# Homework Help: Stress transformation, shear stress state, Mohr's circle c/work

1. Nov 30, 2013

### Brendanmcg

1. The problem statement, all variables and given/known data

DIAGRAM ATTACHED AT BOTTOM

Q. The following statements are true for an element in plane stress state. (this is 2D)
(1) one of the principle stresses is 40Mpa;
(2) σx= -2τxy; (the algebraic values)
(3) in x'oy' with θ=30°, the two normal stresses σx'=σy'

Determine:
(a)σx,σy,τxy;
(b)the principle stress state, and sketch it on an element;
(c)the maximum in-plane(xoy) shear stress state and sketch it on an element;
(d)the absolute maximum shear stress.

2. Relevant equations

ok i haven't been given a set of equations specifically for these questions because its a course work, but my tutor heavily implied i should use Mohr's circle, ive been trying and i have identified im going to have to obviously use Trig relationships and Algebra but apart from that im totally lost. il insert the transformation equations i think might be useful. But obviously doing this with Mohr's circle the majority aren't really needed.

Pinciple Stresses

σx'= (σx+σy)/2 +(σx-σy)/2 cos2θ+τxysin2θ

τx'y'= -(σx-σy)/2 sin2θ+τxycos2θ

max/min in-plane normal stress:

σ1,2=(σx+σy)/2 +/-√((σx-σy)/2)^2 + τxy^2
the bit thats taking the sq root is obviously R in Mohr's circle

Max in-plane Shear Stress

τmax=R

Max in-Plane shear stress orientation:

tan 2θs= -(σx-σy)/2τxy θs1=θp1-45°

3. The attempt at a solution

Any question i have done like this before all principle stresses have been given, im pretty lost without them... trail an error has led me to believe σy=40Mpa but apart from that i've been drawing Mohr's circles and getting lost in algebra and trig... HELP PLEASE!! if any more info is needed just ask. its due in on tuesday and i've wasted 5days on dead ends... it surely couldn't be as hard as im making it.
1. The problem statement, all variables and given/known data

2. Relevant equations

3. The attempt at a solution

#### Attached Files:

• ###### stress diagram.jpg
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Last edited: Nov 30, 2013
2. Nov 30, 2013

### Staff: Mentor

What is your equation for σy'?
I wouldn't use Mohr's circle. This can be done algebraically. From (2) and (3) you have enough information to express σx and σy in terms of τxy. You can then use (1) to solve for τxy.

Last edited: Dec 1, 2013
3. Dec 2, 2013

### Brendanmcg

i didnt have an equation for σy but...

if σx=-2τxy then τxy= σx/-2

if one of the principle stresses are 40Mpa, if we assume σx is 40Mpa then τxy= -20Mpa

so, knowing the angle and assuming τxy is 40Mpa, using mohrs circle and trig functions it just wasn't looking correct.

i then started using transformation equations and this is what i've came up with..... anyone care to shed any light, if im going wrong???

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4. Dec 2, 2013

### Brendanmcg

p.s. sorry about the pics being side ways... cheers

5. Dec 2, 2013

### Staff: Mentor

I wasn't able to follow what you did on the attachments, but here's what I considered. You would like to express (σxy) and (σxy) in terms of τxy. Then you can substitute these into the equation for the principal stresses to solve for τxy.
xy)=(σ'x+σ'y)
From the condition for the stresses at 30 degrees, you have at 30 degrees:
xy)=2σ'x=(σxy)+(σxy)cos(60)+2τxysin(60)
From this, it follows that
xy)=-2τxytan(60)
You also know that σx=-2τxy
From the last two equations, it follows that
xy)=2τxy(tan(60)-2)
Now I would substitute these results for (σxy) and (σxy) in terms of τxy into the equation for the principal stress and solve for τxy. I would do it for both the plus sign and the minus sign.

Chet

6. Dec 2, 2013

### Brendanmcg

Thanks Chet...

realized my maths was terrible in those attachments... lol corrected my maths but thats just left me with a dead end. eughhhhh!!!

cant tell where you said i implied (σx+σy)=(σx'+σy') came from...?

7. Dec 2, 2013

### Staff: Mentor

Check out the second equation in your first attachment.

Also, we know that the sum of the diagonal elements of the stress tensor is an invariant. For example, the sum of the principal stresses is also equal to (σx+σy).

Were you able to use the results I derived in terms of τxy to substitute into the equation for the principal stresses and obtain τxy (together with the rest of the solution)?

Chet