Influence line structural analysis engineering

Lecturer: T.A. Esmatullah Masom
Date : 2023/12/4
Kandahar University
Engineering faculty
Civil department
Structure Analysis I
C h a p t e r E i g h t
Influence Lines

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O R G A N I C S C O M P A N Y
Table Of Content
Introduction
8.1 Influence Lines for Beams and Frames by Equilibrium Method
8.2 Müller-Breslau’s Principle and Qualitative Influence Lines
8.3 Influence Lines for Girders with Floor Systems
8.4 Influence Lines for Trusses
8.5 Influence Lines for Deflections
Assignment
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Introduction
•Influence Lines important concept used when structures subjected to variable
(movable) loads.
•Initially introduced by E. Winkler in 1867.
•An influence line is a graph of a response function of a structure as a function
of the position of a downward unit load moving across the structure.
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Introduction
•Analysis of structures for variable loads consists of two steps:
1. Determining the position(𝑥) of the load at which the response function of
interest becomes maximum.
2. Computing the maximum value of the response function.
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8.1 Influence Lines For Beams And Frames By
Equilibrium Method
5
➢Influence Lines for Reactions:
+ ⤹ ∑𝑀𝑐 = 0
−𝐴𝑦 𝐿 + 1 𝐿 − 𝑥 = 0
𝐴𝑦 =
1 𝐿−𝑥
𝐿
= 1 −
𝑥
𝐿
+ ⤹ ∑𝑀𝐴 = 0
−1 𝑥 + 𝐶𝑦 𝐿 = 0
𝐶𝑦 =
1(𝑥)
𝐿
=
𝑥
𝐿

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Equilibrium Method
6

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Equilibrium Method
7
➢Influence Line for Shear at 𝐵:

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Equilibrium Method
8
➢Influence Line for Shear at 𝐵:
𝑆𝐵 = −𝐶𝑦 0 ≤ 𝑥 < 𝑎
𝑆𝐵 = −𝐶𝑦= −
𝑥
𝐿
0 ≤ 𝑥 < 𝑎
𝑆𝐵 = 𝐴𝑦 𝑎 < 𝑥 ≤ 𝐿
𝑆𝐵 = 𝐴𝑦 = 1 −
𝑥
𝐿
𝑎 < 𝑥 ≤ 𝐿

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Equilibrium Method
9
➢Influence Line for Bending Moment at 𝐵:
𝑀𝐵 = 𝐶𝑦(𝐿 − 𝑎) 0 ≤ 𝑥 ≤ 𝑎
𝑀𝐵 =
𝑥
𝐿
𝐿 − 𝑎 0 ≤ 𝑥 ≤ 𝑎
𝑀𝐵 = 𝐴𝑦(𝑎) 𝑎 ≤ 𝑥 ≤ 𝐿
𝑀𝐵 = 𝐴𝑦 = 1 −
𝑥
𝐿
𝑎 𝑎 ≤ 𝑥 ≤ 𝐿

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Example 8.1
10
Draw the influence lines for the vertical reactions at supports 𝐴 and 𝐶, and the
shear and bending moment at point 𝐵, of the simply supported beam shown in
Fig. 8.3(a).

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Example 8.1
11
Solution:
Influence Line for 𝑨𝒚:
+ ⤹ ∑𝑀𝑐 = 0
−𝐴𝑦 5 + 1 5 − 𝑥 = 0
𝐴𝑦 =
1 5−𝑥
5
= 1 −
𝑥
5
Influence Line for 𝑪𝒚:
+ ⤹ ∑𝑀𝐴 = 0
−1 𝑥 + 𝐶𝑦 5 = 0
𝐶𝑦 =
1(𝑥)
5
=
𝑥
5

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Example 8.1
12
Solution:
Influence Line for 𝑺𝑩:
𝑆𝐵 = −𝐶𝑦= −
𝑥
𝐿
= −
𝑥
5
0 ≤ 𝑥 < 3𝑚
𝑆𝐵 = 𝐴𝑦 = 1 −
𝑥
𝐿
= 1 −
𝑥
5
3𝑚 ≤ 𝑥 ≤ 5𝑚

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Example 8.1
13
Solution:
Influence Line for 𝑴𝑩:
𝑀𝐵 = 2𝐶𝑦 =
2𝑥
5
0 ≤ 𝑥 < 3𝑚
𝑀𝐵 = 3𝐴𝑦 = 3 −
3𝑥
5
3𝑚 ≤ 𝑥 ≤ 5𝑚

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Influence line for girder, trusses
& deflection
14
Lecturer: T.A. Esmatullah Masom
Date : 2023/12/11

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8.3 Influence Lines For Girders With Floor Systems
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17

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18
+ ⤹ ∑𝑀𝐹 = 0 − 𝐴𝑦 𝐿 + 1 𝐿 − 𝑥 𝐴𝑦 = 1 −
𝑥
𝐿
+ ⤹ ∑𝑀𝐴 = 0 𝐹𝑦 𝐿 − 1 𝑥 𝐹𝑦 =
𝑥
𝐿

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➢Influence Line for Shear in Panel 𝐵𝐶:

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➢Influence Line for Shear in Panel 𝐵𝐶:
𝑆𝐵𝐶 = −𝐹𝑦= −
𝑥
𝐿
0 ≤ 𝑥 ≤
𝐿
5
𝑆𝐵𝐶 = 𝐴𝑦 = 1 −
𝑥
𝐿
2𝐿
5
≤ 𝑥 ≤ 𝐿
𝑆𝐵𝐶 = 𝐴𝑦−𝐹𝐵= 1 −
𝑥
𝐿
− 2 −
5𝑥
𝐿
= −1 +
4𝑥
𝐿
𝐿
5
≤ 𝑥 ≤
2𝐿
5

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➢Influence Line for Bending Moment at G:
𝑀𝐺 = 𝐹𝑦 𝐿 − 𝑎 =
𝑥
𝐿
𝐿 − 𝑎 0 ≤ 𝑥 ≤
𝐿
5
𝑀𝐺 = 𝐴𝑦 𝑎 = 1 −
𝑥
𝐿
𝑎
2𝐿
5
≤ 𝑥 ≤ 𝐿
𝑀𝐺 = 𝐴𝑦 𝑎 − 𝐹𝐵 𝑎 −
𝐿
5
= 1 −
𝑥
𝐿
𝑎 − 2 −
5𝑥
𝐿
𝑎 −
𝐿
5
=
2𝐿
5
− 𝑎 − 𝑥 1 −
4𝑎
𝐿
𝐿
5
≤ 𝑥 ≤
2𝐿
5

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➢Influence Line for Bending Moment at Panel Point C:
𝑀𝐶 = 𝐹𝑦
3𝐿
5
=
𝑥
𝐿
3𝐿
5
=
3
5
𝑥 0 ≤ 𝑥 ≤
2𝐿
5
𝑀𝑐 = 𝐴𝑦
2𝐿
5
= 1 −
𝑥
𝐿
2𝐿
5
=
2
5
𝐿 − 𝑥
2𝐿
5
≤ 𝑥 ≤ 𝐿

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Example 8.9
23
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. 8.14(a)

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Example 8.9
24
Influence Line for 𝑺𝑩𝑪:
1𝑘𝑁 𝑙𝑜𝑎𝑑 𝑖𝑠 𝑎𝑡 𝐴, 𝐷𝑦 = 0 𝑆𝐵𝐶 = 0
1𝑘𝑁 𝑙𝑜𝑎𝑑 𝑖𝑠 𝑎𝑡 𝐵, 𝐷𝑦 =
1
3
𝑘𝑁 𝑆𝐵𝐶 = −
1
3
kN
1𝑘𝑁 𝑙𝑜𝑎𝑑 𝑖𝑠 𝑎𝑡 𝐶, 𝐴𝑦 = −
1
3
𝑘𝑁 𝑆𝐵𝐶 =
1
3
𝑘𝑁
1𝑘𝑁 𝑙𝑜𝑎𝑑 𝑖𝑠 𝑎𝑡 𝐷, 𝐴𝑦 = 0 𝑆𝐵𝐶 = 0

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Example 8.9
25
Influence Line for 𝑴𝑩:
1𝑘𝑁 𝑙𝑜𝑎𝑑 𝑖𝑠 𝑎𝑡 𝐴, 𝐷𝑦 = 0 𝑀𝐵 = 0
1𝑘𝑁 𝑙𝑜𝑎𝑑 𝑖𝑠 𝑎𝑡 𝐵, 𝐴𝑦 =
2
3
𝑘𝑁 𝑀𝐵 =
2
3
6 = 4kN. m
1𝑘𝑁 𝑙𝑜𝑎𝑑 𝑖𝑠 𝑎𝑡𝐷, 𝐴𝑦 = 0 𝑀𝐵 = 0

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8.4 Influence Lines For Trusses
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+ ⤹ ∑𝑀𝐸 = 0 − 𝐴𝑦 12 + 1 12 − 𝑥 𝐴𝑦 = 1 −
𝑥
12
+ ⤹ ∑𝑀𝐴 = 0 𝐸𝑦 12 − 1 𝑥 𝐸𝑦 =
𝑥
12

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➢Influence Line for Force in Vertical Member 𝐶𝐼:
+↑ ∑𝐹𝑦 = 0 −𝐹𝐶𝐼 +𝐸𝑦 = 0 𝐹𝐶𝐼 = 𝐸𝑦 0 ≤ 𝑥 ≤ 6𝑚
+↑ ∑𝐹𝑦 = 0 𝐹𝐶𝐼 + 𝐴𝑦 = 0 𝐹𝐶𝐼 = −𝐴𝑦 9 ≤ 𝑥 ≤ 18𝑚

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➢Influence Line for Force in Vertical Member 𝐶𝐼:
+↑ ∑𝐹𝑦 = 0 𝐴𝑦 −
9−𝑥
3
+ 𝐹𝐶𝐼 = 0
𝐹𝐶𝐼= −𝐴𝑦 +
9−𝑥
3
6𝑚 ≤ 𝑥 ≤ 9𝑚

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➢Influence Line for Force in Bottom Chord Member 𝐶𝐷:
•Using right portion AC + ⤹ ∑𝑀𝐼 = 0 −𝐹𝐶𝐷 4 + 𝐸𝑦 6 = 0
𝐹𝐶𝐷= 1.5𝐸𝑦 0 ≤ 𝑥 ≤ 6𝑚
•Using left portion AC + ⤹ ∑𝑀𝐼 = 0 𝐹𝐶𝐷 4 − 𝐴𝑦 6 = 0
𝐹𝐶𝐷= 1.5𝐴𝑦 6𝑚 ≤ 𝑥 ≤ 18𝑚

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➢Influence Line for Force in Diagonal Member 𝐷𝐼:

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➢Influence Line for Force in Diagonal Member 𝐷𝐼:
+↑ ∑𝐹𝑦 = 0
4
5
𝐹𝐷𝐼 + 𝐸𝑦 = 0 𝐹𝐷𝐼 = −1.25𝐸𝑦 0 ≤ 𝑥 ≤ 6𝑚
+↑ ∑𝐹𝑦 = 0 −
4
5
𝐹𝐷𝐼 + 𝐴𝑦 = 0 𝐹𝐷𝐼 = 1.25𝐴𝑦 9 ≤ 𝑥 ≤ 18𝑚

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➢Influence Line for Force in Top Chord Member 𝐼𝐽:
+ ⤹ ∑𝑀𝐷 = 0 𝐹𝐼𝐽 4 + 𝐸𝑦 3 = 0
𝐹𝐼𝐽 = −0.75𝐸𝑦 0 ≤ 𝑥 ≤ 9𝑚
+ ⤹ ∑𝑀𝐷 = 0 −𝐹𝐼𝐽 4 − 𝐴𝑦 9 = 0
𝐹𝐼𝐽 = −2.25𝐴𝑦 9𝑚 ≤ 𝑥 ≤ 18𝑚

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➢Influence Line for Force in Vertical Member 𝐹𝐿:

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Example 8.12
36
Draw the influence lines for the forces in members 𝐴𝐹, 𝐶𝐹, and 𝐶𝐺 of the
Parker truss shown in Fig. 8.19(a). Live loads are transmitted to the bottom
chord of the truss.

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Example 8.12
37
Solution:
Influence Lines for Reactions: The influence lines for the reactions 𝐴𝑦 and 𝐸𝑦
obtained by applying the equilibrium equations, ∑𝑀𝐸 = 0 and ∑𝑀𝐴 = 0

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Example 8.12
38
Solution:
Influence Line for 𝑭𝑨𝑭:

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Example 8.12
39
Solution:
Influence Line for 𝑭𝑨𝑭:
+↑ ∑𝐹𝑦 = 0 𝐴𝑦 − 1 +
3
5
𝐹𝐴𝐹 = 0
Because 𝐴𝑦 = 1 𝑘𝑁, 𝐹𝐴𝐹 = 0 𝑓𝑜𝑟 𝑥 = 0
•When the 1𝑘𝑁 load is located to the right of joint B, we write
+↑ ∑𝐹𝑦 = 0 𝐴𝑦 +
3
5
𝐹𝐴𝐹 = 0
𝐹𝐴𝐹 = −1.667𝐴𝑦 𝑓𝑜𝑟 4𝑚 ≤ 𝑥 ≤ 16𝑚

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Example 8.12
40
Solution:

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Example 8.12
41
Solution:
Influence Line for 𝑭𝑪𝑭:
+ ⤹ ∑𝑀𝑂 = 0
3
5
𝐹𝐶𝐹 16 + 𝐸𝑦 24 = 0
𝐹𝐶𝐹 = −2.5𝐸𝑦 0 ≤ 𝑥 ≤ 4𝑚
•When the 1 𝑘𝑁 load is located to the right of C, we consider the equilibrium
of the left portion AB to obtain
+ ⤹ ∑𝑀𝑂 = 0 𝐴𝑦 8 −
4
5
𝐹𝐶𝐹 3 −
3
5
𝐹𝐶𝐹 12 = 0
𝐹𝐶𝐹 = 0.833𝐴𝑦 8𝑚 ≤ 𝑥 ≤ 16𝑚

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Example 8.12
42
Solution:
Influence Line for 𝑭𝑪𝑮:
•We will determine the influence line for 𝐹𝐶𝐺 by considering the equilibrium of
joint 𝐺
+↑ ∑𝐹𝑦 = 0 − 𝐹𝐶𝐺 −
1
17
𝐹𝐹𝐺 −
1
17
𝐹𝐺𝐻 = 0
𝐹𝐶𝐺 = −
1
17
(𝐹𝐹𝐺 + 𝐹𝐺𝐻) (1)

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Example 8.12
43
Solution:
+→ ∑𝐹𝑥 = 0 −
4
17
𝐹𝐹𝐺 +
4
17
𝐹𝐺𝐻 = 0
𝐹𝐹𝐺 = 𝐹𝐺𝐻 (2)
•By substituting Eq. (2) into Eq. (1), we obtain
𝐹𝐶𝐺 = −
2
17
𝐹𝐹𝐺 = −0.485𝐹𝐹𝐺 (3)
•We will first construct the influence line for 𝐹𝐹𝐺 by using section 𝑎𝑎
•Than apply Eq. (3) to obtain the desired influence line for 𝐹𝐶𝐺.

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Example 8.12
44
Solution:

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Example 8.12
45
Solution:
Influence Line for𝑭𝑭𝑮:
+ ⤹ ∑𝑀𝑐 = 0
4
17
𝐹𝐹𝐺 4 + 𝐸𝑦 8 = 0
𝐹𝐹𝐺 = −2.062𝐸𝑦 0 ≤ 𝑥 ≤ 4𝑚
•When the 1 𝑘𝑁 load is located to the right of 𝐶, we consider the equilibrium
of the left portion AB to obtain
+ ⤹ ∑𝑀𝑐 = 0 −
1
17
𝐹𝐹𝐺 4 −
4
17
𝐹𝐹𝐺 3 − 𝐴𝑦 8 = 0
𝐹𝐹𝐺 = −2.062𝐴𝑦 8𝑚 ≤ 𝑥 ≤ 16𝑚

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Example 8.12
46
Solution:
The desired influence line for 𝐹𝐶𝐺 can now be obtained by multiplying the
influence line for 𝐹𝐹𝐺 by −0.485 ,

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8.5 Influence Lines For Deflections
47
•A more efficient procedure for constructing the deflection influence lines can
be devised by the application of Maxwell’s law of reciprocal deflections:
•Considering again the beam of Fig. 8.21(a).
•If 𝑓𝐵𝑋 is the vertical deflection at B when the unit load is placed at an
arbitrary point X,
•Then 𝑓𝑋𝐵 represents the ordinate at X of the influence line for the vertical
deflection at B.
𝑓𝐵𝑋 = 𝑓𝑋𝐵

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8.5 Influence Lines For Deflections
48

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Example 8.14
49
Draw the influence line for the vertical deflection at end B of the cantilever
beam shown in Fig. 8.22(a).

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Example 8.14
50
Solution:
𝑀𝑋 =
1
𝐸𝐼
−3 1 −
𝑥
3
𝑥
𝑥
2
−
1
2
3 − 3 1 −
𝑥
3
𝑥
2𝑥
3
𝑀𝑋 =
1
6𝐸𝐼
(𝑥3 − 9𝑥2)

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Example 8.14
51
Solution:

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Example 8.14
52
Solution:
•Thus, the deflection at X on the real beam is;
𝑓𝑋𝐵 =
1
6𝐸𝐼
(𝑥3
− 9𝑥2
)
•By applying Maxwell’s law of reciprocal deflections;
𝑓𝐵𝑋 = 𝑓𝑋𝐵
𝑓𝐵𝑋 =
1
6𝐸𝐼
(𝑥3 − 9𝑥2)
•By substituting the numerical values of 𝐸 and 𝐼, we get
𝑓𝐵𝑋 =
(𝑥3−9𝑥2)
240000
•The influence line for vertical deflection at B, obtained by plotting the
preceding equation, is shown in Fig. 8.22(e).

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ASSIGNMENT
•Read and exercise with (8.2 Müller-Breslau’s Principle and Qualitative
Influence Lines).
53

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O R G A N I C S C O M P A N Y 54

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