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. P8.10. FIG. P8.1, P8.59

Draw the influence lines for the shear and bending moment at point C.

Draw the influence lines for the shears just to the left and just to the right of support D.

Calculation:

Apply a 1 kN unit moving load at a distance of x from left end A.

Sketch the free body diagram of beam as shown in Figure 1.

Refer Figure 1.

Find the equation of support reaction (By) at B using equilibrium equation:

Take moment about point D.

Consider moment equilibrium at point D.

Consider clockwise moment as positive and anticlockwise moment as negative

Sum of moment at point D is zero.

ΣMD=0By(12)−1(16−x)=012By=16−xBy=43−x12        (1)

Find the equation of support reaction (Dy) at D using equilibrium equation:

Apply vertical equilibrium equation of forces.

Consider upward force as positive (↑+) and downward force as negative (↓−).

By+Dy=1

Substitute 43−x12 for By.

43−x12+Dy=1Dy=1−43+x12Dy=x12−13        (2)

Influence line for the shear at point C.

Apply 1 kN load at just left of C.

Find the equation of shear force at C of portion AB (0≤x≤8 m).

Sketch the free body diagram of the section AC as shown in Figure 2.

Refer Figure 2.

Apply equilibrium equation of forces.

Consider upward force as positive (↑+) and downward force as negative (↓−).

ΣFy=0

By−SC−1=0SC=By−1

Substitute 43−x12 for By.

SC=(43−x12)−1=13−x12

Apply 1 kN load at just right of C.

Find the equation of shear force at C of portion CF (8 m≤x≤22 m).

Sketch the free body diagram of the section AC as shown in Figure 3.

Refer Figure 3.

Apply equilibrium equation of forces.

Consider upward force as positive (↑+) and downward force as negative (↓−).

ΣFy=0

By−SC=0SC=By

Substitute 43−x12 for By.

SC=(43−x12)=43−x12

Thus, the equations of the influence line for SC as follows

SC=13−x12, 0≤x≤8 m        (3)

SC=43−x12, 8 m≤x≤22 m        (4)

Find the value of influence line ordinate of shear force SC at various points of x using the Equations (3) and (4) and summarize the value as in Table 1.

Draw the influence lines for the shear force at point C using Table 1 as shown in Figure 4.

Influence line for moment at point C.

Refer Figure 2.

Consider clockwise moment as positive and anticlockwise moment as negative.

Find the equation of moment at C of portion AC (0≤x≤8 m).

MC=By(4)−(1)(8−x)

Substitute 43−x12 for By.

MC=(43−x12)(4)−(1)(8−x)=163−x3−8+x=2x3−83

Refer Figure 3.

Consider clockwise moment as negative and anticlockwise moment as positive.

Find the equation of moment at C of portion CF (8 m≤x≤22 m).

MC=By(4)

Substitute 43−x12 for By.

MC=(43−x12)(4)=163−x3=16−x3

Thus, the equations of the influence line for MC as follows,

MC=2x3−83, 0≤x≤8 m        (5)

MC=16−x3, 8 m≤x≤22 m        (6)

Find the value of influence line ordinate of moment MC at various points of x using the Equations (5) and (6) and summarize the value as in Table 2