BEAMS - Majed Dabdoub
Download
Report
Transcript BEAMS - Majed Dabdoub
BEAMS
SHEAR AND MOMENT
1
Beam Shear
Shear and Moment
Diagrams
Vertical shear:
tendency for one part
of a beam to move
vertically with respect
to an adjacent part
2
Beam Shear
Magnitude (V) = sum of vertical forces on
either side of the section
– can be determined at any section along the
length of the beam
Upward forces (reactions) = positive
Downward forces (loads) = negative
Vertical Shear = reactions – loads
(to the left of the section)
3
Beam Shear
Why?
necessary to know the maximum value of
the shear
– necessary to locate where the shear changes
from positive to negative
• where the shear passes through zero
Use of shear diagrams give a graphical
representation of vertical shear throughout
the length of a beam
4
Beam Shear –
Simple beam
– Span = 20 feet
– 2 concentrated loads
Construct shear
diagram
5
Beam Shear – Example 1
1)
Determine the reactions
1 Fx 0
2 Fy 0 R1 8000lb. 1200lb. R2
3 M1 0 (8000lb. 6' ) (1200lb. 16' ) ( R2 20' )
Solving equation (3):
20' R2 48,000lb. ft. 19,200lb. ft.
67,200lb. ft.
R2
3,360lb.
ft.
20
Solving equation (2):
R1 8,000lb. 1,200lb. 3,360lb.
R1 5,840lb.
Figure 6.7a =>
6
Beam Shear – Example 1 (pg. 64)
Determine the shear at various
points along the beam
V( x 1) 5,480 0 5,480lb.
V( x 8) 5,480 8,000 2,160lb.
V( x 18) 5,480 8,000 1,200 3,360lb.
7
Beam Shear – Example 1
Conclusions
–
max. vertical shear = 5,840 lb.
•
–
max. vertical shear occurs at greater
reaction and equals the greater reaction
(for simple spans)
shear changes sign under 8,000 lb. load
•
where max. bending occurs
8
Beam Shear – Example 2
Simple beam
– Span = 20 feet
– 1 concentrated load
– 1 uniformly distr. load
Construct shear diagram,
designate maximum shear,
locate where shear passes
through zero
9
Beam Shear – Example 2
Determine the reactions
1 Fx 0
2 Fy 0 R1 (1,000lb ft. 12' ) 6,000lb. R2
3 M1 0 [(1,000lb ft. 12' )(6' )] (6000lb. 16' ) ( R2 24' )
Solving equation (3):
24' R2 72,000lb. ft. 96,000lb. ft.
168,000lb. ft.
lb.
R2
7
,
000
ft.
24
Solving equation (2):
R1 12,000lb. 6,000lb. 7,000lb.
R1 11,000lb.
10
Shear and Moment Diagrams
11
Beam Shear – Example 2
Determine the shear at various
points along the beam
V( x 1) 11,000 (11,000) 10,000lb.
V( x 2) 11,000 (2 1,000) 9,000lb.
V( x 12 ) 11,000 (121,000) 1,000lb.
V( x 16 ) 11,000 (121,000) 1,000lb.
V( x 16 ) 11,000 121,000 6,000 7,000lb.
V( x 24 ) 11,000 121,000 6,000 7,000lb.
12
Beam Shear – Example 2
Conclusions
–
max. vertical shear = 11,000 lb.
•
–
at left reaction
shear passes through zero at some
point between the left end and the
end of the distributed load
•
x = exact location from R1
–
at this location, V = 0
V 0 11,000 ( x 1,000)
x 11 feet
13
Beam Shear – Example 3
Simple beam with
overhanging ends
– Span = 32 feet
– 3 concentrated loads
– 1 uniformly distr. load
acting over the entire beam
Construct shear diagram,
designate maximum shear,
locate where shear passes
through zero
14
Beam Shear – Example 3
1 Fx 0
2 Fy 0 2,000lb. R1 12,000lb. 4,000lb. (500lb ft. 32' ) R2
3 M 1 0 (2,000lb. 8' ) (12,000lb. 14' ) (4,000lb. 24' ) (500lb. ft. 32' )32 2 8 R2 (20' )
4 M 2 0 (4,000lb. 4' ) (12,000lb. 6' ) (2,000lb. 28' ) (500lb. ft. 32' )32 2 4 R1 (20' )
20' R2 16,000lb. ft. 168,000lb. ft. 96,000lb. ft. 128,000lb. ft.
376,000lb. ft.
R2
18,800lb.
ft.
20
20' R1 16,000lb. ft. 72,000lb. ft. 56,000lb. ft. 192,000lb. ft.
304,000lb. ft.
R1
15,200lb.
20'
15
Determine the reactions
Solving equation (3):
Solving equation (4):
16
Beam Shear – Example 3
Determine the shear at various
points along the beam
V( x 8 ) 0 2,000lb. 500lb. ft. 8' 6,000lb.
500
V( x 8 ) 15,200lb. 2,000lb. 500lb. ft. 8' 9,200lb.
V( x 22 ) 15,200lb. 2,000lb.
lb. ft .
22' 2,200lb.
500
28' 12,800
500 28' 6,000
500 32' 4,000
V( x 22 ) 15,200lb. 2,000lb. 12,000lb. 500lb. ft. 22' 9,800lb.
V( x 28 ) 15,200lb. 2,000lb. 12,000lb.
lb. ft .
V( x 28 ) 15,200lb. 18,800lb. 2,000lb. 12,000lb.
V( x 32 ) 15,200lb. 18,800lb. 2,000lb. 12,000lb.
lb.
lb. ft .
lb.
lb. ft .
lb.
17
Beam Shear – Example 3
Conclusions
– max. vertical shear = 12,800
lb.
•
disregard +/- notations
– shear passes through zero at
three points
•
R1, R2, and under the 12,000lb.
load
18
Bending Moment
Bending moment: tendency of a beam to bend due
to forces acting on it
Magnitude (M) = sum of moments of forces on
either side of the section
– can be determined at any section along the length of the
beam
Bending Moment = moments of reactions –
moments of loads
• (to the left of the section)
19
Bending Moment
20
Bending Moment – Example 1
Simple beam
– span = 20 feet
– 2 concentrated loads
– shear diagram from
earlier
Construct moment
diagram
21
Bending Moment – Example 1
1)
Compute moments at
critical locations
–
under 8,000 lb. load &
1,200 lb. load
M ( x6') (5,840lb. 6' ) 0 35,040lb. ft.
M ( x16') (5,840lb. 16' ) (8,000lb. 10' ) 13,440lb. ft.
22
Bending Moment – Example 2
Simple beam
– Span = 20 feet
– 1 concentrated load
– 1 uniformly distr. Load
– Shear diagram
Construct moment
diagram
23
Bending Moment – Example 2
1)
Compute moments at
critical locations
–
When x = 11 ft. and under
6,000 lb. load
M ( x11') (11,000lb. 11' ) 1,000lb ft 11112 60,500lb. ft.
M ( x16') (11,000lb. 16' ) 1,000lb ft 1212 2 4 56,000lb. ft.
24
Negative Bending Moment
Previously, simple beams subjected to positive
bending moments only
– moment diagrams on one side of the base line
• concave upward (compression on top)
Overhanging ends create negative moments
• concave downward (compression on bottom)
25
Negative Bending Moment
deflected shape has
inflection point
– bending moment = 0
See example
26
Negative Bending Moment Example
– Simple beam with
overhanging end on right
side
• Span = 20’
• Overhang = 6’
• Uniformly distributed load
acting over entire span
– Construct the shear and
moment diagram
– Figure 6.12
27
Negative Bending Moment Example
1) Determine the
reactions
1 Fx 0
2 Fy 0 R1 (600lb ft. 26' ) R2
3 M 1 0 (600lb ft. 26' )26 2 ( R2 20' )
-
Solving equation (3):
20' R2 202,800lb. ft.
202,800lb. ft.
R2
10,140lb.
ft.
20
-
Solving equation (4):
R1 15,600 lb. 10,140 lb.
R1 5,460 lb.
28
Negative Bending Moment Example
2) Determine the shear at
various points along
the beam and draw
the shear diagram
V( x 1) 5,460 (1 600) 4,860lb.
V( x 10 ) 5,460 (10 600) 540lb.
V( x 20 ) 5,460 (20 600) 6,540lb.
V( x 20 ) 5,460 10,140 (20 600) 3,600lb.
29
Negative Bending Moment Example
3) Determine where the
shear is at a
maximum and where
it crosses zero
–
max shear occurs at the right
reaction = 6,540 lb.
V 0 5,460 ( x 600)
x 9.1 feet
30
Negative Bending Moment Example
4) Determine the
moments that the
critical shear points
found in step 3) and
draw the moment
diagram
20' ) 600
24,843
2
20'20 10,800
2
M ( x 9.1) (5,460lb. 9.1' ) 600lb ft 9.1' 9.1
lb. ft .
M ( x 20 ) (5,460lb.
lb. ft .
lb ft
31
Negative Bending Moment Example
4) Find the location of the inflection
point (zero moment) and max.
bending moment
2
M 0 (5,460 x) 600lb ft x x
x2
M 0 5,460x 600
2
M 0 5,460x 300x 2 300x 2 5,460x
5,460 (5,460) 2 4(300)(0)
x
2(300)
5460 5,460
x
0 feet;18.2 feet
600
since x cannot =0, then we use x=18.2’
Max. bending moment =24,843 lb.-ft.
32
Rules of Thumb/Review
shear is dependent on the loads and reactions
– when a reaction occurs; the shear “jumps” up by the
amount of the reaction
– when a load occurs; the shear “jumps” down by the
amount of the load
point loads create straight lines on shear diagrams
uniformly distributed loads create sloping lines of
shear diagrams
33
Rules of Thumb/Review
moment is dependent upon the shear diagram
– the area under the shear diagram = change in the
moment (i.e. Ashear diagram = ΔM)
straight lines on shear diagrams create sloping
lines on moment diagrams
sloping lines on shear diagrams create curves on
moment diagrams
positive shear = increasing slope
negative shear = decreasing slope
34
Typical Loadings
In beam design, only
need to know:
–
–
–
–
reactions
max. shear
max. bending moment
max. deflection
35