eb9bg moment area example
DESCRIPTION
...TRANSCRIPT
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Moment Area Examples
For the cantilever shown below calculate
vertical deflection at the tip of the cantilever.
E = 200 kN/mm2
I = 200 000 cm4
EI
8 m
EI
2 m
A B C
100 kN
-
Taking Moment of the area about C
DC = [(800x8/2) x 22/3] / EI = 70400/3EI
E =200 kN/mm2 = 200 x 106 kN/m2
I = 20000 cm4 = 200000x10-8 = 2000x10-6 m4
DC = [70400/(3x200x106x2000x10-6)] x 1000
DC = 58.66 mm
Solution
8 m, EIA BC
DC
2 m, EI100 kN
800
8/3 m 2 + 2x8/3 = 22/3 m
-
=+
10m, E, I
2
100 kN
1
D1
10m, E, I
2
V1
1
D1
10V1
20/3
500
5 +10/3 = 25/3
10m, E, I
2
100 kN
1 5m
D1 = [(500 x 5/2) x 25/3] / EI
D1 = 31250/3EI
D1 = [(10V1 x 10/2) x 20/3] / EI
D1 = 1000V1/3EI(Equate deflections)
1000 V1 = 31250
V1 = 31.25 kN V2 = 100 31.2 = 68.8 kN
-
M2 = 500 10V1 = 500 312.5 = 187.5 kN-m
MUnder the Load = 31.25 x 5 = 156.25 kN-m
Or 312.5/2 = 156.25 kN-m
187.5 kN-m
156.25 kN-m
Bending Moment Dia.
10V1=
312.5
20/3 500
5 +10/3 = 25/3
-
For the structure shown below:
Calculate support reactions
Sketch BMD showing values at critical locations
Calculate vertical deflection at C.
=
L, E, IA B
DB
qB
100 kN/m
VB
+
EI = 80000 kN-m2
VB
DB
There are 4 D.o.F and
only 3 Equations of
Equilibrium:
Remove reaction at B
to change the structure
to a determinate
structure
Apply VB such that the
net deflection at B is
Zero
8m, E, I B
100 kN/m C2m, E, I
VA
0
MA
A
-
Draw two tangents at A and B
DB = (1/3x3200x8) x 6/EI = 51200/EI
Draw two tangents at A and B
DB = (8VBx8/2) x 16/3EI = 512VB/3EI
8VB
16/3 m
VB
DB
L, E, IA
B
DB
qB
100 kN/m3200
+CG
2m 6m 2m
-
51200/EI = 512VB/3EI (No deflection at B)
VB = 300 kN
VA = 800 300 = 500 kN
MA = 3200 8 VB = 3200 2400 = 800 kN-m
Bending Moment Diagram
5 mB C
800
450
A3
20
0
+CG
2m 6m
8V
B=
24
00
16/3 m
Sagging Moment = 300*3/2 = 450kN-m
500
300
Shear Force Diagram
5m3m
-
Deflection calculations
DC1 = (1/3x3200x8) x 8/EI = 68267/EI
DC2 = (2400x8/2)x(16/3 + 2)/EI = 70400/EI
DC = DC1 DC2 = -2133/EI
DC =(-2133.33/80000) * 1000
DC = 26.7 mm Upward
32
00
+CG
2m 6m
8m
24
00
16/3 m
8m
2m
-
Alternative solution
Introduce a hinge at A to get red of a D.o.F
8m, E, IA B
100 kN/m C2m, E, I
VA
0
MA
=
8m, E, IA B
100 kN/m C2m, E, I
VA
0
MA
+
8m, E, IA BC2m, E, I
VA
0
MA
-
qA = 2/3 x 800 x 4 /EI =6400/3EI
MA = 3EI/8 qA = 3EI/8 x 6400/3EI = 800kN-m
VA = 400 + 800/8 = 500 kN
VB = 400 800/8 = 300 kN
8m, E, IA B
100 kN/m C2m, E, I
VA
0
MA
400
0
B
C
800
qA
100 kN/m
400
qA
MA/8
MA
A BC
MA/8
0MA
-
Now student try the following question
For the structure shown below:
Calculate support reactions
Sketch BMD showing values at critical
locations
Calculate vertical deflection at c
8m, E, I BC2m, E, I
VA
0
MA 100 kN
A
EI = 80000 kN-m2
4m
-
VB
DB
There are 4 D.o.F
Take reaction at B out
Apply VB such that the net
deflection at B is Zero
8m, E, I BC2m, E, I
VA
0
MA
A
100 kN
4m
400
A B
8/3 +4 = 20/3A B
DB
qB
100 kN4m
Draw two tangents at A and B
DB = (8VB*8/2) x 16/3EI = 512VB/3EI8V
B
16/3
8m
512VB/3EI = 16000/3EI VB = 31.25kN & VA= 68.75kN
Draw two tangents at A and B
DB = (400*4/2) x 20/3EI = 16000/3EI
-
4m
400
A B
8/3 +4 = 20/3
250
16/3
8m
31.25 *4 = 125 kN-m
Bending Moment Dia.
400-250 =150kNm68.7
5kN
31.2
5kN4m4m
Shear Force Dia.
-
Remove the support at 2V1
100 kN
5m, EI 5m, EI 5m, EI
1 32
V3V2
100 kN
5m, EI
100
100 kN
5m, EI 10m, EI1 32
100
100 kN
5m, EI
D2 D2
Bending Moment Diagram
Moment of area at 3D2 =[(500x5/2) x (2/3)x5 + (500x5) x (5+2.5)]/EI
D2 = 12500/3EI + 18750/EI = 68750/3EI
500500
-
Now apply V2 back to push the beam to
its original position
V3 = V2/210m, EI 10m, EI1 32
V1 = V2/2 V2
D2 D2
(V2/2) x 10 = 5V2
D2 = (5V2x10/2) x 20/3EI =500V2/3EI
500V2/3EI = 68750/3EI V2 = 68750/500 = 137.5 kN
Moment of area at 3
-
V1 = 100 V2/2 = 31.25 kN
V3 = V1 = 31.25 kN
M2 = 5V2 500 = 5*137.5 -500 =187.5 kN-m
31.25x5
=156.25
187.5
156.25
-
Thinking outside the box
Students to work in groups of 2 or 3 and
discuss if they can use the conditions of
symmetry to simplify the structure to work
out the unknown parameters.
Hint: Draw deformed shape of the structure
& think about simplifying it.
V1
100 kN
5m, EI 5m, EI 5m, EI
1 32
V3V2
100 kN
5m, EI
-
Thinking outside the box
Due to symmetrical geometry and loading,
the structure can be simplified to:
At the central support the structure can not
move vertically or horizontally
Due to symmetry there is no rotation either.
This could be replaced to a fixed support.
10m, E, I10m, E, I
2
100 kN 100 kN
21 3
-
10m, E, I
2
100 kN
1
=
+10m, E, I
2
100 kN
1
D1
10m, E, I
2
V1
1
D1
10V1
20/3
500
5+10/3 = 25/3
-
From the first Structure:
D1 = [(500 x 5/2) x 25/3] / EI
D1 = 31250/3EI
From the Second Structure:
D1 = [(10V1 x 10/2) x 20/3] / EI
D1 = 1000V1/3EI
1000 V1 = 3125 (Equate deflections)
V1 = 31.25 kN
(which is the same as before)
-
Challenging Question
For the structure shown:
Calculate support reactions.
Sketch BMD and calculate values at critical location.
The structure is indeterminate as there are 4 variables and 3 equations of equilibrium.
120 kN
2m, EI 4m, EI 6m, EI
1 32
-
Solution
0
120 kN
2m, EI 4m, EI 6m, EI
1 32
V1 V3V2
=
+
120 kN2m, EI 4m, EI 6m, EI
1 32
100 20D2
D1 d2
V3 = V2/22m, EI 4m, EI 6m, EI1
32
V1 = V2/2 V2 = 2V1
D2D2
-
Draw three tangents at 1, 2 and 3.
D1 = [(200x2/2x4/3) + (200x10/2)(2+10/3)]/EI
D1 = [800/3+16000/3] / EI = 16800/3EI = 5600/EI
d2 + D2 = D1 / 2 = 2800/EI
d2= (120x6/2) x 6/3EI = 720/EI D2 = (2800-720)/EI
D2 = 2080/EI
120 kN
2m, EI 4m, EI 6m, EI1 32
100 20D2
D1 d2
200120
D12
-
D2 = (3V2x6/2) x 4/EI =36V3/EI 36V2/EI = 2080/EI V2 = 2080/36 = 57.78kN V1 = 100 57.78/2 = 71.11 kN V3 = 20 - 57.78/2 = -8.89 kN
V3 = V2/22m, EI 4m, EI 6m, EI1
32
V1 = V2/2 V2
D2D2
(V2/2) x 6 = 3V2
53.34
142.22
BMD
8.892m, EI 4m, EI 6m, EI
1 32
71.11 57.78
-
0120 kN
2m, EI 4m, EI 6m, EI
1 32
V1 V3V2
120 kN
2m, EI 4m, EI 6m, EI
1 32
V1 V2
D3
V3
2m, EI 4m, EI 6m, EI
1 32
V1 V2 D3
=
+
Alternative SolutionRemove V3
-
V32m, EI 4m, EI 6m, EI
1 32
V1D3A
V3
2m, EI 4m, EI
6m, EI1 32
V1
D3Bq2
M2 =6V3
V3
2m, EI 4m, EI 6m, EI
1 32
V3 2V3 D3
=
+
-
Draw two tangents at 1 and 2 take moment
of the area at 1
D3 =[(160x2/2) x (2x2/3) + (160x4/2) x (2+4/3)]/EI
D3 =(640/3 +320x10/3)/EI = 3840/3EI
120 kN
2m, EI 4m, EI 6m, EI
1 32
V1 V2
D3
6m, EI
1 32
80 40
2m 4m
160 kN-m BMD
D3
-
D3A =(6V3x6/2) x 4/EI = 72V3 / EI
M2 = (3EI/6) q2 = EI q2 /2
q2 = 2M2/EI = 2x(6V3)/EI = 12V3/EI
D3B =6x q2 = 6 x (12xV3/EI) = 72V3 / EI
D3 = D3A + D3B = 144V3 / EI
144V3 / EI = 3840/3EI V3 = 8.89 kN
6m, EI2 3
D3A
V3
BMD+
4m
6V3
6m, EI
1 32
V1
D3Bq2
M2 =6V36m, EI
-
V1 = 80 8.89 = 71.11 kN
V2 = 40 + 2x8.89 = 57.78 kN
53.34
142.22
Bending Moment Diagram
71.11
48.89
8.89
Shear Force Diagram