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Determine bending moment and bending stress

  1. May 7, 2012 #1
    1. The problem statement, all variables and given/known data
    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 top and a plate of thickness 2t on the bottom 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 ρsteel =7850 kg/m3, ρtimber = 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).

    Question!!: Before the steel plates are connected (ie based on the timber beam only) determine the maximum
    bending moment and bending stress in the beam due to self weight only.

    2. Relevant equations
    h (mm)=300 b (mm)=190 t (mm)=3 L (m)=3.6

    3. The attempt at a solution
    sorry,i dont know how to do this question? normally, we will get bending moment at somewhere, then draw the stress distribution across the section.
    : ( i dont need calculations, just give me some ideas, thank you!
     
    Last edited: May 7, 2012
  2. jcsd
  3. May 8, 2012 #2

    PhanthomJay

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    Re: determine bending momnet and bending stress

    You are just asking about the wood timber max moment and stresses without the steel, under its selfweight. Calculate the weight of the wood beam using the given density and dimensions, and apply it as a uniformly distributed load. Then calculate the maximum moment in the simply supported beam, and from that, the max stress at that section. Please show your work.
     
  4. May 10, 2012 #3
    Re: determine bending momnet and bending stress

    first, the volume of the beam is .19*.19*.3=10.83E-3;the mass is 900*10.83E-3=9.747kg;the weight is 9.747*9.81=95.62N;and load is 95.62N/m;
    so the max bending moment is in the mid, which is 95.62*(.19/2)=9.08N*m, isn't it?
     
  5. May 10, 2012 #4
    i dun know how to get the bending stress...:((
    is it = (M*y)/Ix? M is the bending moment, y is the distance from the central axis,Ix is the second moment area? if so,
    what is y in this question?
     
  6. May 10, 2012 #5
    oh wait now i think i get it;
    the volume is .3*.19*3.6=.2052 m^3, then the mass is 184.68 kg,and weight is 1811.7N. Therefore, the max bending moment is 1811.7*1.8( which is in the middle) =3261.08 Nm=3261060 Nmm. and the bending is (using the eqn f=My/Ix)= 3261060*(300/2)/(3600*300^3/12)=0.06 MPa..
    but would it be too small??
     
  7. May 10, 2012 #6

    PhanthomJay

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    Yes, that is the total weight of the beam. Since the weight is uniformly distributed along the entire length of the beam, the uniformly distributed load , w, is then w= 1811.7/3.6 = 503.25 N/m.
    No, you are not calculating the moment correctly. If you are allowed to use beam tables, the max moment at the center is wL^2/8, where w is the uniformly distributed load of 503.25 N/m. If you are not supposed to use beam tables, you will have to calcualte support reactions and then find the max moment at the center using a free body diagram of the left or right half of the beam, to achieve the same cookbook result.
    Correct your value for M. Your y value is correct. Your calculation for Ix is not correct. Ix is a property of the beam's cross sectional area. Ix = bh^3/12.
     
  8. May 10, 2012 #7
    morning jay; i did my question neatly again;
    v=0.3*0.19*3.6=0.2052 m^3
    m=pV=0.2052*900=184.68 Kg
    W=mg=184.68*9.81=1811.7 N;
    since the weight is uniformly distributed along the entire length, the uniformly distributed load, w, is then w= 1811.7/3.6=503.25 N/m
    Max moment=w*L^2/8=503.25*3.6^2/8=815.27Nm=815269.86 Nmm
    So M=815269.89 Nmm, y=300/2=150 mm, Ix=190*300^3/12=427.5E6;
    therefore, f=(815269.89 *150)/427.5E6=0.2861 MPa
    but, i think f is too small, and i've checked my calcs twice, nth wrong...
    what do u think?
     
  9. May 10, 2012 #8

    PhanthomJay

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    math looks OK i guess, but since i live in the US where metric units are never used in structural engineering, I don't have a good feel for the numbers, nor the math conversions with too many zero's .But the stress value should be small because you have a hefty timber there (we call it an 8 X 12) which can take a sizeable uniform load, and it is only being loaded by its self (dead) weight, which is small compared to the load it can handle safely.
     
  10. May 11, 2012 #9
    hi jay, thanks your help and your patience. the answer i got is very similar to my frd's.
    now comes the 2nd part of this question.
    the question is b)
    The steel plates are then added (this increases the self weight). A point load, P, is applied at midspan. What value of P (in conjunction with self weight) will induce the critical bending stress(10 MPa (for the timber) or 275 MPa (for the steel))?
     
  11. May 11, 2012 #10
    here is my work; would u check if i got it right or not? thanks : ), the method i used is transforming the timber into a I section steel;
    1) find the ration of E-values of these 2 materials, which is
    12500/200000=1/16;
    2) find the width of the transformed I section steel, which is the width of timber * the ratio of E-values, which is
    190*1/16=11.875 mm;
    3) find the centroid (only care about the y) of the I section; which is y=
    (3*190*307.5+6*190*3+11.875*300*156)/(3*190+6*190+11.875*300)=139.3 mm;
    4) find the 2nd moment of area, which is
    (190*6^3/12+6*190*(y-3)^2)+(190*3^3/12+190*3*(139.3-307.5)^2)+(11.875*300^3/12+11.875*300*(139.3-156)^2)=65.02E4 mm^4;
     
  12. May 11, 2012 #11
    5) set up the equations for the bending stress for f1-steel(top),f2-steel(bottom),f3-timber(top),f4-timber(bottom);
    f1=M*y/Ix=M*(309-139.3)/(65.02E6)=2.61E-6*M;
    f2=M*(139.3-0)/(65.02E6)=2.14E-6*M;
    f3=M*(306-139.3)/(65.02E6)=160.24E-9*M;
    f4=M*(139.3-6).(65.02E6)=128.13E-9*M;
    since max bending stress for steel is 275 MPa, and 10 MPa for timber;
    therefor the max bending moment for steel(top)=105.36E6, steel(bottom)=128.5E6, timber(top)=62.41E6,timber(bottom)=78.05E6;
    and we have to choose the min one, so the max bending moment for the whole section is going to be 62.41E6;
     
  13. May 11, 2012 #12
    we need to find p
    and i find the max bending moment is 62.41E6 = wL^2/8+pL/4;
    because steel was added on the timber, w is gonna change, and i got 634.934 N/m for the new w, the distributed load;
    in the equation, w=634.934, L=3.6, after cals,
    P is = 69.34E6 N !!
    am i right, jay?
    is my method correct?
     
  14. May 11, 2012 #13
    sorry ,p=68.20E6 N
     
  15. May 11, 2012 #14

    PhanthomJay

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    I think your method was good, i assume you multipied by n when you had to, in calcualting equivalent area and stress in wood, but the load P seems very high, i don't know maybe check your numbers because SI leaves me cold.
     
  16. May 11, 2012 #15
    yep, i got the same answer with my friend. hopefully, it is correct and i'll get full mark for this assignment.
    thanks very much, jay. u r such a nice guy !!
    have a nice weekend!
     
  17. May 16, 2012 #16
    hi,chris, m doin' this too...

    do you want to check for first question..the Ix here..i think you swapped the b and h coz tim was doin' that with bh^3 and here it's hb^3..
     
  18. May 16, 2012 #17
    ix is bh^3/12,yah? b is the width,h is the height
     
    Last edited: May 16, 2012
  19. May 17, 2012 #18

    PhanthomJay

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    Yes, b is 190mm per problem wording that the minor axis is vertically oriented.

    But please check your math again, 60 million newtons is ridiculously high, maybe you used 3.6 mm as length instead of 3600 mm? Gotta be careful using SI with so many darn zeros before or after the decimal point, which defies even the best of calculators.:yuck:
     
  20. May 17, 2012 #19
    thanks jay, i knew it, i did use 3.6 instead of 3600 mm, and i ve changed p=68KN , thats seems reasonable.
     
  21. Oct 18, 2013 #20
    Hello mates,

    How do we balance the cargo load on the cargo carrier steel frame with four connecting points.
    the cargo load is 100T
    Dimensions of the frame for instance length 21 meters, height 2 meters and width is 3 meters.
    connection points on the frame are in x-direction is at 0m, 12m, 16m, 21m. the center of gravity of the cargo and the frames are different.
    cog of the cargo is (14m, 1.5m, 2m)
    My question is what is the force which we ll get at the four connecting points from the cargo.

    Thanks in advance

    Cheers
    Karthik
     
    Last edited: Oct 18, 2013
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