Mechanics of materials: Statically Indeterminate member

[fayan77]

Homework Statement

Find reactions at ends and deflection at C

Homework Equations

R_A + R_E = 115kN
$\delta$_Total = $\sum$ (P_iL_i)/(A_iE_i)

The Attempt at a Solution

R_A + R_E = 115kN
Then apply superposition: (Get rid of R_E) divide whole member into 4 elements and add.
$\delta$_Total = $\sum$ (P_iL_i)/(A_iE_i)
$\delta$_Total = $\delta$_RE for steel and another for aluminum ...Solve for R_E
Then 115kN-R_E=R_A

Deflection at C
Assuming the above is correct

$\delta$_C = (R_AL)/AE

 

[Chestermiller]

Homework Statement

View attachment 220492 Find reactions at ends and deflection at C

Homework Equations

R_A + R_E = 115kN
$\delta$_Total = $\sum$ (P_iL_i)/(A_iE_i)

The Attempt at a Solution

R_A + R_E = 115kN
Then apply superposition: (Get rid of R_E) divide whole member into 4 elements and add.
$\delta$_Total = $\sum$ (P_iL_i)/(A_iE_i)
$\delta$_Total = $\delta$_RE for steel and another for aluminum ...Solve for R_E
Then 115kN-R_E=R_A

Deflection at C
Assuming the above is correct

$\delta$_C = (R_AL)/AE

Much of this looks correct, but I'm not sure if all of it is. Can you provide the detailed calculations? I would have set this analysis up a little differently.

 

[fayan77]

when dividing element into 4 section I mean DE, CD, BC, AB (assuming R_E is not there)
$\delta$_DE = 0 because there is no force affecting that area
$\delta$_CD = [(40000)(.1)]/[($\pi$/4)(.03)²(E)]
$\delta$_BC = [(40000)(.12)]/[($\pi$/4)(.04)²(E)]
$\delta$_AB = [(115000)(.18)]/[($\pi$/4)(.04)²(E)]
After adding the deltas = $\delta$_Total, I equate it to the elongation caused by R_E...
$\delta$_Total = $\delta$_RE = [R_E(.2)]/[($\pi$/4)(.03)²(E)] + [R_E(.3)]/[($\pi$/4)(.04)²(E)]
I solve for R_E = 40kN Therefore R_A = 75kN
Assuming that is correct I then
$\delta$_C = [(75000)(.3)]/[($\pi$/4)(.04)²(E)]

 

[Chestermiller]

I'm not able to follow what you did, but I don't think this is correct. Here is how I would have done it.

Let $R_A$ be the tension in the bar at A. Then to the left of P1, the tension is $R_A$. In the section to the right of $P_1$, but to the left of $P_2$, the tension is $R_A-P_1$. And, in the section to the right of $P_2$, the tension is $R_A-P_1-P_2$. So, the changes in length of the various sections are:
$$\delta_{AB}=\frac{R_AL_{AB}}{E_SA_S}$$
$$\delta_{BC}=\frac{(R_A-P_1)L_{BC}}{E_SA_S}$$
$$\delta_{CD}=\frac{(R_A-P_1)L_{CD}}{E_BA_B}$$
$$\delta_{DE}=\frac{(R_A-P_1-P_2)L_{DE}}{E_BA_B}$$
The sum of these length changes must be equal to zero, since the total length of the bar is constrained. So,
$$\frac{R_AL_{AB}}{E_SA_S}+\frac{(R_A-P_1)L_{BC}}{E_SA_S}+\frac{(R_A-P_1)L_{CD}}{E_BA_B}+\frac{(R_A-P_1-P_2)L_{DE}}{E_BA_B}=0$$
or equivalently
$$R_A\left[\frac{L_{AB}}{E_SA_S}+\frac{L_{BC}}{E_SA_S}+\frac{L_{CD}}{E_BA_B}+\frac{L_{DE}}{E_BA_B}\right]=\frac{P_1L_{BC}}{E_SA_S}+\frac{P_1L_{CD}}{E_BA_B}+\frac{(P_1+P_2)L_{DE}}{E_BA_B}$$
This is solved for $R_A$

 

[fayan77]

I solved it for RE which would be,

$R_E\left[\frac{L_{AB}}{E_SA_S}+\frac{L_{BC}}{E_SA_S}+\frac{L_{CD}}{E_BA_B}+\frac{L_{DE}}{E_BA_B}\right]=\frac{P_2L_{BC}}{E_SA_S}+\frac{P_2L_{CD}}{E_BA_B}+\frac{(P_1+P_2)L_{AB}}{E_SA_S}$

This is what I got when solving for RE

 

[Chestermiller]

OK. But, you know that RE is compressive (negative tensile) and RA is tensile, correct?

Your result for $\delta_C$ is not correct. The tensile stress is equal to RA only up to the location of P1. Between there and the location of P2, the tension is essentially zero. So you would only be using 180 mm in the calculation of RA, not 300 mm

 

[fayan77]

Yes, I got it my set up is correct it was just computational error, this was driving me nuts. Anyways, $\delta_B$ I'm having trouble with. Is it acceptable to draw a free body diagram of only that element and have Ra pointing to the left, and P₁ to the right this divides it into 2 elements

$\delta_B$ = [R_AL_AB] / [AE] or [(R_A-75)L_AB] / [AE]

 

[Chestermiller]

Yes, I got it my set up is correct it was just computational error, this was driving me nuts. Anyways, $\delta_B$ I'm having trouble with. Is it acceptable to draw a free body diagram of only that element and have Ra pointing to the left, and P₁ to the right this divides it into 2 elements

$\delta_B$ = [R_AL_AB] / [AE] or [(R_A-75)L_AB] / [AE]

What you do to get this right is first to determine the tension (or compression) in each of the 4 segments. Then you calculate how much each segment stretches (or compresses).

What is the tension in segment AB? How much does this segment increase in length?
What is the tension in segment BC? How much does this segment increase in length?
What is the tension in segment CD? How much does this segment increase in length?
What is the tension in segment DE? How much does this segment increase (or decrease) in length?

 

[fayan77]

I thought I could just consider element AC

I understand that $\delta_C$ = $\delta$_AB + $\delta$_BC

But I do not know what forces are acting on each element

 

[Chestermiller]

I thought I could just consider element AC

I understand that $\delta_C$ = $\delta$_AB + $\delta$_BC

But I do not know what forces are acting on each element

Are you aware that the tensions in the 4 segments are not equal? There is a jump change in tension on each side of the concentrated loads. To get the tension on a segment, you cut the segment in the middle and do a force balance on the entire part of the bar between either end and the location where the cut was made. If you do this to get the tension in segment AB, you find that it is 75 kN. Do you see how I got this value? Can you do the same kind of force balance to get the tensions in the other segments?

 

[fayan77]

Do you get 75kN and section AB because RA=75kN? so there is no tension force in Section AB? There is a 40kN force in CD? And no force in section DE?

 

[Chestermiller]

Do you get 75kN and section AB because RA=75kN? so there is no tension force in Section AB? There is a 40kN force in CD? And no force in section DE?

No. But, you're on the right track. Your tension of 75kN in AB is correct, and your tension force of 0 in BC (I think you meant BC in your post, not AB) is also correct. But the tension in CD is likewise 0, and the tension in DE is -40 kN (i.e., section DE is under compression). So, in this particular (and peculiar) problem, section AB stretches, and section DE compresses by an equal amount. Sections BC and CD do not change in length.

 

[michael page]

I feel like this one is in the same category. I have only the data shown, no E, no d. I need to find the P1, P2 loads purely from stiffness considerations. Obviously P1 + P2 = 10000 but what outside equation or field of knowledge can I use to get a second equation from which I can help to solve this first equation?
Thanks for your help,
Michael

 

[Chestermiller]

Please start this in a new thread, and please provide the exact statement of the problem,...and please use the homework template.

 

[michael page]

Thanks, Chestermiller,
https://www.physicsforums.com/threads/pin-in-double-shear-mechanics-of-materials.943188/
This is the new link. Any advice is appreciated.
Michael