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Connorm1
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Homework Statement
A beam of rectangular cross section 200 mm deep and 100 mm wide. If the beam is 3m long, simply supported at either end. And carries point loads R1 to far left, R2 to far right, 1m to left 5kN load, 2m to left 10kN load.
(a) Calculate the maximum bending moment
(b) Calculate the maximum stress in the beam
(c) At the point of maximum stress sketch a graph of the stress distribution through the thickness of the beam, indicating which are tensile and compressive stresses.
(d) Determine the dimensions of the cross section which will minimise the maximum stress value if: • the cross sectional area of the beam can be increased by 20% • the beam section is to remain a solid rectangle • neither the breadth or depth of the beam section can be reduced below their original dimensions. Show the dimensions of the proposed beam cross section with the aid of a sketch.
(e) Determine the percentage reduction of the maximum stress value when the new cross section is used.
Homework Equations
M/I=σ/y
Ixx=bd^3/12
The Attempt at a Solution
a) Firstly i started by finding the bending moment. so taking moments at Point A (R1) i have 3R2=(5x1)+(2x10). R2=25/3kN. As upward forces must equal downwards forces. 10+5=R1+25/3 so R1=20/3. I then moved on to find the max bending moment. so at M0 (0meters from left) =0kNm. At M1 at 1m from left = (20/3*1) = 20/3kNm . At M2 at 2m from left = (20/3*2)-(5*1)= (25/3) = 8.333 kNm. At M3 at 3m from left = (20/3*3)-(5*2)-(10*1)=0kNm. Therefore maximum bending moment is 8.333kNm.
b) using the formula i rearrange to make maximum stress the subject. So σ=My/I. Where M=8.333kNm (8.333*10^3) y= d/2 = 0.1m (100mm) and I we need to find using bd^3/12. so I(NA)=0.1*0.2^3/12 = 6.667*10^-5 m^4. Plugging our values for M, y & I we get σ=(8.333*10^3*0.1) / (6.667*10^-5) = 12.5 *10^6 or 12.5 MPa (or Nm^-2 unsure on SI).
c) I have to admit i don't really know what diagram this is referring too and have no idea of what to draw nor figures to use.
d) So my first assumption was that to minimise the stress value in a beam, we would need to increase the depth (again i don't know 100% why but i could have trial and error both width and depth). As we can only increase it by 20% for the cross sectional area. I'd make this 240mm*100mm. The maximum stress now would be using the same formula. except now I=0.1*0.24^3/12 so I(NA)= 1.152*10^-4 and new y=d/2 0.24/2=0.12m (120mm). So σmax = (8.333*10^3*0.12) / (1.152*10^-4) = 6.58 * 10^6 or 6.58MPa (Nm^-2).
e) I'm assuming to show the percentage reduction of maximum stress across the new cross section used i just take my previously calculated max stresses and do 6.58 *10^6 /12.5 *10^6 =52.6%-100% which is a 47.4% reduction.
Am i heading in the right directions. Bare in mind it's hard to type it out and work it on here so if I've made a mistake on figures i apologise!
Thanks for your help.
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