Download Bending Stress p375

Bending Stress p375
Bending-Stress Page 1
Bending-Stress Page 2
Bending Stress example
Reactions:
Force = 0.193*9.81 = 1.8933 N
Measure the torque of the hinge...
260mm
Spring: Moment (Torque)
260*2.943 = 765.18 Nmm
(0.765 Nm)
Predict the load on the beam...
Bending-Stress Page 3
300g = 0.3*9.81
= 2.943 N
765 Nmm
Reaction Force = 765 / 170 = 4.5 N
Force at centre = 4.5 * 2 = 9 N
In kg = 9/9.81 = 0.9174 kg
Hinge = 386g
Extra load = 917-386 = 531 g
Experiment: 18*50 = 900 g (Missed!)
Double Check:
Mass = 386+18*50 = 1,286 g
Force = 1.286*9.81 = 12.6157 N
Bending 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 g
Force = 0.786*9.81 = 7.711 N
Bending Moment = 0.5*7.711*485/2 = 935 Nmm
765.18 Nmm
Could it be friction?
Masses were not in the centre.
Bending-Stress Page 4
Q2 Simply Supported Beam
Wednesday, 4 May 2016
6:48 PM
Q2: If distance a=3.7m, b=2.1m and force
W=5.1kN, find the maximum bending
moment. Sag=(+), Hogg=(-)
If we have not done MEM30005A Forces,
there is a shortcut way to find the max
bending moment M.
From <http://www.learneasy.info/MDME/iTester/tests/10306
_Bending_Moment/images/beam_simple.jpg>
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)
Solution: a=3.7m, b=2.1m and force
W=5.1kN, L=a+b = 3.7+2.1 = 5.8 m
Max BM = Wab/L
= 5100*3700*2100/5800
= 6.8322E6 Nmm
From <http://www.learneasy.info/MDME/modules/FEA/Beams.htm>
Q2: Distance a=5m, b=1.6m: Dimensions
c=67mm, d=190mm: Force W=28kN.
Find maximum bending stress.
Solution: a=5m, b=1.6m and force
W=28kN, L=a+b = 5+1.6 = 6.6 m
Max M = Wab/L
= 28000*5000*1600/6600
= 3.3939E7 Nmm
Now find Bending Stress:
 = My/I
We need to find I;
I = bh3/12
= 67*190^3 / 12
= 3.8296E7 mm4
Bending-Stress Page 5
From <http://www.learneasy.info/MDME/iTester/tests/10307
_Bending_Stress/images/beam_simple_rectangular.jpg>
y = 190/2 = 95 mm (half depth)
 = My/I
= 3.3939E7 * 95 / 3.8296E7
= 84.1917 MPa
Bending-Stress Page 6
Cantilever
Wednesday, 4 May 2016
7:09 PM
Q9: Force a=7.3kN and force b=8.9kN.
Find the maximum bending moment.
Sag=(+), Hogg=(-)
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.
From <http://www.learneasy.info/MDME/iTester/tests/10306
_Bending_Moment/images/cant01.jpg>
Solution:
M = WL
= a*2 + b*6
= 7300*2000+8900*6000
= 68000000 Nmm
From <http://www.learneasy.info/MDME/modules/FEA/Beams.htm>
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
Bending-Stress Page 7
Now find Bending Stress:
 = My/I
We 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
Bending-Stress Page 8
From <http://www.learneasy.info/MDME/iTester/tests/10307_Bending_Stress/images/cant01
_stress.jpg>
Bicep question
Wednesday, 4 May 2016
7:26 PM
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
Bending-Stress Page 9
Bending Stress: Shaft Diameter
Friday, 13 May 2011
11:11 AM
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 = d4/64
y = d/2
M = a*b = 5900*110 = 649000 Nmm
 = My/I
= M * d/(2*d4/64)
= 64*M /(2*d3)
= 32*M/(pi*d3)
= 6610660 / d3
So now find diameter;
d = (6.61066E6 / )(1/3)
= (6.61066E6/270)^(1/3) = 29.038 mm
Bending-Stress Page 10
Worked example.
Tuesday, 2 August 2011
5:39 PM
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.
Process:
1. Find Centroid and Ixx
2. Find volume and mass
3. Find M
4. Find Stress
Bending-Stress Page 11
Worked ex. 2
Tuesday, 2 August 2011
5:39 PM
Spec gravity of 0.7.
Beam span is 8m
1. Find Centroid and Ixx
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
Elemt
Total
mm2
mm
mm3
mm
mm4
mm4
mm4
Area
y
Ay
d
Ad2
Ic
I
1
15000
275 4125000
75 84375000
2
15000
125 1875000
75 84375000 78125000 162500000
30000
6000000
Ixx = 250000000
yc =
200
Bending-Stress Page 12
3125000
87500000
Worked ex. 3
Tuesday, 2 August 2011
5:39 PM
Spec gravity of 0.7.
Beam span is 8m
Yc = 200mm
I = 250E6 mm4
Find volume
V=A*L
= 30E3*8E3
= 240E6 mm3
Find Mass:
m=*V
= 700*240E6/1E9
= 168 kg
Reactions = 9.81*168/2
= 824.04 N
Distributed load = 9.81*
168/8 = 206.01 N/m
824.04 N
824.04 N
BM max = 0.5*4*824 = 1648 Nm
Bending-Stress Page 13
Worked ex. 4
Tuesday, 2 August 2011
5:39 PM
Spec gravity of 0.7.
Beam span is 8m
Yc = 200mm
I = 250E6 mm4
BM 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 pine
F5 F7 structural pine
F14 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
Bending-Stress Page 14