Urban and Civil

Nov 25, 2013 (3 years and 6 months ago)

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Bending Copyright Prof Schierle 2011 1
Bending
Bending Copyright Prof Schierle 2011 2
Bending resisting elements
1 Beam
2 Slab (analyze a strip as beam)
3 Folded plate
4 Cylindrical shell
5 Frame
6 Vierendeel girder
(named after the inventor,
19
th
century Belgian engineer)
Bending Copyright Prof Schierle 2011 3
Beams
1 Simple beam
2 Cantilever beam
3 Beam with overhang
4 Beam with two overhangs
5 Beam with fixed end support
6 Continuous beam / girder
2
Bending Copyright Prof Schierle 2011 4
Slab/joist/beam/girder
1 Slab
2 Joists
3 Beams
4 2-layer system:
Joists supported by beams
5 3-layer system:
Joists supported by beams,
beams supported by girders
Bending Copyright Prof Schierle 2011 5
Steel joist / girder
IIT building, Chicago
Architect: Mies Van der Rohe
• Joists subject to bending
• Girders, subject to bending
Bending Copyright Prof Schierle 2011 6
Bending
1 Simple beam
3 Bending stress:
Top shortens in compression
Neutral Axis = 0 stress
Bottom elongates in tension
3
Bending Copyright Prof Schierle 2011 7
Beam shear
2 Vertical shear effect
3 Horizontal shear effect
4 Shear diagram (max at supports)
5 Beam with square markings
6 Square markings deformed
7 Shear effect on square marking
8 Tensile/compressive effect of shear
Bending Copyright Prof Schierle 2011 8
Equilibrium Method
Assume:
L = 10’
P = 2 k
V Shear diagram
M Bending moment diagram
 Deflection diagram
Reactions ( Vb = 0)
R – P = 0, R – 2 = 0 R = 2 k
M – 10 P = 0, M = 10 x 2 M = 20 k’
Shear V ( V = 0)
V
al
= 0 Val = 0
V
ar
= 0 – 2k V
ar
= -2 k
V
bl
= -2k +- 0 V
bl
= -2 k
V
br
= -2k + R = -2k + 2k V
br
= 0
Bending ( M = 0)
@ a: M = 0 P M = 0
@ 5’: M = – 5 P M = -10 k’
@ b: M = –10 P M = -20 k’
Note: Point load = linear bending moment
Bending Copyright Prof Schierle 2011 9
1 Beam diagram: L = 20’, w = 100 plf
2 Free-body diagram of partial beam
3 Shear diagram
4 Bending diagram
Reactions
R = w L /2 = 100 x 20 / 2 R = 1000 #
Shear forces @ distance x from a
Vx = 0; R – wx-Vx = 0 Vx = R–w x
V0 = 0+1000 V0 = 1000 #
V5 = 1000 – 5 x 100 V5 = 500 #
V10 = 500 – 5 x 100 V10 = 0 #
V15 = 0 – 5 x 100 V15 = -500 #
V20 = – 500 -5 x 100 V20 = -1000 #
Note: linear shear distribution
Bending moments @ x
Mx = 0; Rx – wx (x/2) – Mx = 0 Mx = Rx-wx
2
/2
M0 = 1000 x 0’ M0 = 0 #’
M5 = 1000x5-100x5
2
/2 M5 = 3750 #’
M10 = 1000x10-100x10
2
/2 M10 = 5000 #’
M15 = 1000x15-100x15
2
/2 M15 = 3750 #’
M20 = 1000x20-100x20
2
/2 M20 = 0 #’
Note: parabolic bending distribution
4
Bending Copyright Prof Schierle 2011 10
Simple beam formulas
Reactions R = w L /2
Shear force V
x
= R – w x
Max. shear at supports
Max. V = R - 0
Max. shear V = R
Bending moment M
x
= Rx-wx
2
/2
Max. M at x = L/2
Max. M= (wL/2) L/2 - (wL/2)L/4
Max. M= 2wL
2
/8 - wL
2
/8
Max. bending M= wL
2
/8
Note:
Formulas for simple beams & uniform load only !
Verify last example (L= 20’, w = 100plf)
M= wL
2
/8 = 100 x 20
2
/8 M = 5000#’, ok
Bending Copyright Prof Schierle 2011 11
Bending members are common
Hence: Bening is important 