Concept explainers
Draw the influence lines for the shear and bending moment at point B.
Draw the influence lines for the shear at internal hinge C.
Explanation of Solution
Calculation:
Influence line for shear at point B:
Apply a 1 kN unit moving load at a distance of x from left end A.
Sketch the free body diagram of frame as shown in Figure 1.
Refer Figure 1.
Consider the unit load at a variable position x to the left hinge C. (placed portion AC of the beam 0≤x<12 m).
Find the vertical support reaction (Ay) at A for portion AC.
Take Moment at hinge C from left end A.
Consider clockwise moment as positive and anticlockwise moment as negative.
ΣMACC=0Ay(12)−1(12−x)=012Ay=12−xAy=1−x12
Consider the unit load at a variable position x to the right hinge C. (Place portion CG of the beam 12 m<x≤30 m).
Find the vertical support reaction (Ay) at A for portion CG.
Take Moment at hinge C from left end A.
Consider clockwise moment as positive and anticlockwise moment as negative.
ΣMACC=0Ay(12)=0Ay=0
Thus, the equations of the influence line ordinate for Ay are,
Ay=1−x12,0≤x<12 m (1)
Ay=0,12 m<x≤30 m (2)
Refer Figure 1.
Find the equation of influence line ordinate for the vertical reaction (Dy) using equilibrium equation.
Apply moment equilibrium at F.
Consider clockwise moment as positive and anticlockwise moment as negative.
ΣMF=0Ay(24)+Dy(6)−1(24−x)=06Dy=24−x−24AyDy=4−x6−4Ay (3)
Find the influence line ordinate of vertical reaction (Dy) for portion AC (0≤x<12 m).
Substitute 1−x12 for Ay in Equation (3).
Dy=4−x6−4(1−x12)=4−x6−4+x3=−x+2x6=x6
Find the influence line ordinate of vertical reaction (Dy) for portion CG(12 m<x≤30 m).
Substitute 0 for Ay in Equation (3).
Dy=4−x6−4(0)=4−x6
Thus, the equations of the influence line ordinate for Dy are,
Dy=x6,0≤x<12 m (4)
Dy=4−x6,12 m<x≤30 m (5)
Refer Figure 1.
Find the equation of influence line ordinate for the vertical reaction (Fy) using force equilibrium equation.
Consider the vertical forces equilibrium condition, take the upward force as positive (↑+) and downward force as negative (↓−).
ΣFy=0Ay+Dy+Fy=1Fy=1−Ay−Dy (6)
Find the influence line ordinate of vertical reaction (Fy) for portion AC(0≤x<12 m).
Substitute 1−x12 for Ay and x6 for Dy in Equation (6).
Fy=1−(1−x12)−(x6)=1−1+x12−x6=−x12
Find the influence line ordinate of vertical reaction (Fy) for portion AC(12 m<x≤30 m).
Substitute 0 for Ay and 4−x6 for Dy in Equation (6).
Fy=1−(0)−(4−x6)=1−4+x6=x6−3
Thus, the equations of the influence line ordinate for Fy are,
Fy=−x12,0≤x<12 m (7)
Fy=x6−3,12 m<x≤30 m (8)
Find the equation of shear at B of portion AB(0≤x<6 m).
Sketch the free body diagram of the section AB 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
Ay−1−SB=0SB=Ay−1
Substitute 1−x12 for Ay.
SB=(1−x12)−1=−x12
Find the equation of shear at B of portion BG(6 m<x≤30 m).
Sketch the free body diagram of the section BG 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
SB−1+Dy+Fy=0SB=1−Dy−Fy (9)
Find the influence line ordinate of shear at B(SB) for portion BC (6 m<x<12 m).
Substitute x6 for Dy and −x12 for Fy in Equation (9).
SB=1−(x6)−(−x12)=1−x6+x12=1−x12
Find the influence line ordinate of shear at B(SB) for portion CG(12 m<x≤30 m).
Substitute 4−x6 for Dy and x6−3 for Fy in Equation (9).
SB=1−(4−x6)−(x6−3)=1−4+x6−x6+3=−3+0+3=0
Thus, the equations of the influence line ordinate for SB are,
SB=−x12,0≤x<6 m (10)
SB=1−x12,6 m<x<12 m (11)
SB=0,12 m<x≤30 m (12)
Find the influence line ordinate of SB using Equation (10), (11), and (12) and summarize the values in Table 1.
| x (m) | Points | Influence line ordinate of SB(kN/kN) |
| 0 | A | 0 |
| 6 | B− | ‑0.5 |
| 6 | B+ | 0.5 |
| 12 | C | 0 |
| 18 | D | 0 |
| 21 | E | 0 |
| 24 | F | 0 |
| 30 | G | 0 |
Sketch the influence line diagram for the shear at pointB as shown in Figure 4.
Influence line for moment at point B:
Refer Figure 2.
Find the equation of bending moment at B of portion AB(0≤x<6 m).
Take moment at B.
Consider clockwise moment as positive and anticlockwise moment as negative.
MB=Ay(6)−1(6−x)
Substitute 1−x12 for Ay.
MB=(1−x12)(6)−1(6−x)=6−x2+6+x=x2
Refer Figure 3.
Find the equation of bending moment at B of portion BG(6 m<x≤30 m).
Take moment at B.
Consider clockwise moment as negative and anticlockwise moment as positive.
MB=−1[24−(30−x)]+Dy(12)+Fy(18)=−24+30−x+12Dy+18Fy=6−x+12Dy+18Fy (13)
Find the influence line ordinate of moment at B(MB) for portion BC(6 m<x<12 m).
Substitute x6 for Dy and −x12 for Fy in Equation (13).
MB=6−x+12(x6)+18(−x12)=6−x+2x−3x2=6−x2
Find the influence line ordinate of moment at B (MB) for portion CG(12 m<x≤30 m).
Substitute 4−x6 for Dy and x6−3 for Fy in Equation (13).
MB=−6+x−12(4−x6)−18(x6−3)=−6+x−48+2x−3x+54=0
Thus, the equations of the influence line ordinate for MB are,
MB=x2,0≤x<6 m (14)
MB=6−x2,6 m<x<12 m (15)
MB=0,12 m<x≤30 m (16)
Find the influence line ordinate of MB using Equation (14), (15), and (16) and summarize the values in Table 2.
| x (m) | Points | Influence line ordinate of MB(kN-m/kN) |
| 0 | A | 0 |
| 6 | B | ‑3 |
| 12 | C | 0 |
| 18 | D | 0 |
| 21 | E | 0 |
| 24 | F | 0 |
| 30 | G | 0 |
Sketch the influence line diagram for the bending moment at point B as shown in Figure 5.
Influence line for shear at hinge C:
Apply a 1 kN unit moving load at a distance of x from left end A.
Sketch the free body diagram of frame as shown in Figure 6.
Refer Figure 6.
Consider the unit load at a variable position x to the left hinge C. (placed portion AC of the beam 0≤x<12 m).
Find the vertical support reaction (Ay) at A for portion AC.
Take Moment at hinge C from left end A.
Consider clockwise moment as positive and anticlockwise moment as negative.
ΣMACC=0Ay(12)−1(12−x)=012Ay=12−xAy=1−x12
Consider the unit load at a variable position x to the right hinge C. (Place portion CG of the beam 12 m<x≤30 m).
Find the vertical support reaction (Ay) at A for portion CG.
Take Moment at hinge C from left end A.
Consider clockwise moment as positive and anticlockwise moment as negative.
ΣMACC=0Ay(12)=0Ay=0
Thus, the equations of the influence line ordinate for Ay are,
Ay=1−x12,0≤x<12 m (17)
Ay=0,12 m<x≤30 m (18)
Refer Figure 1.
Find the equation of influence line ordinate for the vertical reaction (Dy) using equilibrium equation.
Apply moment equilibrium at F.
Consider clockwise moment as positive and anticlockwise moment as negative.
ΣMF=0Ay(24)+Dy(6)−1(24−x)=06Dy=24−x−24AyDy=4−x6−4Ay (19)
Find the influence line ordinate of vertical reaction (Dy) for portion AC(0≤x<12 m).
Substitute 1−x12 for Ay in Equation (19).
Dy=4−x6−4(1−x12)=4−x6−4+x3=−x+2x6=x6
Find the influence line ordinate of vertical reaction (Dy) for portion CG(12 m<x≤30 m).
Substitute 0 for Ay in Equation (19).
Dy=4−x6−4(0)=4−x6
Thus, the equations of the influence line ordinate for Dy are,
Dy=x6,0≤x<12 m (20)
Dy=4−x6,12 m<x≤30 m (21)
Refer Figure 1.
Find the equation of influence line ordinate for the vertical reaction (Fy) using force equilibrium equation.
Consider the vertical forces equilibrium condition, take the upward force as positive (↑+) and downward force as negative (↓−).
ΣFy=0Ay+Dy+Fy=1Fy=1−Ay−Dy (22)
Find the influence line ordinate of vertical reaction (Fy) for portion AC(0≤x<12 m).
Substitute 1−x12 for Ay and x6 for Dy in Equation (22).
Fy=1−(1−x12)−(x6)=1−1+x12−x6=−x12
Find the influence line ordinate of vertical reaction (Fy) for portion AC(12 m<x≤30 m).
Substitute 0 for Ay and 4−x6 for Dy in Equation (22).
Fy=1−(0)−(4−x6)=1−4+x6=x6−3
Thus, the equations of the influence line ordinate for Fy are,
Fy=−x12,0≤x<12 m (23)
Fy=x6−3,12 m<x≤30 m (24)
Find the equation of shear at C of portion AC(0≤x<12 m).
Sketch the free body diagram of the section AC as shown in Figure 7.
Refer Figure 7.
Apply equilibrium equation of forces.
Consider upward force as positive (↑+) and downward force as negative (↓−).
ΣFy=0
Ay−1−SC=0SC=Ay−1
Substitute 1−x12 for Ay.
SC=(1−x12)−1=−x12
Find the equation of shear at C of portion CG (12 m<x≤30 m).
Sketch the free body diagram of the section CG as shown in Figure 8.
Refer Figure 8.
Apply equilibrium equation of forces.
Consider upward force as positive (↑+) and downward force as negative (↓−).
ΣFy=0
SC−1+Dy+Fy=0SC=1−Dy−Fy (25)
Substitute 4−x6 for Dy and x6−3 for Fy in Equation (25).
SC=1−(4−x6)−(x6−3)=1−4+x6−x6+3=−3+0+3=0
Thus, the equations of the influence line ordinate for SC are,
SC=−x12,0≤x<12 m (26)
SC=0,12 m<x≤30 m (27)
Find the influence line ordinate of SC using Equation (26) and (27) and summarize the values in Table 3.
| x (m) | Points | Influence line ordinate of SC(kN/kN) |
| 0 | A | 0 |
| 6 | B | ‑0.5 |
| 12 | C− | ‑1 |
| 12 | C+ | 0 |
| 18 | D | 0 |
| 21 | E | 0 |
| 24 | F | 0 |
| 30 | G | 0 |
Sketch the influence line diagram for the shear at point C as shown in Figure 9.
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