Fig-6

5.     Draw the influence lines for vertical reactions at supports B and D and the shear and bending moment at point C of the beams shown in Fig-7

Fig-7

6.     Draw the influence lines for the shear and bending moment at point C and the shears just to the left and just to the right of support D of the beam shown in Fig.9

Fig-9

7.     Draw the influence lines for the vertical reactions at supports A and E and the reaction moment at support E of the beam shown in Fig.10

Fig-10

8.     Draw the influence lines for the vertical reactions at supports A, D, and F of the beam shown in Fig.11

Fig-11

9.     Draw the influence lines for the vertical reactions at supports A, C, E, and G of the beams shown in Figs. 12 and 13

Fig-12

Fig-13

10. Draw the influence lines for the reaction moment at support A and the vertical reactions at supports A, D, and F of the beam shown in Fig.14

Fig-14

11. Draw the influence lines for the vertical reactions at supports A and B of the beam shown in Fig.15

Fig-15

12. Determine the absolute maximum bending moment in a 75 ft long simply supported beam due to the wheel loads of the moving HS15-44 truck shown in Fig.16

Fig-16

13. Determine the absolute maximum bending moment in a 15 m long simply supported beam due to the series of three moving concentrated loads shown in Fig.17

Fig-17

14. Determine the absolute maximum bending moment in a 60 ft long simply supported beam due to the series of four moving concentrated loads shown in Fig.18

Fig-18