How to Solve Mechanics Problems Involving Framed Structures

[D44]

I think I have calculated the moment about the rhs point. I have this as 68.112Nm. Does this look about right?

When calculating shear forces and bending moments, do I still need to keep the rectangle and triangle separate? This part is what's causing trouble!

 

[PhanthomJay]

I think I have calculated the moment about the rhs point. I have this as 68.112Nm. Does this look about right?

Yes, clockwise on the beam and counterclockwise on the joint. Round it off to 68.1 N-m

When calculating shear forces and bending moments, do I still need to keep the rectangle and triangle separate? This part is what's causing trouble!

It's easiest to do it this way, since it helps to fing the load at a given point and the cg of the load. Start with the shear diagram, working left to right. There's nothing going on between 0 and 0.1 m, so the shear is 0 in this portion. Now look at the next section between 0.1 m and 2.15 m. Call the 0.1 m mark the '0' point along the x axis. Now draw a free body diagram of the beam around the left end that cuts through the distributed loading at a distance 'x' from the '0' point. What do you get as a total (shear) load at that cut, as a function of x? What do you get for the bending moment at that cut? You should proceed in the same manner as you did when calculating the shear and moment at the knee, except you will have your answer in terms of x.

 

[D44]

I understand what you said about zero shear force in the 0.1m section. When drawing the free body diagram for the cut, I still need to include this 0.1m section though, right?

I've had a go at this bit already and got 3 possibilities:
Qx=-4.65(x-0.1)^2
Qx=-4.44(x-0.1)^2
Qx=-4.26(x-0.1)^2
When x=0.1, Qx=0 and when x=2.25, Qx=20.52N?

For the rectangular section:
Qx=-20(x-0.1)
When x=0.1, Qx=0 and when x=2.25, Qx=-43N?
For the bending moments I have 0Nm at 0.1m from the left, -46.23Nm at the 2.25m point and -50.53Nm at the very rhs?

Cheers

 

[PhanthomJay]

I understand what you said about zero shear force in the 0.1m section. When drawing the free body diagram for the cut, I still need to include this 0.1m section though, right?

well you could, and did, but it just gets in the way and makes calculations more difficult than they need be. Just cut out that piece and chuck it. Pretend its not there. It's at the free end of the beam with no loading, so it doesn't do anything.

I've had a go at this bit already and got 3 possibilities:
Qx=-4.65(x-0.1)^2
Qx=-4.44(x-0.1)^2
Qx=-4.26(x-0.1)^2
When x=0.1, Qx=0 and when x=2.25, Qx=20.52N?

Your first possibility is correct for the triangular piece, in which case the shear (Qx?) in this section at x = 2.25 is 21.5 N under this load. I don't know where your other possibilities came from, since you did not show your work. Also, you can get rid of that 0.1 and just say Qx = -4.65x^2, where x is 0 at the start and 2.15 m at the end of the distributed loading , instead of 0.1 and 2.25 m.
, respectively.

For the rectangular section:
Qx=-20(x-0.1)
When x=0.1, Qx=0 and when x=2.25, Qx=-43N?

Yes, good. Now add up the shear from each loading to get 64.5 N at your x =2.25 m, whch agrees with the shear you got earlier at the knee (the shear is constant over the last 0.1 m). In between (your) x =0.1 and x= 2.25 , the shear is Qx= (-4.65(x-0.1)^2) + (-20(x-0.1)). The slope of the shear diagram at a given point is the negative of the loading intensity (N/m) at that point.

For the bending moments I have 0Nm at 0.1m from the left, -46.23Nm at the 2.25m point and -50.53Nm at the very rhs?

Cheers

Something's amiss here because the moment at the very rhs is -68.1 which we established earlier. Please show your work, as you are making me work hard for my money

 

[D44]

Hmmm... but then surely I already knew the shear forces of both the triangle and the rectangle when I calculated the resultant force in order to work out the 64.5N force as the reaction force at the right hand side?

The moment of the triangle I got as:
Mx=-(4.65*(x-0.1)^2)*((x-0.1)/3)
At 2.25, Mx=-13.35Nm
At 0.1, Mx=0

As for the rectangle...
Mx= -10(x-0.1)^2

 

[D44]

Ooops, Mx = -4.65x^3 for the triangle

 

[D44]

Sorry

For the rectangle I have between 0.1 and 2.25m:
Mx=-10(x-0.1)^2
at 0.1 Mx = 0
at 2.25 Mx = -46.23Nm

Then between 2.25m and the rhs I have:
Mx = -43(x-1.175)
at 2.25m Mx = -46.23Nm
at 2.35m Mx = -50.5Nm

I now understand that I can remove the 0.1 from my equations. I'm just keeping them in for consistency on here :)

 

[D44]

Right...

I've been looking at it some more...

Rectangular section:
Mx= -43*(x-1.175)
at x=2.25m, Mx=-46.23Nm
at x=2.35m, Mx=-50.5

Triangular section:
Mx= -21.5*(x-((2/3)*2.15))
at x=2.25m, Mx=-15.48Nm
at x=2.35m, Mx=-17.63Nm

Looking any better?

 

[PhanthomJay]

Hmmm... but then surely I already knew the shear forces of both the triangle and the rectangle when I calculated the resultant force in order to work out the 64.5N force as the reaction force at the right hand side?

The moment of the triangle I got as:
Mx=-(4.65*(x-0.1)^2)*((x-0.1)/3)
At 2.25, Mx=-13.35Nm

math error...Mx = - 15.4

At 0.1, Mx=0
As for the rectangle...
Mx= -10(x-0.1)^2

yes...

 

[PhanthomJay]

Right...

I've been looking at it some more...

Rectangular section:
Mx= -43*(x-1.175)
at x=2.25m, Mx=-46.23Nm
at x=2.35m, Mx=-50.5

Triangular section:
Mx= -21.5*(x-((2/3)*2.15))
at x=2.25m, Mx=-15.48Nm
at x=2.35m, Mx=-17.63Nm

Looking any better?

Yes, that looks about right...good work. Now you have to (roughly) draw the shear and moment diagrams... watch your curvatures...then proceed to the slanted member.

 

[D44]

This is where I now don't really know what to do next.

So...

The shear force diagram is going to look something like...a straight line starting at the lhs out to the 0.1m point, then would it be a cubic curve (terminology?) down to -64.5, horizontal line to the very rhs and then a vertical line up to the 0 line?

As for the bending moment diagram, zero until 0.1m then a steady curve down to -61.1 at...the very rhs? Not really sure on this one.

For this particular part, do I need to draw the whole shape back together and draw the respective diagrams underneath? I shouldn't leave them apart, right?

 

[PhanthomJay]

This is where I now don't really know what to do next.

So...

The shear force diagram is going to look something like...a straight line starting at the lhs out to the 0.1m point,

, yes, and the value of the shear is 0 in this section

then would it be a cubic curve (terminology?)

the shear is a quadratictc (parabolic) curve in this section of the problem from x =0.1 to x =2.25, (it is in the form of Ax^2 + Bx + C)

down to -64.5,

yes, at x = 2,25 m

horizontal line to the very rhs and then a vertical line up to the 0 line?

Yes from x = 2.25 to x =2.35 m, it is a horizontal line with a value of -64.5 to the very rhs, then a straight vertical line up to 0 line. Very good! Now you have to draw the shape of the parabola correctly...remembering that the slope of the parabolic shear at a given point is the negative of the load intensity at that point, that is, -20 at x =0.1, and -40 at x=2.25, so the slope gets increasingly negative as you move from left to right.

As for the bending moment diagram, zero until 0.1m then a steady curve down to -61.1 at...the very rhs? Not really sure on this one.

For this particular part, do I need to draw the whole shape back together and draw the respective diagrams underneath? I shouldn't leave them apart, right?

I'll get back you as soon as I can...

 

[D44]

Ok brilliant, I'll get that bit done then.

Thanks so much, you've been a massive help so far.

 

[jobz1]

how did u get 68.112

I think I have calculated the moment about the rhs point. I have this as 68.112Nm. Does this look about right?

When calculating shear forces and bending moments, do I still need to keep the rectangle and triangle separate? This part is what's causing trouble!

 

[jobz1]

when u say cut left of the knee, do u mean cutting the l shaped frame work or cutting it at the start of the horizontal section

This problem seems to be a popular one. This response is for you and jobz1yes, correct, 64.5 N up Start with the horizontal member first. Draw a free body diagram of that member, isolating it (cutting it) just to the left of the knee. The vertical (shear) force at that cut is 64.5 N, as you have noted. That is the shear at that point. There is also a moment at that point. Can you calculate the moment at that point? It acts opposite and equal to the moment about that point from the trapezoidal load. The moment from the trapezoidal load can be found by breaking that load into a uniformly distributed load and a triangularly distributed load, and summing the moments of the resutants of each about that point.

 

[PhanthomJay]

how did u get 68.112

The rectangular load has a resultant load of 20(2.15) = 43 N acting at the cg of it's load, which is (2.15/2) + 0.1 = 1.175 m away from the knee. Thus its moment is 43(1.175) = 50.5 N-m ccw. The triangular load has a resultant load of (20/2)(2.15) = 21.5 N acting at the cg of the load (1/3 its length away from the fat end), which is (2.15/3) + 0.1 = 0.82 m away from the knee. Thus its moment is 21.5(0.82) = 17.6 N-m ccw. And the total of the moments from both those loadings is 50.5 + 17.6 = 68.1 N-m ccw. So the moment just to the left of the knee acting on the horizontal beam is 68.1 N-m cw.

when u say cut left of the knee, do u mean cutting the l shaped frame work or cutting it at the start of the horizontal section

Cut the horizontal beam with a saw just to the left of the knee. That exposes the internal forces and moments at that knee, which we are solving for. They act up and cw respectively, on the beam, and down and ccw, respectively, on the slanted member, per Newton 3.

 

[PhanthomJay]

As for the bending moment diagram, zero until 0.1m then a steady curve down to -61.1 at...the very rhs? Not really sure on this one.

You have to proceed in the same manner; you have established earlier that the moment of the loading about a point x distance from the start is Mx=-(4.65*(x-0.1)^2)*((x-0.1)/3) + (-10(x-0.1)^2). This is a cubic curve. The slope of the moment diagram at a given point is equal to the shear at that point. You can also plug in different values of x to get the numerical values at those different points. After x = 2.25, then the moment diagram continues as a straight slanted line with a slope having the shear value of -64.5 in that short 0.1 m section.

For this particular part, do I need to draw the whole shape back together and draw the respective diagrams underneath? I shouldn't leave them apart, right?

Well, the moment diagram as I explained above includes both the triangular and rectangular loading in the same diagram.

 

[jobz1]

right i get it know... this took like 4 hours to understand ... i would recommend to look through http://books.google.co.uk/books?id=...#v=onepage&q=free body diagram moment&f=false
Engineering Mechanics By Sharma D. P.

because u have have to understand how the moment and the force act on the triangular part

any how thanks d44 and jay , much appreciated

Yes, that looks about right...good work. Now you have to (roughly) draw the shear and moment diagrams... watch your curvatures...then proceed to the slanted member.

 

[D44]

So in order to get the curve right I can just enter values between 0.1 and 2.15 for x into that equation and whatever the value is then I plot that on the graph?

As for the next part with the slanted section, am I meant to turn that so it's horizontal?

 

[PhanthomJay]

So in order to get the curve right I can just enter values between 0.1 and 2.15 for x into that equation and whatever the value is then I plot that on the graph?

yes, you can do it that way, and get exact points, and draw the curve...and as a check, the slope of the shear diagram at a point is equal to the negative of the distributed load intensity at that point, and the slope of the momemt diagram at a point is equal to the shear at that point.

As for the next part with the slanted section, am I meant to turn that so it's horizontal?

yes, resolve the end forces into components perpendicular and parallel to the slanted member...leave the end momemt as is...and proceed. The parallel force components produce an axial compressive force...it's the perp force components and end moment that produce the shears and moments in that member.

 

[D44]

Can I just recheck, when I'm drawing these bending moment diagrams, combined, it's a cubic curve? The curve starts at 0Nm at 0.1m and curves down to -61Nm at 2.25m, then a straight diagonal line down to -68.1Nm then back up to 0Nm.

For the reaction forces of the ground on the sloping section I have Fy=64.5N, Fx=17.28N and FR=66.78N. Obviously there's an anti clockwise moment at the top of the slope and a clockwise moment at the bottom. Are they both equal to 68.1Nm?

So when the section is turned to the horizontal position, would there be a force of 66.76N acting vertically down (ground side)? Are there only moments at the top part then? What are the x1 and x2 values needed for that you mentioned on the posts from months ago? I've calculated these, but I'm not entirely sure what they are for? Moments?

 

[PhanthomJay]

Can I just recheck, when I'm drawing these bending moment diagrams, combined, it's a cubic curve?

yes, the moment is in the form of Ax^3 + Bx^2 + Cx + D, in the distributed load piece only

The curve starts at 0Nm at 0.1m and curves down to -61Nm at 2.25m, then a straight diagonal line down to -68.1Nm then back up to 0Nm.

yes, more or less, there is a round off error somewhere, Mx = -61.7 at x = 2.25.

For the reaction forces of the ground on the sloping section I have Fy=64.5N, Fx=17.28N and FR=66.78N.

Be careful how you define your axes and components. In the vertical Y direction, Fy = 64.5 N up . There is no horizontal x component. But when you are looking at the slanted member and want to draw a shear and moment diagram, you want to break up the 64.5 N vertical force into its components perpendicular and parallel to the member. The perp comp is 64.5 sin15 (that's the shear), and the axial compressive load is 64.5 cos 15. See post #11 on page 1.

Obviously there's an anti clockwise moment at the top of the slope and a clockwise moment at the bottom. Are they both equal to 68.1Nm?

No, the cw moment at the bot is larger...the shear force at the top adds moment.

So when the section is turned to the horizontal position, would there be a force of 66.76N acting vertically down (ground side)? Are there only moments at the top part then? What are the x1 and x2 values needed for that you mentioned on the posts from months ago? I've calculated these, but I'm not entirely sure what they are for? Moments?

When you turn the member horizontal by rotating your computer 75 degrees ccw, the shear force at the left is 64.5 sin 15 down,and you know the moment there. Calculate the shear and moment at the fixed (right) using the equilibrium equations, and draw the shear and moment diagrams. The moment at the fixed end should be the P1x1 + P2x2 values you have calculated.

 

[D44]

So the ground reaction force is the hypotenuse of the triangle used to calculate the other forces? I'd previously written it as the adjacent. Just for future reference, how would you know which it was?

The shear force, 64.5sin15, does this not act at the rhs (ground)? If not, so far at the lhs I have a moment of 68.1Nm, a shear force of 64.5sin15N - at the rhs I have a horizontal force of 64.5cos15N and an moment of 128.4Nm? Am I also trying to find a vertical force at the rhs too?

 

[D44]

Using the equillibrium equations, the vertical up force at the rhs is equal to the 64.5sin15N force?

What happens with the horizontal force at the rhs? Because the sum of Fx=0, but where does the other horizontal force come from to balance the rhs force? 64.5cos15 from lhs from splitting the 64.5N force up?

 

[PhanthomJay]

So the ground reaction force is the hypotenuse of the triangle used to calculate the other forces? I'd previously written it as the adjacent. Just for future reference, how would you know which it was?

When you first look at the frame in its entirety, always find the reactions first, using sum of y forces =0, sum of x forces = 0, and sum of moments about any point = 0. When you look in the x direction, there are no applied forces in that direction, hence, no horizontal reaction at the support. There is only a vertical reaction P1 +P2, and a moment, P1x1 + P2x2. The vertical reaction is the resultant hypotenuse if you wish to break it up into comonents parallel and perpendicular to the sloped member.

The shear force, 64.5sin15, does this not act at the rhs (ground)?

yes...

If not, so far at the lhs I have a moment of 68.1Nm, a shear force of 64.5sin15N - at the rhs I have a horizontal force of 64.5cos15N and an moment of 128.4Nm? Am I also trying to find a vertical force at the rhs too?

yes, the shear force at the right hand side as you noted. Direction of force please at the left and right hand sides?

Using the equillibrium equations, the vertical up force at the rhs is equal to the 64.5sin15N force?

yes...

What happens with the horizontal force at the rhs? Because the sum of Fx=0, but where does the other horizontal force come from to balance the rhs force? 64.5cos15 from lhs from splitting the 64.5N force up?

yes...the horizontal (axial) force in the member is 64.5 cos 15. You are doing very well so far...now just draw the shear and moment diagrams for this slanted member and you've got it nailed...

 

[D44]

Shear forces at both ends then, lhs acting downwards at 64.5sin15 and rhs acting upwards at 64.5sin15?

So the axial force of 64.5cos15 is what equals the ground reaction force to make the sum of Fx=0?

When drawing the shear and moment diagrams, how is the axial force incorporated? What effect does it have?

 

[PhanthomJay]

Shear forces at both ends then, lhs acting downwards at 64.5sin15 and rhs acting upwards at 64.5sin15?

Yes! Since the sum of all forces perp to the member must be 0, if you have the shear force down at the left end, then it must act up at the right end,equal in mgnitude, because there are no other external forces applied on the beam in that direction.

So the axial force of 64.5cos15 is what equals the ground reaction force to make the sum of Fx=0?

The axial compressive force of 64.5cos15 at the lhs must be balanced by an equal and opposite axial component of the reaction force at the ground.

When drawing the shear and moment diagrams, how is the axial force incorporated? What effect does it have?

It has no effect when drawing the shear and moment diagrams. You could draw an axial force digram separately, which would be a constant force throughout. When determining the axial and bending memberstresses later, then you combine those stresses for the final results, per P/A +/- Mc/I.

 

[D44]

Brill, thank you!

My shear force diagram is looking like a vertical line down to -16.69N, horizontal line straight along to the rhs then back up to 0?

As for the BM diagram, I'm not so sure, but it has to go vertically staight down to -68.1Nm and then up to 128.4Nm at the rhs? Is this a curve crossing at the centre point of the 0 line, not a linear line?

The induced stresses are a combination of which, sorry? The shear forces and the axial force? Although I know what the P/A +/- Mc/I means, how am I to apply this? How do you mean per?

 

[PhanthomJay]

Brill, thank you!

My shear force diagram is looking like a vertical line down to -16.69N, horizontal line straight along to the rhs then back up to 0?

Yes, you are quite correct.

As for the BM diagram, I'm not so sure, but it has to go vertically staight down to -68.1Nm

yes

and then up to 128.4Nm at the rhs? Is this a curve crossing at the centre point of the 0 line, not a linear line?

It doesn't cross the line. The moment at the lhs and the shear at the lhs, both produce ccw moments about the support, so the straight line (not a curved line) from left to right slants down, not up. You must remember that the slope of the moment diagram at a given point is equal to the shear at that point. Thus, its slope is -16.69, a negative slope, and the moment at the right end is thus -68.1 - 16.69(4) = about -134 ( there is a round off error somwhere, as it should agee with the 128.4 you calculated earlier).

The induced stresses are a combination of which, sorry? The shear forces and the axial force? Although I know what the P/A +/- Mc/I means, how am I to apply this? How do you mean per?

It looks like the problem as given on page 1 did not ask for stresses..that's probably the 2nd part of the question.. If you haven't got to stresses yet, don't go any further until you study that topic.

 

[jobz1]

i am confused about the moment at the slanting bit... i have worked it out as 134 .. is that right

Yes, you are quite correct. yes It doesn't cross the line. The moment at the lhs and the shear at the lhs, both produce ccw moments about the support, so the straight line (not a curved line) from left to right slants down, not up. You must remember that the slope of the moment diagram at a given point is equal to the shear at that point. Thus, its slope is -16.69, a negative slope, and the moment at the right end is thus -68.1 - 16.69(4) = about -134 ( there is a round off error somwhere, as it should agee with the 128.4 you calculated earlier). It looks like the problem as given on page 1 did not ask for stresses..that's probably the 2nd part of the question.. If you haven't got to stresses yet, don't go any further until you study that topic.