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Pin shear question

  • Thread starter kieranl
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Homework Statement


The beam is supported by a pin at A and a short link BC, as shown in Figure 3.0. Determine the maximum magnitude P of the loads the beam will support if the average shear stress in each pin is not to exceed 80 MPa. The diameter for each of the pins is 18 mm.

The Attempt at a Solution



diagram aswell as working are shown in the attached file.. Just want to know if by my free body diagram the shear stresses for each pin are correct??

thanks

also how do you include pictures with your text?? I only know how to attach it as a pdf as image files are to big??
 

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Answers and Replies

  • #2
PhanthomJay
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Yer, FBD looks good; you need to calculate the angle at which Fa is applied (or its components), then after calculating T and Fa, see which force controls the shear design of the pins, which, as you note, are in double shear.
 
  • #3
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ok. does that mean that the shear force acting at pin C would be 0.5(vertical component of T) and at pin B be 0.5(horizontal component of T) and at pin A be 0.5(horizontal component of Fa)????
 
  • #4
PhanthomJay
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ok. does that mean that the shear force acting at pin C would be 0.5(vertical component of T) and at pin B be 0.5(horizontal component of T) and at pin A be 0.5(horizontal component of Fa)????
No, you had it right the first time in your work shown in your attachment (for example, shear force at pin C = 0.5T, etc.).
 
  • #5
Hello PhanthomJay

Could you run us through all the calculations concerning the amount of force acting on all 3 pins, given that the short link BC is acting at an angle?

I am assuming pin ‘A’ carries more of the total load than pin B & C; or am I incorrect? (I.e. F=M*A*Cos angle. Thus the proportion of the total load acting on pin A & B is not 50/50 but maybe 60/40?)
 
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  • #6
PhanthomJay
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Hello PhanthomJay

Could you run us through all the calculations concerning the amount of force acting on all 3 pins, given that the short link BC is acting at an angle?

I am assuming pin ‘A’ carries more of the total load than pin B & C; or am I incorrect? (I.e. F=M*A*Cos angle. Thus the proportion of the total load acting on pin A & B is not 50/50 but maybe 60/40?)
In re-examining the OP's working and my response, both were a bit vague. The key here is that link BC is a 2 force member, and as such, it can only take axial loads along the direction of BC, no shear or bending stresses. Thus, the reaction at C must be along the direction of BC, and the internal reaction at B must be equal and opposite to the reaction at C. You need now to use the 3 equilibrium equations to solve for Cx, Cy, Bx, By, Ax, and Ay, in terms of P, then use the allowable shear stress equations for a pin in double shear, to solve for P.
 
  • #7
Hi Jay

Excellent, I am with you so far.

So after calculating Bx & By (horizontal and vertical loads and arbitrarily lets say they are 5N/m and 10N/m ) Then when I use them to calculate the shear load acting on pin B do I keep them separate by saying we have two shear loads of 5N/m and 10N/m. Thus the maximum shear load acting on the pin B is 10N/m (P=10N/m), or do I add them together to calculate the shear on Pin B as 15N/m? (P=15N/m)

And for Pin C I think Cx & Cy should be added together to give a total shear load of 15N/m? (P=15N/m)

And the stress/strain load in member BC = 15N/m

Thank you very much for time, it is appreciated
Willy
 
  • #8
Hi Jay

Op's sorry; I think I can answer my last question.

The shear load acting on Pin B is the total vector load, I.e. Bx + Bc= P. Thus 5N/m + 10N/m = 15N/m of shear; (divided by two for a double shear pin)

Willy
 
  • #9
PhanthomJay
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Hi Jay

Excellent, I am with you so far.

So after calculating Bx & By (horizontal and vertical loads and arbitrarily lets say they are 5N/m and 10N/m )
They are not arbitrary, if By works out to be 10 N then Bx is related to it using Bx = By/tan 30. Forces are in units of N, not N/m.
Then when I use them to calculate the shear load acting on pin B do I keep them separate by saying we have two shear loads of 5N/m and 10N/m. Thus the maximum shear load acting on the pin B is 10N/m (P=10N/m), or do I add them together to calculate the shear on Pin B as 15N/m? (P=15N/m)
You add them together vectorially to get the total force, but since they are vector components at right angles to each other, you can't just add them algebraically, use the vector addition laws.
And the stress/strain load in member BC = 15N/m
The loading in BC and on the bolt at B is a force unit in N, not a stress/strain unit. The bolt shear stress is Force/area, which has units of N/m2, or Pa.
Thank you very much for time, it is appreciated
Willy
You are very welcome.
 
  • #10
Hi Jay

Sorry that's not quite what I meant so I'll rephrase.

If we calculate the tension in member BC to be 15N and that is made up of a horizontal component of 5N and a vertical component of 10N and a beam angle of 60 degrees

(I know these aren’t the actual equilibrium equations forces and factors, but they will do for now)

Then the shear force acting on pin C = 15n (which I think is right). Or do we have to work out a vector some based on 5N horizontal force @ 900 + 10N vertically @ 600? (Also, since the pins are in double shear, so / 2)


NEXT EXAMPLE
We have a sign hanging outside of a shop. The sign weighs 50kg and is supported by a rigid diagonal bar and a rigid horizontal bar (the sign is attached at the end of the horizontal bar).

The internal angle of the bar to the wall is 600 and 300 at the horizontal bar end. The tension in the vertical bar is 980N

sine (30 degrees) = (490 N ) / (Ftens) = Ftens = (490 N) / [ sine 30 (degrees) ] = 980 N.

My question is: What are the shear, stress/strain and torque loads acting on the bolts attaching the bars to the wall? (By stress/strain I mean the force trying to elongate the bolts by stretching them).

To answer my own question, I think:

SHEAR LOAD (which acts evenly on both the bars bolts) = Is simply the weight of the sign + the weight of the horizontal bar + the weight of the diagonal bar?. Divided by the number of bolts. (And als converting the weights to Newtons)

THE HORIZONTAL BAR
- Stress/Strain = Horizontal component of diagonal bar which either = 0 or because it is pushing into the wall due to the effect of the diagonal bar = 0
- Torque = 0, because the diagonal bar holds the vertical force, or at the least, it is very minimal.

THE DIAGONAL BAR
- Stress/ Strain = Tension in the diagonal bar? But if the torque of the diagonal bar is greater than tention force we should just consider the torque / bending force?

NOTE: For stress/strain; we are only interested in either the torque or tension, which ever is largest, and we don't add them together right?

Willy
 
Last edited:
  • #11
Hi Jay

Hi hope I'm not being to annoying,

In reference to my last posting

Quote: “We have a sign hanging outside of a shop. The sign weighs 50kg and is supported by a rigid diagonal bar and a rigid horizontal bar (the sign is attached at the end of the horizontal bar”.

So if we

1) Bring the diagonal bar straight out from the wall at 90 degrees (so it to is horizontal).
Then there is no tension in the bar? And, the only thing we are concerned with is the bending moment of the bar, and, torque & Shear acting on the bolts?,

--------sine ( 0 degrees) = (490 N ) / (Ftens) = Ftens = (490 N) / [ sine 0 (degrees) ] = 0 N.-------


2) Next if increase the angle of the diagonal bar a little to 0.5 degrees (where it meets the end of the horizontal bar I.e. almost flat but not quite)

-------Ftens = (490 N) / [ sine 0.5 (degrees) ] ---- (490 N) / [ sine 0.5 (degrees) ] = 56150N-------

This number is so big it just can’t be right? And also considering that it is almost horizontal?

All the best
Willy
 
  • #12
PhanthomJay
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Hi Jay

Sorry that's not quite what I meant so I'll rephrase.

If we calculate the tension in member BC to be 15N and that is made up of a horizontal component of 5N and a vertical component of 10N and a beam angle of 60 degrees

(I know these aren’t the actual equilibrium equations forces and factors, but they will do for now)

Then the shear force acting on pin C = 15n (which I think is right). Or do we have to work out a vector some based on 5N horizontal force @ 900 + 10N vertically @ 600? (Also, since the pins are in double shear, so / 2)
As I mentioned earlier, Link BC is a 2-force member, meaning it can take axial load only directed along its length. If the tension in BC is 15 N, then yes, the shear force on the bolt in double shear is 15/2 = 7.5 N. Its x and y components are related ...Bx = 15 cos 30 = 13 N, and By = 15 sin30 = 7.5 N. Now if you use vector addition of Bx and By, you get the resultant force on the bolt is sq rt (13^2 + 7.5^2) = 15 N; thus shear on bolt in double shear is 15/2 = 7.5 N...the same answer.
NEXT EXAMPLE
Help coming later. (see response below)
 
Last edited:
  • #13
PhanthomJay
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Hi Jay

NEXT EXAMPLE
We have a sign hanging outside of a shop. The sign weighs 50kg and is supported by a rigid diagonal bar and a rigid horizontal bar (the sign is attached at the end of the horizontal bar).

The internal angle of the bar to the wall is 600 and 300 at the horizontal bar end. The tension in the vertical bar is 980N

sine (30 degrees) = (490 N ) / (Ftens) = Ftens = (490 N) / [ sine 30 (degrees) ] = 980 N.
The force in the diagonal is, yes, 980 N, but it is not a tension force it is a compression force (the member gets squished, not extended).
My question is: What are the shear, stress/strain and torque loads acting on the bolts attaching the bars to the wall? (By stress/strain I mean the force trying to elongate the bolts by stretching them).
Don't confuse the terminology...the only force on the bolts will be a shear force acting in a plane along the bolts circular cross section..that is the definition of a shear force...there is no tension on the bolts in this example...nor torque, under the applied load...and forget that incorrect stress/strain terminology...and the bolts do not elongate under the applied load.
To answer my own question, I think:

SHEAR LOAD (which acts evenly on both the bars bolts) = Is simply the weight of the sign + the weight of the horizontal bar + the weight of the diagonal bar?. Divided by the number of bolts. (And als converting the weights to Newtons)
there is no shear on either bar...they are 2 force members and can't take shear...they only take axial compression or tension...no shear...no bending...a 2-force member is a member where loads are applied to each end but not in between the ends....this is a truss.
THE HORIZONTAL BAR
- Stress/Strain = Horizontal component of diagonal bar which either = 0 or because it is pushing into the wall due to the effect of the diagonal bar = 0
- Torque = 0, because the diagonal bar holds the vertical force, or at the least, it is very minimal.
you need to study trusses...the horizontal bar is in axial tension only..draw a free body diagram of the joint where the members and sign meet.
THE DIAGONAL BAR
- Stress/ Strain = Tension in the diagonal bar? But if the torque of the diagonal bar is greater than tention force we should just consider the torque / bending force?

NOTE: For stress/strain; we are only interested in either the torque or tension, which ever is largest, and we don't add them together right?

Willy
Please digest earlier responses and post again.
 

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