Calculating Shear Force for a Linearly Increasing Load on a Rod

[Patdon10]

Homework Statement

A rod of length L is attached to a wall. The load on the rod increases linearly (as shown by the arrows in the figure) from zero at the left end to W Newtons per unit length at the right end.

(a) Find the shear force at the right end. (Use any variable or symbol stated above as necessary.)

(b) Find the shear force at the center. (Use any variable or symbol stated above as necessary.)

(c) Find the shear force at the left end. (Use any variable or symbol stated above as necessary.)

Homework Equations

Shear force = Force * Cross sectional area

The Attempt at a Solution

I really don't know where to start. I don't know how I can find the shear force without the diameter of the rod. Having it all in variables makes it tougher. Can anyone point me in the right direction?

 

[radou]

It seems you're mixing up shear forces and strains.

Take a piece of the rod from its end and write down the equations of equilibrium.

 

[Patdon10]

ok. So from the middle:

summation of x component force = 0
Summation of y component forces = a uniformly increasing force
Moment of y: the uniformly increasing force centered between the middle and the wall, pushing the negative direction.

I don't know how to set up the equations.

 

[Beaks]

I have this same problem for my Phys. homework and I'm also having trouble with it. I think I got it, but I want to be sure before I commit to it and lose points. Could someone please verify my work? Here's what I have:

Starting with the wall as the origin, the function for the force on the rod is:
F(x) = W(x/L)

If the right end of the rod is chosen as the pivot point, then I can use the torque resulting from F(x) to find the shear force, right? So I set up an integral to calculate this torque:

moment arm = (L-x)

∫(L-x)F(x)dx = (W/L)∫(L-x)xdx = (W/L)∫(Lx - x²)dx =

(W/L)[(Lx²/2) - (x³/3)]

Evaluating from 0 to L, I get:

WL²/6

So the force required to produce this torque with a moment arm of L would be...

(WL²)/(6L) = WL/6

... which would also be the shear force, correct?

 

[PhanthomJay]

It is not necessary in this problem to look at torques at all, since it just asks for the shear forces. Use the equilibrium equation for forces in the vertical y direction. First solve for the vertical reaction force at the wall to get the shear force in the beam at that point. Then draw free body diagrams of a beam section cut thru the requested points and use the same equation of equilibrium of forces in the y direction to solve for the shear at those points. Note that the area under the distributed loading curve gives the resultant force of that load distribution

 

[Beaks]

So the resultant force of the load distribution is:
(W/L)∫xdx = (Wx²)/(2L)

Limits of 0 to L produces:
WL/2

\SigmaF_y = 0 = resultant force - reaction force

So the reaction force at the wall = WL/2 ?

I must be missing something here because I tried this earlier and it was incorrect.

 

[Beaks]

Any other brave souls? The professor said he will go over it Monday, but it will be haunting me until then.

 

[PhanthomJay]

So the resultant force of the load distribution is:
(W/L)∫xdx = (Wx²)/(2L)

Limits of 0 to L produces:
WL/2

\SigmaF_y = 0 = resultant force - reaction force

So the reaction force at the wall = WL/2 ?

I must be missing something here because I tried this earlier and it was incorrect.

It is correct, it is the reaction force at the wall. Part a asks for the shear at the right (free) end. Part c asks for the shear at the left (wall ) end. You have answered part c.

 

[Beaks]

That makes sense to me, it seems like it would have to support the entirety of the force, or else where is the rest of the reaction force coming from? That was the first thing I tried, I entered it for part c but WebAssign says I'm wrong.

The force W would tend to produce a torque at the wall, as opposed to a shear force, so maybe that has something to do with it. Lol, or maybe WebAssign is wrong, I don't know.

 

[PhanthomJay]

Are you sure you have the exact same problem as the OP? Usually the force distribution is represented in terms of w (that's a lower case w) in units of force per unit length. The resultant load is often represemtred by a W (upper case letter). There is both a torque and a shear force at the wall, the value of the shear force (and the value of the reaction) being WL/2 for the problem as given, where W is the max value of the distributed load, in units of force per unit length.

 

[Beaks]

You were right right! I entered it in WebAssign as (WL/2) N, but it counted it wrong. I saw that someone else entered it as just (WL/2) and they got it right. The problem says "W Newtons," so if the units aren't included in W they should be included in the answer, at least by my figuring.