Bending Stress p375

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Bending Stress example

Reactions:Force = 0.193*9.81 = 1.8933 N

Measure the torque of the hinge...

300g = 0.3*9.81 = 2.943 N

260mm

Spring: Moment (Torque)260*2.943 = 765.18 Nmm(0.765 Nm)

Predict the load on the beam...

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765 Nmm

Reaction Force = 765 / 170 = 4.5 NForce at centre = 4.5 * 2 = 9 NIn kg = 9/9.81 = 0.9174 kgHinge = 386gExtra load = 917-386 = 531 g

Experiment: 18*50 = 900 g (Missed!)

Double Check:Mass = 386+18*50 = 1,286 gForce = 1.286*9.81 = 12.6157 NBending Moment = 0.5*12.6157*170 = 1072 Nmm

Could it be friction?Masses were not in the centre.

Double Check 2:Mass = 386+8*50 = 786 gForce = 0.786*9.81 = 7.711 NBending Moment = 0.5*7.711*485/2 = 935 Nmm765.18 Nmm

Could it be friction?Masses were not in the centre.

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From <http://www.learneasy.info/MDME/iTester/tests/10306_Bending_Moment/images/beam_simple.jpg>

Solution: a=3.7m, b=2.1m and force W=5.1kN, L=a+b = 3.7+2.1 = 5.8 m

= 5100*3700*2100/5800 = 6.8322E6 Nmm

Max BM = Wab/L

Q2: If distance a=3.7m, b=2.1m and force W=5.1kN, find the maximum bending moment. Sag=(+), Hogg=(-)

From <http://www.learneasy.info/MDME/modules/FEA/Beams.htm>

If we have not done MEM30005A Forces, there is a shortcut way to find the max bending moment M.

Engineers often use beam bending tables.In this question the same TYPE of loading must be found in the bending table.

Looking for a simply supported beam where the load is offset from the centre. (Last row of table)

Q2: Distance a=5m, b=1.6m: Dimensions c=67mm, d=190mm: Force W=28kN. Find maximum bending stress.

From <http://www.learneasy.info/MDME/iTester/tests/10307_Bending_Stress/images/beam_simple_rectangular.jpg>

Solution: a=5m, b=1.6m and force W=28kN, L=a+b = 5+1.6 = 6.6 m

= 28000*5000*1600/6600 = 3.3939E7 Nmm

Max M = Wab/L

Now find Bending Stress:

= My/IWe need to find I;I = bh3/12 = 67*190^3 / 12 = 3.8296E7 mm4

Q2 Simply Supported BeamWednesday, 4 May 2016 6:48 PM

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y = 190/2 = 95 mm (half depth)

= My/I = 3.3939E7 * 95 / 3.8296E7 = 84.1917 MPa

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Solution: M = WL = a*2 + b*6 = 7300*2000+8900*6000 = 68000000 Nmm

From <http://www.learneasy.info/MDME/modules/FEA/Beams.htm>

Looking for a cantilevered beam where there are 2 loads.Since the max BM occurs at the wall for BOTH forces, then we can add them up separately.

Q7: Force a=5kN and force b=3.3kN. The beam is c=106mm wide by d=223mm deep. Find the highest stress..

Solution: M = 5000*2000+3300*6000 = 29800000 Nmm

Now find Bending Stress:

= My/I

From <http://www.learneasy.info/MDME/iTester/tests/10306_Bending_Moment/images/cant01.jpg>

Q9: Force a=7.3kN and force b=8.9kN. Find the maximum bending moment. Sag=(+), Hogg=(-)

CantileverWednesday, 4 May 2016 7:09 PM

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Now find Bending Stress:

= My/IWe need to find I;I = bh3/12 = 106*223^3 / 12 = 9.7958E7 mm4

y = 223/2 = 111.5 mm (half depth)

= My/I = 29800000 * 111.5 / 9.7958E7 = 33.9196 MPa

From <http://www.learneasy.info/MDME/iTester/tests/10307_Bending_Stress/images/cant01_stress.jpg>

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Q12: Freddy's forearm is 378mm long, his bicep attaches 45.9mm from the elbow. Assuming a bone diameter of 34mm, what stress would 25kg cause?

From <http://www.learneasy.info/MDME/iTester/tests/10307_Bending_Stress/images/bicep_stress.jpg>

This is really just a simply supported beam - upside-down!

Max M = Wab/L

Bicep questionWednesday, 4 May 2016 7:26 PM

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A classic problem is to find the diameter of a shaft. The problem is, we can’t find the stress or even Ixx until we know the diameter.A typical "engineeringy" way to do this is to guess a shaft size, calculate it, then adjust size, recalculate, etc.

But a simple bending problem can be solved by algebra.

Solution: Set up the problem based on diameter, d.(Do it in Nmm)

I = d4/64y = d/2M = a*b = 5900*110 = 649000 Nmm

= M * d/(2*d4/64)

= 64*M /(2*d3)= 32*M/(pi*d3) = 6610660 / d3

= My/I

So now find diameter;

d = (6.61066E6 / )(1/3)

= (6.61066E6/270)^(1/3) = 29.038 mm

Bending Stress: Shaft DiameterFriday, 13 May 2011 11:11 AM

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Calculate the maximum tensile stress for the beam shown below. It is a simply supported beam made of wood that has a spec gravity of 0.7. Beam span is 8m, find the stress due to it's own weight.

Find Centroid and Ixx1.Find volume and mass2.Find M3.Find Stress4.

Process:

Worked example.Tuesday, 2 August 2011 5:39 PM

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Spec gravity of 0.7. Beam span is 8m

Find Centroid and Ixx1.

Centroid:

Yc = (Ay) / (A) = 6E6/3E4 = 200 mm

Parallel Axis Theorem: (p372)(The I effect of each element = It's own Ic + Area * d2 from neutral plane.)

I = Ic + Ad2

mm2 mm mm3 mm mm4 mm4 mm4

Elemt Area y Ay d Ad2 Ic I

1 15000 275 4125000 75 84375000 3125000 87500000

2 15000 125 1875000 75 84375000 78125000 162500000

Total 30000 6000000 Ixx = 250000000

yc = 200

Worked ex. 2Tuesday, 2 August 2011 5:39 PM

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Spec gravity of 0.7. Beam span is 8m Yc = 200mmI = 250E6 mm4

Find volumeV=A*L= 30E3*8E3 = 240E6 mm3

Find Mass:

m=*V= 700*240E6/1E9 = 168 kgReactions = 9.81*168/2 = 824.04 NDistributed load = 9.81*168/8 = 206.01 N/m

824.04 N 824.04 N

BM max = 0.5*4*824 = 1648 Nm

Worked ex. 3Tuesday, 2 August 2011 5:39 PM

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Spec gravity of 0.7. Beam span is 8m Yc = 200mmI = 250E6 mm4BM max = 1648 Nm= 1648*1000 = 1648E3 Nmm

Find Stress

= My / I = 1648E3 * 200 / 250E6 = 1.3184 MPa

Wood is rated (bending tensile stress - fibre stress) like this...F4 soft pineF5 F7 structural pineF14 F17 hardwood (like eucalypts, not balsa) F4 = 4 MPa (rated fibre stress)

Compare to steel: Steel could handle a stress of...(a) 2 MPa(b) 20 MPa(c) 200 MPa(d) 2 GPa(e) 20 Gpa

Worked ex. 4Tuesday, 2 August 2011 5:39 PM

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