Moving Load & Its Effects on Structural Members - 2 - Structural Analysis

Influence Line for Moment

An influence line for a response function (reaction, shear, or bending moment) gives the value of that function at a specified point of a structure as a unit load moves across the structure. Influence lines for moment are constructed in the same way as for shear or reactions: place a unit load at various positions, compute the response at the point of interest and join the ordinates to obtain the influence line. Influence lines are essential to determine the critical position of moving loads (concentrated or distributed) for design checks.

Example 4

Construct the influence line for the moment at point C of the beam shown in Figure 37.14.

Solution: Method and tabulation

Place a unit load at different locations along the span between supports and compute the support reactions. For each unit-load position, take a section at C and compute the bending moment at C. Tabulate the ordinate of the influence line (the moment at C under a unit load) for the several load positions and then plot the ordinates to obtain the influence line for M_c.

For example, place the unit load at x = 2.5 m measured from support A (see Figure 37.15).

The support reactions for a unit load at x = 2.5 m are

R_A = 0.833 and R_B = 0.167.

Take a section at C and write moment equilibrium about C to obtain M_c.

ΣM_c = 0 : - M_c + R_B × 7.5 = 0

M_c = R_B × 7.5 = 0.167 × 7.5 = 1.25

Similarly compute M_c for different unit-load positions and tabulate the values of M_c.

The graphical representation of the influence line for M_c is shown in Figure 37.16.

Influence line equations for M_c

There are two linear expressions (one for the load to the left of C and one for the load to the right of C) because the bending moment at C under a moving unit load is a piecewise-linear function of the load position.

When the unit load is located before point C (to the left), summing moments about C gives

ΣM_c = 0 : M_c + 1(7.5 - x) - (1 - x/15) × 7.5 = 0

Rearrange to obtain

M_c = x/2, where 0 ≤ x ≤ 7.5.

When the unit load is located after point C (to the right), summing moments about C gives

ΣM_c = 0 : M_c - (1 - x/15) × 7.5 = 0

Rearrange to obtain

M_c = 7.5 - x/2, where 7.5 < x ≤ 15.0.

The two linear equations are plotted in Figure 37.16 to give the complete influence line for the moment at C.

Example 5

Construct the influence line for the moment at point C of the beam shown in Figure 37.19.

Solution: Method and tabulation

Use the same procedure as in Example 4: place a unit load at various positions, compute support reactions, take a section at C and compute M_c for each unit-load position, then tabulate and plot.

For instance, when the unit load is at x = 2.5 m from A (see Figure 37.20), the support reactions are

R_A = 0.75 and R_B = 0.25.

Taking moments about C for this position gives

ΣM_c = 0 : - M_c + R_B × 5.0 = 0

M_c = R_B × 5.0 = 0.25 × 5.0 = 1.25

Compute M_c for other unit-load positions, tabulate, and plot. The tabulated values are shown below.

The graphical influence line for M_c is shown in Figure 37.21.

Influence line equations for M_c

When the unit load is placed before C (left side), summing moments about C yields

ΣM_c = 0 : M_c + 1(5.0 - x) - (1 - x/10) × 5.0 = 0

Thus

M_c = x/2, where 0 ≤ x ≤ 5.0.

When the unit load is placed after C (right side), summing moments about C yields

ΣM_c = 0 : M_c - (1 - x/10) × 5.0 = 0

Thus

M_c = 5 - x/2, where 5 < x ≤ 15.

The two linear expressions are plotted in Figure 37.21.

Influence Line for a Beam with Simultaneous Point Load and UDL

Beams and girders are primary load-carrying elements in structures; hence it is necessary to construct influence lines for the reaction, shear or moment at any specified point in a beam to determine critical positions of live loads. In service conditions a beam may be subject to a moving concentrated load and a moving uniformly distributed load (UDL) simultaneously. The total effect at a point is the superposition of the individual effects due to the concentrated loads and the UDL.

Concentrated load

Consider a point load P moving from A to B on a simply supported span. If the influence line ordinate for a response at a point (for example reaction A) gives y when a unit load is at the load position, then the response due to the point load P at that position is y × P. For example, if the influence line ordinate at the centre position is 0.5, then the reaction at A for a point load P at centre is 0.5 × P. Thus for varying positions of P, the reaction (or other response) is found by multiplying the influence line ordinate at each position by P.

Uniformly distributed load (UDL)

For a UDL of intensity w (force per unit length) spread over the span, the effect at the point under consideration can be found by integrating the contribution of infinitesimal concentrated loads over the loaded length. Consider an infinitesimal segment of UDL of length dx carrying a concentrated load dP = w · dx acting at that segment. If the influence line ordinate at that segment is y(x), the contribution of this elemental load to the response is d(effect) = (w · dx) · y(x). Integrating over the length of the span where UDL acts gives the total response due to the UDL:

Effect due to UDL = ∫ (w · y(x) · dx) = w · ∫ y(x) dx.

The integral ∫ y(x) dx is the area under the influence line over the portion of span where the UDL is applied. Thus the response from a UDL is equal to the UDL intensity times the area under the influence line over the loaded region.

For a simple example, if the influence line for a reaction A is triangular with peak 1 at A and area under the line equal to 0.5 · 1 · l = 0.5 l, then for a uniformly distributed load w across the entire span the reaction at A is

Reaction A = w × 0.5 l = 0.5 w l.

Numerical Example

Problem: Find the maximum positive live shear at point C when the beam (Figure 37.30) is loaded with a concentrated moving load of 10 kN and a uniformly distributed moving load of 5 kN/m.

Solution

Use the influence line for shear at C and superpose the effect of the concentrated load and the UDL. The influence line for shear at C when a unit load moves from A to B is shown in Figure 37.31.

Concentrated load:

The maximum positive shear at C due to the concentrated load will occur when the 10 kN load is placed just to the right of C (so that the jump in the shear influence line is positive at C).

For a unit load just right of C the influence ordinate is 0.5.

Therefore the shear at C due to the 10 kN load is

V_c (from 10 kN) = 0.5 × 10 = 5 kN.

Uniformly distributed load (UDL):

The maximum positive shear at C due to the moving UDL of intensity 5 kN/m occurs when the UDL is located on the span section corresponding to positive ordinates of the influence line (for this beam between x = 7.5 m and x = 15 m as shown in Figure 37.31).

Compute the contribution using the area under the influence line over the loaded portion.

The relevant triangular area under the influence line has base (15 - 7.5) and height 0.5, so the area is

Area = 0.5 × (15 - 7.5) × 0.5 = 0.5 × 7.5 × 0.5 = 1.875 m.

Multiply the area by the UDL intensity w = 5 kN/m to obtain the shear contribution:

V_c (from UDL) = 5 × 1.875 = 9.375 kN.

Total maximum positive shear at C by superposition is

(V_c)_max = 5 + 9.375 = 14.375 kN.

The positions of the concentrated load and the UDL that produce this maximum positive shear at C are shown in Figure 37.32. For verification one may compute the shear at C directly by statics for those loading positions.