Use of influence lines

In the first guide, influence lines, I wrote about the use of influence lines to find favorable or non-deployable forces, maximum or minimum force load on a structure. In this guide, I will present two examples of how it looks. It is very possible that you will calculate such things in class.

The first example.
Let us assume that a car with the pressure of the left wheel 45kN and the right car 30 kN moves on the bridge with the static scheme given below.

Our task will be to find the greatest and smallest value of force that the wheels in the A support will cause.

We’re looking for VAmax. and VAmin..

First, we start with drawing LI for support A. The graph below will immediately contain the setting of forces to the maximum and minimum values.
Before calculating, we need a formula to calculate the maximum force and minimal, it looks as follows.

V_{A} = \Sigma P_{i} * \eta _{i},

P – working force
η – the ordinate of the impact line graph under force (we can calculate the ratio)

Let’s go to calculations. Here is a IW Va diagram immediately with visible P1 and P2 strength settings.

We’re looking for  V_{A,max}, so
V_{A,max} = \Sigma P_{i} * \eta _{i},
P_{1} = 45,00kN,\; \; \eta _{1} = 1
P_{2} = 30,00 kN,\; \; n_{2} = 0,60

V_{A,max} = 45,00 * 1 + 30,00 * 0,6 = 63,00 kN – the largest force.


Now, V_{A,min}, so
V_{A,min} = \Sigma P_{i} * \eta _{i},

P_{1} = 45,00kN,\; \; \eta _{1} = -0,1
P_{2} = 30,00 kN,\; \; n_{2} = -0,50

V_{A,min }= 45,00 * (-0,10) + 30,00 * (-0,50) = -19,50 kN – the smallest force.

Second example.
When using Winkler’s unevenness, calculate the maximum bending moment in the α-α cross-section in a free-domed beam loaded with conjugate forces. Beam diagram below.

We start by drawing Li for the bending moment in the α-α cross-section, but first I will present the formula for Winkler’s inequality, it looks as follows.

\frac{W_{P}}{W_{l}+P_{m}}< \frac{X_{a}^{'}}{X_{a}}< \frac{W_{p}+P_{m}}{W_{l}}

Wp – forces on the right side
Wl – forces on the left side
Pmax – maximum force
X’α – distance on the right side
Xα – distance on the left side

To find the maximum bending moment, we first need to find the most unfavorable force setting visible in the diagram above. That’s what Winkler’s inequality will do. However, in order to find this unfavorable setting, these forces must be shifted. While solving this example, I was lucky because I found the solution the first time.
The largest of the center group’s forces is set above ηmax. It looks like this.

We have to fulfill Winkler’s inequality, let’s check.

W_{p} \mapsto 4 + 1 + 3
W_{l} \mapsto 2 + 3
P_{max} \mapsto 4
X_{\alpha }^{'} \mapsto 7
X_{\alpha } \mapsto 3

\frac{4+1+3}{2+3+4}< \frac{7}{3}< \frac{4+1+3+4}{2+3}

0,888 <2,3 <2,4

Inequality is met, this is the most unfavorable setting of forces!

Now we can go to the calculation of the maximum bending moment in the α-α cross-section.
As in the previous task, we will use the formula:

M_{max}^{\alpha -\alpha } = \Sigma P_{i} * \eta _{i}

\eta _{max} = \frac{(X_{\alpha } * X_{\alpha }^{'})}{l} = \frac{(3 * 7)}{10} = 2,10m

Then we calculate the remaining η from Talesa.

\eta _{1} = \frac{1}{3}* \eta _{max} = \frac{1}{3} * 2,1 = 0,70m

\eta _{2} = \frac{2}{3}* \eta _{max} = \frac{2}{3} * 2,1 = 1,40m
\eta _{3} = \eta _{max}
\eta _{4} = 1,80m
\eta _{5} = 1,50m
\eta _{6} = 1,20m

Bending moment M_{max}^{\alpha -\alpha }  from the influence line.

M_{max}^{\alpha -\alpha } = P_{1} * \eta _{1} + P_{2} * \eta_{2} + P_{3} * \eta_{3} + P_{4} * \eta_{4} + P_{5} * \eta_{5} + P_{6} * \eta_{6}

M_{max}^{\alpha -\alpha } = 2,00 * 0,70 + 3,00 * 1,40 + 4,00 * 2,10 + 4,00 * 1,80 + 1,00 * 1,50 + 3,00 * 1,20

M_{max}^{\alpha -\alpha } = 26,30 kNm

At the end I will add that you have to practice, practice and rehearse to master particular issues.
The above two examples are a drop in the sea of tasks that you can get, but knowing well in the lines of influence, you can handle each task. The tasks I have presented are examples to show you how to use impact lines.