Calculating Elastic Curve for a given Section

[member 392791]

Homework Statement

Homework Equations

The Attempt at a Solution

Hello,

Comparing these two problems, 9.6 says to calculate the elastic curve for portion BC. However, in the solution they include the contribution from P which is not contained in portion BC in the representative cut to calculate the moment at X.

Yet, in the next problem 9.7 they ask to calculate the elastic curve for portion AB, and they don't include the loading distribution in the calculating of the moment at X for the representative cut.

This seems to be contradictory and now I am confused.

Also, for problem 9.4, I am not understanding why the loading of a right triangle is 1/2w(x/L)x and not 1/2wx. After all, for a straight loading the force on the representative cut is wx, not w(x/L)x

 

[SteamKing]

For Prob. 9.4, the distributed load at point A is 0 and at point B is w0. The ordinate of the load at any point in between A and B is thus going to be x*(w0/L). When x = L, the ordinate of the load becomes just w0. The total force at a point x from point A is:

F = \int x* (w0 / L) dx = (w0 / L) * (x^2) / 2

The total load on the entire beam is found by letting x = L in the expression above,

F [@ x = L] = (w0 / L) * L^2 / 2 = w0 * L / 2, which is also the area under the triangle.

 

[member 392791]

So that's why the w0 is hovering above point B? I thought that w0 was just the load distribution

 

[SteamKing]

The Attempt at a Solution

Hello,

Comparing these two problems, 9.6 says to calculate the elastic curve for portion BC. However, in the solution they include the contribution from P which is not contained in portion BC in the representative cut to calculate the moment at X.

Yet, in the next problem 9.7 they ask to calculate the elastic curve for portion AB, and they don't include the loading distribution in the calculating of the moment at X for the representative cut.

This seems to be contradictory and now I am confused.

I beg to differ with your interpretation of problems 9.6 and 9.7.

The calculation of the reactions of the beams in each of the problems does include all of the applied loads. All of the loads are present in the moment equation, which is then integrated twice to obtain the deflection equation. The loads may not appear to you that they have been included, but they have been.

 

[member 392791]

I beg to differ with your interpretation of problems 9.6 and 9.7.

The calculation of the reactions of the beams in each of the problems does include all of the applied loads. All of the loads are present in the moment equation, which is then integrated twice to obtain the deflection equation. The loads may not appear to you that they have been included, but they have been.

Look at problem 9.7 moment equation for the representative cut,

ƩM = -wL^2/4 + wx(x/2) + M = 0

Nowhere do I see the load distribution in segment BC included. The wx(x/2) term is from the loading on segment AB, and the -wL^2/4 is from the calculation of the moment at the wall.

Now looking at problem 9.6

ƩM= wL/5(L + x/2) - 4/5wLx + wx(x/2) + M = 0

The wL/5 term is from the loading at the edge P

 

[member 392791]

Also, does F(x) mean the equivalent concentrated load at x, or is it the total load at a given point? Meaning, a summation of the load from all points up to and including x

 

[SteamKing]

Also, does F(x) mean the equivalent concentrated load at x, or is it the total load at a given point? Meaning, a summation of the load from all points up to and including x

It's not clear what you mean by F(x); I don't see that used in these problems. Is this question relating to another problem?

 

[SteamKing]

With regard to problem 9.6: the overhang to the left where load P is applied can be omitted entirely from the beam by replacing it with an equivalent force and couple located at point B. This has been done when the reaction at B was calculated. Since you are interested in the deflection between points B and C, you have no further need to worry about P because its effects on the beam have already been taken into account.

With regard to problem 9.7: the beam is essentially loaded by the couple produced by the two distributed loads of equal magnitude operating in the opposite direction. You have no reaction at point A except for the moment wL^2/4. Since you are interested only in the deflections between points A and B, you make a cut thru the beam at the arbitrary point J and make a free body diagram as shown in the solution. The moment M located at point J is supposed to account for the moments produced by all of the loads which are applied between point J and point C, in order to keep the segment A-J in static equilibrium.

 

[member 392791]

For Prob. 9.4, the distributed load at point A is 0 and at point B is w0. The ordinate of the load at any point in between A and B is thus going to be x*(w0/L). When x = L, the ordinate of the load becomes just w0. The total force at a point x from point A is:

F = \int x* (w0 / L) dx = (w0 / L) * (x^2) / 2

The total load on the entire beam is found by letting x = L in the expression above,

F [@ x = L] = (w0 / L) * L^2 / 2 = w0 * L / 2, which is also the area under the triangle.

I am referring to this

 

[SteamKing]

That's what the equivalent point load is for the triangular load distribution which has the max. ordinate of w0 at
x = L.

Look at the free body diagrams for Prob. 9.4.