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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

2 Bending deformation under load

3 Bending stress:

Top shortens in compression

Neutral Axis = 0 stress

Bottom elongates in tension

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Bending Copyright Prof Schierle 2011 7

Beam shear

1 Beam with uniform load

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

Cantilever beam with point load

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

Simple beam with uniform load

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

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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

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