# Homework Help: Maximum bending stress in wooden beam

1. Jan 12, 2012

### mh1985

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

A timber beam will be used to support part of the ceiling of a house. It is
going to be attached to the walls as a simply supported beam, Figure 6.

If three forces are applied to it and considering the weight of the beam as a
uniformly distributed load of magnitude w = ρAg (which you will need to
calculate and include in the FBD), determine the maximum bending stress

The properties of the wooden beam are ρ = 470
kg/m^3 , σMax = 10 MPa. Can the beam support these loads or will it fail?

2. Relevant equations

σ = y* M / I

3. The attempt at a solution

I'm struggling to get the Moment about the axis, because of the other loads, not sure if I'm supposed to section the beam or not.

470 kg/m^3 x 0.165 m x 0.250 m x 9.81 = 190.191375 N /m

So far I have:
I = 1/12 * bh^3 = 2.1484 * 10^-4 m^4

The reactions at the bases :
I took the distributed load and put it as a resultant

ƩM @A = (RB*5 m) -(10 kN *1 m) -(5.95096 kN*2.5 m) - (10 kN * 4 m)

RB = 64.8774/5 = 12.97548 kN

Now I'm stuck as to what the next step is...

2. Jan 12, 2012

### nvn

mh1985: Hint 1: Would you need to compute bending moment at midspan? And then compute bending stress at midspan?

3. Jan 12, 2012

### mh1985

would that be at x = 2.5 m?

4. Jan 12, 2012

### nvn

Yes.

5. Jan 12, 2012

### mh1985

Not sure if this is right but:

Bending moment at x = 2.5,
(12.98 * 2.5) - (10 * 1.5) + (10 * 1.5) - (12.98 * 2.5) = 0 ??

Again, dont know if I need to do this but I have sectioned the beam and done the bending moment diagrams, which has given me the equation:

12.98x - 0.0951x^2, for x= 2.5 this is 31.855625 kNm and 12.98x - 0.0951x^2 + 10 -10x, for x = 2.5 this is 16.855625 kNm

I don't know if this is the right way to do it, also not sure how to get the bending stress at midspan?

6. Jan 12, 2012

### nvn

mh1985: That currently looks correct. Your equation for bending stress is in post 1.

7. Jan 12, 2012

### mh1985

Thanks.
how do I include the weight of the beam in this though?
Which value for M do I use in the bending stress equation?

8. Jan 12, 2012

### nvn

mh1985: The beam self weight is already included in your moment calculation. You included it in your calculation. Use the moment you computed in post 5.

9. Jan 12, 2012

### mh1985

but aren't there two depending on which side of 2.5 x is? ie. 31.855625 kNm and 18.855625 kNm

EDIT:

I have done tau_max = Mc / I for both:

(31.85 x 10^3 N x 0.125 m)/ 2.1484 x 10^-4 m = 18.53 MPa
(18.856 x 10^3 N x 0.125 m) / 2.1484 x 10^-4m = 10.97 Mpa

So the beam cannot support these loads? I'm not sure I've answered the question properly...

Last edited: Jan 12, 2012
10. Jan 12, 2012

### pongo38

quote but aren't there two depending on which side of 2.5 x is? ie. 31.855625 kNm and 18.855625 kNm end quote

There is obviously an error because the bending moment to the right of centre should be exactly the same as the bending moment to the left. I suggest (1) that there is a central 5 kN load in your diagram that is not the self weight of the beam, and which you ignored when getting the reactions. (2) To lump the distributed load at the centre is Ok for getting reactions but totally wrong when calculating bending moments.

11. Jan 12, 2012

### nvn

mh1985: In post 6, when I said, "That currently looks correct," I was referring to 16.84 kN*m. And I forgot to say, your equation for the 31.86 kN*m in post 5 is not applicable beyond x = 1.0 m; therefore, ignore your second equation in post 5.

I do not know why you mistyped your moment in post 9, when it was correct in post 6. Try again. Also, the stress should not be called tau_max in post 9; it should be called sigma_max.

(1) By the way, the unit symbol for megapascal is spelled MPa, not Mpa. Lowercase p means pico.

(2) Two units (or quantities) multiplied together must be separated by an asterisk, middle dot (·), space, or parentheses. E.g., kN*m, not kNm.

(3) Generally always maintain at least four significant digits throughout all your intermediate calculations, then round only the final answer to three significant digits, unless the first significant digit of the final answer is 1, in which case round the final answer to four significant digits.​

Last edited: Jan 13, 2012
12. Jan 13, 2012

### mh1985

many thanks for pointing those out!

With regard to the moment being different depending on which equation I used, what about when I use the one for 2.5< x <4, shouldn't this give the same value as the equation for 1 < x<2.5 provided I used x = 2.5 ?

I've redone it and come up with this,

σ (max) = Mc / I = [(16.855625 kN*m) * (0.125 m)]/ 2.1484*10^-4m = 9.807 MPa

13. Jan 13, 2012

### nvn

mh1985: Nice work. Your answer is correct. See item 3 in post 11. Somewhere along the way, you may have rounded a number too much (?). If you do not, I think you would end up with 9.80 MPa, whereas you currently ended up with 9.81 MPa.
Yes.

14. Jan 13, 2012

### mh1985

Thanks, well at least I have the right answer.

I think I'm doing something wrong with the bending moment equations in that case because I get something different for each one:

for 1 < x < 2.5
M = 12.98x - 0.0951x^2 - 10(x-1)
x = 2.5, M = 16.86 kN*m

for 2.5 < x < 4
M = 12.98x - 0.0951x^2 - 10(x-1) - 5(x-2.5)
x = 2.5, M = 26.86 kN*m

15. Jan 13, 2012

### nvn

mh1985: You did not check your arithmetic yet. Try computing M a few more times, and check your arithmetic. If you are having trouble with your calculator, try an HP calculator (probably preferably HP 35s), and use RPN (reverse Polish notation).

Last edited: Jan 13, 2012
16. Jan 15, 2012

### mh1985

I have just seen it now, I was expanding the brackets out before, but if I don't do that, on the 2nd equation where we have

5(x-2.5) for x=2.5, this is 5(2.5-2.5) = 0...must've made an error when expanding them out. thanks!

17. Jan 24, 2012

### 060411bt

where have you used ρ=470?