# Homework Help: Structural Mechanics. bending moment and bending stress

1. May 20, 2014

### jacko12

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

A timber beam, with rectangular cross section (h × b) is reinforced with additional full width (b) steel plates. There is a plate of thickness t securely connected on the bottom and a plate of thickness 2t on the top of the timber to ensure composite action in bending. The section is then used as a simply supported beam of length L. The beam is oriented such that the minor principal axis is vertical. It should be taken that Psteel = 7850 kg/m3, Ptimber = 900 kg/m3, Etimber = 12500 MPa and Esteel = 200000 MPa, and that both materials exhibit linear elastic behaviour. The design engineer needs to ensure that the normal bending stress (tension or compression) does not exceed 10 MPa (for the timber) or 275 MPa (for the steel). The beam experiences a vertically downwards point load of magnitude P acting mimdspan.
It can be assumed that the self-weight of the steel-timber beam is negligible compared to the value of P.

a)Before the steel plates are connected (ie based on the timber beam only) what is the maximum value of P that can be applied before the timber reaches its maximum stress.
b)For the value of P calculated in (a), what is the maximum deflection of the beam.
c)The steel plates are then added. What new value of P will induce the critical bending stress (10 MPa (for the timber) or 275 MPa (for the steel))? Draw the stress and strain distributions (values required) for both the steel and the timber on the critical cross-section when this occurs.

2. Relevant equations

h(mm)=300
b(mm)=190
t(mm)=3
L(m)= 3.6

3. The attempt at a solution

FOR question A**:
Found the following variables so far.
V= 0.2052m^3
m= 184.68 kg
W= 1881.7N
Distributed weight= 503.25N/m
Max moment= 815269.86Nm = 815268.89Nmm
then f= 0.281MPa
How do I find the maximum value of P from the stress value of 0.291MPa. and then the corresponding deflection?

2. May 20, 2014

### paisiello2

How did you get 0.281MPa and why is it relevant when the maximum allowable is given as 10MPa?

3. May 20, 2014

### jacko12

by using the equation f= My/Ix
Yeah Iam not sure how I approach the question so I worked out serval unknowns. I thought the 10Mpa was relevant for question C.. As the steel plates are not added yet?

4. May 20, 2014

### SteamKing

Staff Emeritus
1. The maximum stress of 10 MPa is for the timber, so it applies in both conditions, before and after the steel plates are attached. IMO, you don't want a situation where the timber has failed and all of the load is being supported by the steel plates.

2. The problem specifically states that the mass of the steel-timber combination beam is to be considered negligible w.r.t. the central load P. It's not clear why you have nevertheless calculated a distributed load of 503.25 N/m. (In other words, if the weight of the steel-timber beam is negligible, then the weight of the timber beam alone should be negligible, as well.)

3. In any event, a maximum central load P has not been determined. (Hint: the value of P is the answer the problem seeks for parts A. and C.)

5. May 21, 2014

### jacko12

mm im still bit confused will have to sit down again and work on question.

6. May 21, 2014

### jacko12

so what formula will be used to find the Load? thats whats stunted me

7. May 21, 2014

### paisiello2

What formula did you use to find the bending moment M?

8. May 21, 2014

### jacko12

thanks

Last edited: May 21, 2014
9. May 21, 2014

### jacko12

hhhh

Last edited: May 21, 2014
10. May 21, 2014

### jacko12

oh yes. That makes total sense I just skip the step of self weight to find the max moment then to find the stress. Instead I sub in the given max stress of 10MPa in f= My/Ix equation to find the max moment.

11. May 21, 2014

### jacko12

Only question is now i think I found formula to use. stress (f) = P/A to find the load. what is the A area? of just the timber obviously but the cross section area of just L*w?

12. May 21, 2014

### SteamKing

Staff Emeritus
You are overlooking several important areas in the analysis of the beam to determine the max. central load P:
1. The load P is going to produce a certain bending moment in the beam, given that the beam is simply supported at the ends. What is the maximum bending moment of the beam with a central load P?
2. What is the max. bending stress in the beam? How do you calculate it?
3. Is stress = P/A applicable for calculating bending stress? Or is it applicable to some other loading?

This is for the easy part of the problem. You still have to consider the effect of attaching the steel plates to the timber beam.

This problem has many different elements which should test how well a student has learned basic strength of materials. IMO, you have much work ahead of you to master this material. I hope a final exam is not around the corner for this course.

13. May 21, 2014

### paisiello2

That's wrong. Remember you are trying to find the bending moment M to stick into your first equation.

14. May 21, 2014

### jacko12

can i just use the Mmax= PL/4? Also for question b/c im not sure what formulas to use cause all the deflection involve self weight.

15. May 21, 2014

### jacko12

mm okay i will have to have good look. but I mean i can do these questions. Im just not sure how to approach it cause im use to using the self weight formula in the process. Havent done a question like this. Hence why I posted it for help not to get another question from my own questions

16. May 22, 2014

### paisiello2

That's the right formula!

See if you can find the one for deflection.

17. May 22, 2014

### jacko12

Okay I think I have figured it all out. So part A. Use Ix=bh^3/12 with corresponding values. The use M=f*Ix/y to find moment then M=PL/4 to find load. Then part B. to find the deflection use D=PL^3/48EI.
Then part c. Was bit tricky not 100% frist of all found the ration of E values of the 2 materials. The transfer timber section in an I beam steel. So width= original width*ration. then find the centeroid of y. followed by 2nd moment of Inertia. Then use Mx=fIx/y formula for each of steel (top and bottom) and timber (top and bottom). Pick the smallest moment out of the 4 which i found was timber top. The input this moment value into the Mmax= PL/4 to find the new corresponding P value. I got a P value of around 40kN. Please let me know if you thing this method sounds okay. Only thing now is too draw the stress and strain distributions.

18. May 22, 2014

### paisiello2

Your approach looks right to me.