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Moving Loads and
Influence Line Diagram
Beam
Mrs. S. M. Madhale
Assistant professor
Civil Engineering Department
ATS’s SBGI Miraj
Solved Problem on BEAM
ILD BY: Mrs. S. M. Madhale 2
• A Simple beam AB having span12m. And a 20Kn
load is act at its center. Calculate the Reaction at
support, maximum shear force and bending
moment at a section of 3m from left support A.
ILD BY: Mrs. S. M. Madhale 3
Reaction at support RA
• Remove the support reaction RA and Lift end A in
the direction of Ra using some amount of force F =
1 and End B is free to rotate. Hence due to force the
beam will deflection as shown in Fig
• Make ordinate at A equal to one.
• Calculate the ordinate X1
•
1
12
=
𝑋1
9
• 𝑋1 =
9
12
= 0.75
• Reaction At support A = Load * Load factor X1
= 20 * 0.75
= 15 Kn
ILD BY: Mrs. S. M. Madhale 4
A B
20 Kn
3 3 6
F
RB
1 X1
3 9
12
Reaction at support RB
• Remove the support reaction RB and Lift end B in the
direction of Rb using some amount of force F = 1
and End A is free to rotate. Hence due to force the
beam will deflection as shown in Fig
• Make ordinate at B equal to one.
• Calculate the ordinate X2
•
1
12
=
𝑋2
3
• 𝑋2 =
3
12
= 0.25
• Reaction At support A = Load * Load factor X2
= 20 * 0.25
= 5 Kn
ILD BY: Mrs. S. M. Madhale 5
A B
20 Kn
3 3 6
F
RA
1
X2
3 9
12
Shear Force at Section
• Remove the ability to resist shear force the beam and
providing rollers in vertical direction at section and
apply unit deflection in the direction of shear force.
• Make ordinate at section equal to one.
• Calculate the ordinate X1 and X2
1
12
=
𝑋2
3
𝑋2 =
3
12
= 0.25
1
12
=
𝑋1
9
𝑋1 =
9
12
= 0.75
Shear force at section
= [ Load * Load factor X1] + [ Load * Load factor X2]
= { + [20 * 0.75] } + { - [20 * 0.25] }
= +15 -5 = 10 Kn
ILD BY: Mrs. S. M. Madhale
6
A B
20 Kn
3 3 6
3 9
F
F
X1
X2
1
1
Bending moment at Section
• Remove the ability to resist moment by providing
hinge at that section and Apply unit rotation equal
to one.
• Calculate the ordinate ∅, 𝜃 𝑎𝑛𝑑 𝛽
• 𝜃 = 1 = 𝜑 + 𝛽
=
𝑎 ∗𝑏
𝐿
=
3 ∗9
12
= 2.25
Bending Moment at section
= [ Load * Load factor]
= 20 * 2.25
= 45 Kn-m
ILD BY: Mrs. S. M. Madhale
7
A B
20 Kn
3 3 6
a b
𝜃
∅ 𝛽
a * b / L
L
• Two point loads of 100 Kn and 200 Kn spaced 3 m apart cross a
girder of span 15 m from left to right with 100 Kn load leading.
Draw a ILD for shear force and bending moment and find the
value of maximum shear and bending moment at a section D, 6 m
from left support. Also, find out the absolute maximum bending
moment due to given loading system
ILD BY: Mrs. S. M. Madhale 8
• Here are 2 conditions where
• 100Kn load lies on section
• 200Kn load lies on section
ILD BY: Mrs. S. M. Madhale 9
D
200 Kn 100 Kn
B
A
6 9
15
Reaction at support RA 100Kn load lies on section
• Remove the support reaction RA and Lift end A in the
direction of Ra using some amount of force F = 1 and
End B is free to rotate. Hence due to force the beam will
deflection as shown in Fig
• Make ordinate at A equal to one.
• Calculate the ordinate X1 and X2
•
1
15
=
𝑋1
12
1
15
=
𝑋2
9
• 𝑋1 =
12
15
= 0.8 𝑋2 =
9
15
= 0.6
• Reaction At support A = Load * Load factor X
={ Load * Load factor X1}+{ Load * Load factor X2}
= {200 * 0.8} + {100 * 0.6} =160+60 =220 Kn
ILD BY: Mrs. S. M. Madhale
10
A B
6 9
F
RB
1 X1
6 9
15
3
200 Kn 100 Kn
X2
3
Reaction at support RA 200Kn load lies on section
• Remove the support reaction RA and Lift end A in the
direction of Ra using some amount of force F = 1 and
End B is free to rotate. Hence due to force the beam will
deflection as shown in Fig
• Make ordinate at A equal to one.
• Calculate the ordinate X1 and X2
•
1
15
=
𝑋1
9
1
15
=
𝑋2
6
• 𝑋1 =
9
15
= 0.6 𝑋2 =
6
15
= 0.4
• Reaction At support A = Load * Load factor X
={ Load * Load factor X1}+{ Load * Load factor X2}
= {200 * 0.6} + {100 * 0.4} =120+40 =160 Kn
ILD BY: Mrs. S. M. Madhale
11
A B
6 9
F
RB
1 X1
6
9
15
3
200 Kn 100 Kn
X2
3
X2
Reaction at support RB 100Kn load lies on section
• Remove the support reaction RB and Lift end B in the
direction of Rb using some amount of force F = 1 and
End A is free to rotate. Hence due to force the beam will
deflection as shown in Fig
• Make ordinate at B equal to one.
• Calculate the ordinate X1 and X2
•
1
15
=
𝑋1
3
1
15
=
𝑋2
6
• 𝑋1 =
3
15
= 0.2 𝑋2 =
6
15
= 0.4
• Reaction At support B = Load * Load factor X
={ Load * Load factor X1}+{ Load * Load factor X2}
= {200 * 0.2} + {100 * 0.4} =40+40 =80 Kn
ILD BY: Mrs. S. M. Madhale
12
A B
6 9
F
RA
1
X1
6 9
15
3
200 Kn 100 Kn
X2
3
Reaction at support RB 200Kn load lies on section
• Remove the support reaction RB and Lift end B in the
direction of Rb using some amount of force F = 1 and
End A is free to rotate. Hence due to force the beam will
deflection as shown in Fig
• Make ordinate at B equal to one.
• Calculate the ordinate X1 and X2
•
1
15
=
𝑋2
9
1
15
=
𝑋1
6
• 𝑋1 =
9
15
= 0.6 𝑋2 =
6
15
= 0.4
• Reaction At support B = Load * Load factor X
={ Load * Load factor X1}+{ Load * Load factor X2}
= {200 * 0.4} + {100 * 0.6} =80+60 =140 Kn
ILD BY: Mrs. S. M. Madhale
13
A B
6 9
F
RB
1 X1
6
9
15
3
200 Kn 100 Kn
X2
3
Shear Force at Section 100Kn load lies on section
• Remove the ability to resist shear force the beam and
providing rollers in vertical direction at section and
apply unit deflection in the direction of shear force.
Make ordinate at section equal to one.
• Calculate the ordinate X1 and X3 [for negative
consider loads before section]
•
1
15
=
𝑋1
6
𝑋1 =
6
15
= 0.4
•
1
15
=
𝑋3
3
𝑋3 =
3
15
= 0.2
Shear force at section
= [ Load * Load factor X1] + [ Load * Load factor X3]
= { - [100 * 0.4] } +{-[200*0.2]}
= - 40 -40 = - 80 Kn
ILD BY: Mrs. S. M. Madhale 14
A
6 9
1
X1
6 9
3
200 Kn 100 Kn
X2
3
B
1
X3
Shear Force at Section 200Kn load lies on section
• Remove the ability to resist shear force the beam and
providing rollers in vertical direction at section and
apply unit deflection in the direction of shear force.
Make ordinate at section equal to one.
• Calculate the ordinate X1 and X3 [for positive consider
load after section]
•
1
15
=
𝑋2
9
𝑋2 =
9
15
= 0.6
•
1
15
=
𝑋3
6
𝑋3 =
6
15
= 0.4
Shear force at section
= [ Load * Load factor X3] + [ Load * Load factor X2]
= { + [100 * 0.4] } +{+[200*0.6]}
= + 40 + 120 = 160 Kn
ILD BY: Mrs. S. M. Madhale 15
A
6 9
1
X1
6 9
3
200 Kn
100 Kn
X2
3
B
1 X3
Bending moment at Section 100Kn load lies on section
• Remove the ability to resist moment by providing
hinge at that section and Apply unit rotation equal to
one. 𝜃=1= 𝜑+ 𝛽 Calculate the ordinate ∅, 𝜃 𝑎𝑛𝑑 𝛽
• Calculate the ordinate X1 and X3 [for negative
consider loads before section] (𝑎 ∗𝑏 )/𝐿
• 𝑋1 =
𝑎 ∗𝑏
𝐿
=
6 ∗9
15
= 3.6
• 𝑋2 =
3.6
6
∗ 3 = 1.8
Bending moment at section
= [ Load * Load factor X1] + [ Load * Load factor X3]
= [100 * 3.6] +[200 * 1.8]
= 360 + 480 = 720 Kn-m
ILD BY: Mrs. S. M. Madhale 16
A
6 9
1
X1
6 9
200 Kn 100 Kn
X2
3
B
1
3
3
Bending moment at Section 200Kn load lies on
section
• Remove the ability to resist moment by providing
hinge at that section and Apply unit rotation equal to
one. 𝜃=1= 𝜑+ 𝛽 Calculate the ordinate ∅, 𝜃 𝑎𝑛𝑑 𝛽
• Calculate the ordinate X1 and X3 [for negative
consider loads before section] (𝑎 ∗𝑏 )/𝐿
• 𝑋1 =
𝑎 ∗𝑏
𝐿
=
6 ∗9
15
= 3.6
• 𝑋2 =
3.6
9
∗ 6 = 2.4
Bending moment at section
= [ Load * Load factor X1] + [ Load * Load factor X3]
= [200 * 3.6] +[100 * 2.4]
= 720 + 240 = 960 Kn-m
ILD BY: Mrs. S. M. Madhale 17
A
6 9
1
X1
6 9
200 Kn
100 Kn
X2
3
B
1
3 6
THANK YOU !!!
ILD BY: Mrs. S. M. Madhale 18

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ILD - beam.pptx

  • 1. Moving Loads and Influence Line Diagram Beam Mrs. S. M. Madhale Assistant professor Civil Engineering Department ATS’s SBGI Miraj
  • 2. Solved Problem on BEAM ILD BY: Mrs. S. M. Madhale 2
  • 3. • A Simple beam AB having span12m. And a 20Kn load is act at its center. Calculate the Reaction at support, maximum shear force and bending moment at a section of 3m from left support A. ILD BY: Mrs. S. M. Madhale 3
  • 4. Reaction at support RA • Remove the support reaction RA and Lift end A in the direction of Ra using some amount of force F = 1 and End B is free to rotate. Hence due to force the beam will deflection as shown in Fig • Make ordinate at A equal to one. • Calculate the ordinate X1 • 1 12 = 𝑋1 9 • 𝑋1 = 9 12 = 0.75 • Reaction At support A = Load * Load factor X1 = 20 * 0.75 = 15 Kn ILD BY: Mrs. S. M. Madhale 4 A B 20 Kn 3 3 6 F RB 1 X1 3 9 12
  • 5. Reaction at support RB • Remove the support reaction RB and Lift end B in the direction of Rb using some amount of force F = 1 and End A is free to rotate. Hence due to force the beam will deflection as shown in Fig • Make ordinate at B equal to one. • Calculate the ordinate X2 • 1 12 = 𝑋2 3 • 𝑋2 = 3 12 = 0.25 • Reaction At support A = Load * Load factor X2 = 20 * 0.25 = 5 Kn ILD BY: Mrs. S. M. Madhale 5 A B 20 Kn 3 3 6 F RA 1 X2 3 9 12
  • 6. Shear Force at Section • Remove the ability to resist shear force the beam and providing rollers in vertical direction at section and apply unit deflection in the direction of shear force. • Make ordinate at section equal to one. • Calculate the ordinate X1 and X2 1 12 = 𝑋2 3 𝑋2 = 3 12 = 0.25 1 12 = 𝑋1 9 𝑋1 = 9 12 = 0.75 Shear force at section = [ Load * Load factor X1] + [ Load * Load factor X2] = { + [20 * 0.75] } + { - [20 * 0.25] } = +15 -5 = 10 Kn ILD BY: Mrs. S. M. Madhale 6 A B 20 Kn 3 3 6 3 9 F F X1 X2 1 1
  • 7. Bending moment at Section • Remove the ability to resist moment by providing hinge at that section and Apply unit rotation equal to one. • Calculate the ordinate ∅, 𝜃 𝑎𝑛𝑑 𝛽 • 𝜃 = 1 = 𝜑 + 𝛽 = 𝑎 ∗𝑏 𝐿 = 3 ∗9 12 = 2.25 Bending Moment at section = [ Load * Load factor] = 20 * 2.25 = 45 Kn-m ILD BY: Mrs. S. M. Madhale 7 A B 20 Kn 3 3 6 a b 𝜃 ∅ 𝛽 a * b / L L
  • 8. • Two point loads of 100 Kn and 200 Kn spaced 3 m apart cross a girder of span 15 m from left to right with 100 Kn load leading. Draw a ILD for shear force and bending moment and find the value of maximum shear and bending moment at a section D, 6 m from left support. Also, find out the absolute maximum bending moment due to given loading system ILD BY: Mrs. S. M. Madhale 8
  • 9. • Here are 2 conditions where • 100Kn load lies on section • 200Kn load lies on section ILD BY: Mrs. S. M. Madhale 9 D 200 Kn 100 Kn B A 6 9 15
  • 10. Reaction at support RA 100Kn load lies on section • Remove the support reaction RA and Lift end A in the direction of Ra using some amount of force F = 1 and End B is free to rotate. Hence due to force the beam will deflection as shown in Fig • Make ordinate at A equal to one. • Calculate the ordinate X1 and X2 • 1 15 = 𝑋1 12 1 15 = 𝑋2 9 • 𝑋1 = 12 15 = 0.8 𝑋2 = 9 15 = 0.6 • Reaction At support A = Load * Load factor X ={ Load * Load factor X1}+{ Load * Load factor X2} = {200 * 0.8} + {100 * 0.6} =160+60 =220 Kn ILD BY: Mrs. S. M. Madhale 10 A B 6 9 F RB 1 X1 6 9 15 3 200 Kn 100 Kn X2 3
  • 11. Reaction at support RA 200Kn load lies on section • Remove the support reaction RA and Lift end A in the direction of Ra using some amount of force F = 1 and End B is free to rotate. Hence due to force the beam will deflection as shown in Fig • Make ordinate at A equal to one. • Calculate the ordinate X1 and X2 • 1 15 = 𝑋1 9 1 15 = 𝑋2 6 • 𝑋1 = 9 15 = 0.6 𝑋2 = 6 15 = 0.4 • Reaction At support A = Load * Load factor X ={ Load * Load factor X1}+{ Load * Load factor X2} = {200 * 0.6} + {100 * 0.4} =120+40 =160 Kn ILD BY: Mrs. S. M. Madhale 11 A B 6 9 F RB 1 X1 6 9 15 3 200 Kn 100 Kn X2 3 X2
  • 12. Reaction at support RB 100Kn load lies on section • Remove the support reaction RB and Lift end B in the direction of Rb using some amount of force F = 1 and End A is free to rotate. Hence due to force the beam will deflection as shown in Fig • Make ordinate at B equal to one. • Calculate the ordinate X1 and X2 • 1 15 = 𝑋1 3 1 15 = 𝑋2 6 • 𝑋1 = 3 15 = 0.2 𝑋2 = 6 15 = 0.4 • Reaction At support B = Load * Load factor X ={ Load * Load factor X1}+{ Load * Load factor X2} = {200 * 0.2} + {100 * 0.4} =40+40 =80 Kn ILD BY: Mrs. S. M. Madhale 12 A B 6 9 F RA 1 X1 6 9 15 3 200 Kn 100 Kn X2 3
  • 13. Reaction at support RB 200Kn load lies on section • Remove the support reaction RB and Lift end B in the direction of Rb using some amount of force F = 1 and End A is free to rotate. Hence due to force the beam will deflection as shown in Fig • Make ordinate at B equal to one. • Calculate the ordinate X1 and X2 • 1 15 = 𝑋2 9 1 15 = 𝑋1 6 • 𝑋1 = 9 15 = 0.6 𝑋2 = 6 15 = 0.4 • Reaction At support B = Load * Load factor X ={ Load * Load factor X1}+{ Load * Load factor X2} = {200 * 0.4} + {100 * 0.6} =80+60 =140 Kn ILD BY: Mrs. S. M. Madhale 13 A B 6 9 F RB 1 X1 6 9 15 3 200 Kn 100 Kn X2 3
  • 14. Shear Force at Section 100Kn load lies on section • Remove the ability to resist shear force the beam and providing rollers in vertical direction at section and apply unit deflection in the direction of shear force. Make ordinate at section equal to one. • Calculate the ordinate X1 and X3 [for negative consider loads before section] • 1 15 = 𝑋1 6 𝑋1 = 6 15 = 0.4 • 1 15 = 𝑋3 3 𝑋3 = 3 15 = 0.2 Shear force at section = [ Load * Load factor X1] + [ Load * Load factor X3] = { - [100 * 0.4] } +{-[200*0.2]} = - 40 -40 = - 80 Kn ILD BY: Mrs. S. M. Madhale 14 A 6 9 1 X1 6 9 3 200 Kn 100 Kn X2 3 B 1 X3
  • 15. Shear Force at Section 200Kn load lies on section • Remove the ability to resist shear force the beam and providing rollers in vertical direction at section and apply unit deflection in the direction of shear force. Make ordinate at section equal to one. • Calculate the ordinate X1 and X3 [for positive consider load after section] • 1 15 = 𝑋2 9 𝑋2 = 9 15 = 0.6 • 1 15 = 𝑋3 6 𝑋3 = 6 15 = 0.4 Shear force at section = [ Load * Load factor X3] + [ Load * Load factor X2] = { + [100 * 0.4] } +{+[200*0.6]} = + 40 + 120 = 160 Kn ILD BY: Mrs. S. M. Madhale 15 A 6 9 1 X1 6 9 3 200 Kn 100 Kn X2 3 B 1 X3
  • 16. Bending moment at Section 100Kn load lies on section • Remove the ability to resist moment by providing hinge at that section and Apply unit rotation equal to one. 𝜃=1= 𝜑+ 𝛽 Calculate the ordinate ∅, 𝜃 𝑎𝑛𝑑 𝛽 • Calculate the ordinate X1 and X3 [for negative consider loads before section] (𝑎 ∗𝑏 )/𝐿 • 𝑋1 = 𝑎 ∗𝑏 𝐿 = 6 ∗9 15 = 3.6 • 𝑋2 = 3.6 6 ∗ 3 = 1.8 Bending moment at section = [ Load * Load factor X1] + [ Load * Load factor X3] = [100 * 3.6] +[200 * 1.8] = 360 + 480 = 720 Kn-m ILD BY: Mrs. S. M. Madhale 16 A 6 9 1 X1 6 9 200 Kn 100 Kn X2 3 B 1 3 3
  • 17. Bending moment at Section 200Kn load lies on section • Remove the ability to resist moment by providing hinge at that section and Apply unit rotation equal to one. 𝜃=1= 𝜑+ 𝛽 Calculate the ordinate ∅, 𝜃 𝑎𝑛𝑑 𝛽 • Calculate the ordinate X1 and X3 [for negative consider loads before section] (𝑎 ∗𝑏 )/𝐿 • 𝑋1 = 𝑎 ∗𝑏 𝐿 = 6 ∗9 15 = 3.6 • 𝑋2 = 3.6 9 ∗ 6 = 2.4 Bending moment at section = [ Load * Load factor X1] + [ Load * Load factor X3] = [200 * 3.6] +[100 * 2.4] = 720 + 240 = 960 Kn-m ILD BY: Mrs. S. M. Madhale 17 A 6 9 1 X1 6 9 200 Kn 100 Kn X2 3 B 1 3 6
  • 18. THANK YOU !!! ILD BY: Mrs. S. M. Madhale 18