Using Macaulay's method of deflection on a cantilever beam

[Kyle Grayston]

Homework Statement

i) Determine the slope of the beam 1m from the wall.
ii) Calculate the deflection at the free end of the beam.

Homework Equations

Macauly's method of deflection: M = -RX + W[X-a]
Integrate once for slope
Integrate twice for deflection

The Attempt at a Solution

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It is my understanding that a positive deflection is downwards and a negative deflection is upwards.

I am getting a negative answer for the slope which would indicate an upwards slope?

When I use my initial equation to calculate the moment before integrating, I get the correct moment value (27.84kNm) so I feel like that is correct, I am worried that maybe some of the values in my equation should be negative which would change the value of my constants A and B.

Any help would be greatly appreciated.

 

[haruspex]

Your hand drawn diagram does not match the printed one. You have 0.5m to the first load where the printed diagram has 1m.

Also, shouldn't there be a torque and vertical force from the support in the expression for M?

 

[Kyle Grayston]

Your hand drawn diagram does not match the printed one. You have 0.5m to the first load where the printed diagram has 1m.

Also, shouldn't there be a torque and vertical force from the support in the expression for M?

Hi, sorry I've drawn the beam out mirrored as I was trying to compare it to a similar one done in class. My diagram is shown with the fixed end on the right.

There is 19.2kN reaction force at the fixed end, I am just not sure whether this is incorporated into the equation.

 

[haruspex]

My diagram is shown with the fixed end on the right.

Ok, so you are measuring x from the free end, but the question asks for the slope 1m from the wall, and you have substituted x=1.
Also, viewed from the free end, your loads will have clockwise moments, so should have negative torques. That is why you got a negative slope.

 

[Kyle Grayston]

Ah yes, I see where I have put the incorrect x value in, thanks!

As for the negative slope, is this what you mean by having a negative slope when looking at the free end?

 

[haruspex]

As for the negative slope,

Ignore what I wrote before... not saying it was wrong, but here's a simpler way..
A downward force left of x leads to an anticlockwise torque at x and makes a negative contribution to y". So if you are taking torques as positive clockwise then your M=... equations are correct but y"=-kM.

 

[Kyle Grayston]

Ah I see what you're saying, that makes more sense. Thanks very much for your help, I really appreciate it.