Influence Lines

source: https://www.udemy.com/course/structural-analysis-i

notes-pdf: LectureNotes_6.1+ReactionInfluenceLine.pdf

Remember this about hinges

Hinges aren't ordinary points —some change always happens around the. Consider this diagram:

When drawing the influene line diagram for the reaction at A: Ay, we know the ordinate will be 1 at the point A, and 0 at the 4m point between B and C since it can't move, but what about the hinge between A and B?

You might be tempted to draw a straight line from A to B, thinking the might behave like just another point, but that's wrong. Some kinda change will happen at the hinge and the members are likely to have different slopes. So the resulting diagram looks like this:

Note how whole member moved up by 1, rather than following the style of the red line.

Firstly, the technique described here works only for drawing influence line for stable structures.

Reaction influence lines

There are some key steps involved in drawing influence lines:

1. Identifying the members
2. Removing the restrictions imposed by the support at the point
3. Immediately drawing a straight line of magnitude 1 at the point
4. Think about how the members move

I) Identifying the members

questionanswer

To understand the importance of this, look at the figure below and ask what are the members?

Think about and determine you answer before checking the answer tab.

The members are AA' (yellow), AC' (green) and C'D (blue).

The identification of these members are important because they force the straight lines that constitute our influence line diagram.

II) Removing the restriction

Suppose we wanted to find the influence line diagram for the reaction at C. For that remove the restriction at C:

III) Drawing straight line of magnitude 1

One thing is certain when drawing influence lines: at the point where influence line diagram is drawn the height of the influence line is 1.

IV) Think about how the members move

Here A'C' is the member that contains our influence line at C —no matter what influence line diagram ends up as, A'C' will be a straight line.

Even though we have removed the restriction at C, the restrictions from support everywhere else will stay: so, the point B is still bound to its support. A'C' can only rotate with B as its fulcrum. Knowing this (and that the hinges A' and C' and the rest of the structure attached to those hinges allow it), we can see that A'C' will end up looking like this:

Drawing the other two member lines are easy. So at the end, we get this influence line diagram:

resultsolutions

Shear Influence Lines

Technique

The process for drawing shear influence lines is the same, except with shear we actually "cut" the statically determinate stable structure to get the shear influence line diagram while in the Support Reaction influence diagrams we just removed the restraints imposed by the support and just kept the reaction force to analyze it.

Quiz

Q1Q2Q3Q4Q5

The below figure can't be an influence line diagram for the shear force at D because the right half, after D, can't go up because the member/segment DC restricts both rotation about B or translation upwards.

Here, its given that the influence line diagram is for shear at D, so we know the total height of the line there is gonna be 1.

Read the question below carefully. We are given the influence line for the shear at D. This means, given loading pattern (distributed/concentrated load on the beam overall) we can find out what the shear on point D will be.

Now. 10K is placed on the midpoint of BC. It is not going to create any shear on point D. So the answer is zero.

Now this question is actually quite easy. We are given the shear influence line at D. So long as the load applied is somewhere where there is gonna be an effect, we will have some force on D.

a 10Kn applied on D will yield 10Kn force on D. an 8Kn force's effect will be determined by the graph:

8kN is 1.5m away from 10kN, the height of influence line diagram at 10kN point is 1, while the height at 8kN is 0.5: so from 8kN the effect on the point D is 4kN. The total load therefore is 10 + 4 = 14kN.

Now read the question carefully. Its the influence line for the shear at midpoint of BC. Given this info, we can tell what the ordinates are:

To get the maximum positive shear force from a distributed load of some length, at the midpoint of BC, as the question asks, we should place the UDL 3kN/m.

If the load applied is gonna be 3kN/m, the diagram for resulting load on BC midpoint is gonna look like this:

Note

Because of the nature of influence line diagram, the 3kN/m force is unable to have the effect of 3kN/m even at point D.

So the maximum positive shear force will be the area under the figure:

0.5 * 3 * 1.5 + 0.5 * 2 * 1.5 = 3.75 kN

...or so I thought. I assumed the load had to be one continus stretch. But it turns out its ok to divide it among the two positive regions. In that case we need to find out the area on the right too:

0.5 * 2 * 1.5 = 1.5 kN.

Left + Right = 3.75 + 1.5 = 5.25.

Moment influence lines

Technique

The technique to draw moment influence line is also very similar to that of shear.

In order to find moment influence line at a particular point, we replace that point with a hinge. This hinge means that:

1. We are essentially creating two members: one to the left of the hinge and one to the right of the hinge.
2. The member to the right of the hinge is given an anti-clockwise moment and to the left of the hinge is given a clockwise moment.

Left member: anti-clockwise moment

Right member: clockwise moment

In the above figure, we see that the "member" DF can't rotate because of constraints. This leaves us to think about 2 other members: CD and CA, which leads us to realise that C will down. So we know the maximum negative moment at D will happen when load is at C and maximum positive moment when load is at A.

Challening

Firstly, in order to determine the maximum negative moment at G, we cut at G and apply a anticlockwise moment on the left member and a clockwise moment on the right member. We can see that as a result of this F will go down the most and E will go up. So immediately we can see that the most negative moment at G will happen when the load is at F.

Now let's place the load of 2kN on F and examine the structure:

Given the load is at F, we can immediately see that Ay will be zero. We can also see that By too will be zero. This is because we can easily visualize the influence line diagram for those support reactions:

Influence line diagram for AyInfluence line diagram for By

So we know both Ay and By are gonna be 0. This greatly simplifies our analysis:

Now we just need to find moment at G.

So to find the moment at G, we just need to consider either the left or right side.

Cy + Dy - 2 = 0 (Since whole thing isn't moving upwards)

Summation of moment about D = 0 (anticlockwise positive)

2kN * 12 - Cy * 8 = 0 24kN = 8Cy Cy = 24kN/8 Cy = 3.

Cy + Dy = 2 3 + Dy = 2

Dy = -1 Cy = 3

Calculating moment at G, from support D:

clockwise: 1 * 4 = 4kNm.

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