7/29/2019 13 Bending

1/11

Bending Copyright Prof Schierle 2011 1

7/29/2019 13 Bending

2/11

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,19th century Belgian engineer)

7/29/2019 13 Bending

3/11

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

7/29/2019 13 Bending

4/11

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

7/29/2019 13 Bending

5/11

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

7/29/2019 13 Bending

6/11Bending Copyright Prof Schierle 2011 6

Bending

1 Simple beam

2 Bending deformation under load

3 Bending stress:

Top shortens in compression

Neutral Axis= 0 stressBottom elongates in tension

7/29/2019 13 Bending

7/11Bending 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

7/29/2019 13 Bending

8/11Bending Copyright Prof Schierle 2011 8

Equilibrium MethodCanti lever 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 kShear V ( V = 0)

Val = 0 Val = 0

Var= 0 2k Var= -2 k

Vbl = -2k +- 0 Vbl = -2 k

Vbr= -2k + R = -2k + 2k Vbr= 0Bending ( M = 0)

@ a: M = 0 P M = 0

@ 5: M = 5 P M = -10 k

@ b: M = 10 P M = -20 kNote: Point load = linear bending moment

7/29/2019 13 Bending

9/11Bending Copyright Prof Schierle 2011 9

Simple beam with uniform load1 Beam diagram: L = 20, w = 100 plf

2 Free-body diagram of partial beam

3 Shear diagram4 Bending diagram

Reactions

R = w L /2 = 100 x 20 / 2 R = 1000 #

Shear forces @ distance x from aVx = 0; R wx-Vx = 0 Vx = Rw 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-wx2

/2M0 = 1000 x 0 M0= 0 #

M5 = 1000x5-100x52/2 M5= 3750 #

M10= 1000x10-100x102/2 M10= 5000 #

M15= 1000x15-100x152/2 M15= 3750 #

M20 = 1000x20-100x202/2 M20= 0 #

Note: parabolic bending distribution

7/29/2019 13 Bending

10/11Bending Copyright Prof Schierle 2011 10

Simple beam formulas

Reactions R = w L /2

Shear force Vx = R w x

Max. shear at supports

Max. V = R - 0

Max. shear V = R

Bending moment Mx = Rx-wx2/2

Max. M at x = L/2

Max. M= (wL/2) L/2 - (wL/2)L/4Max. M= 2wL2/8 - wL2/8

Max. bending M = wL2/8

Note:Formulas for simple beams & uniform load only !

Verify last example (L= 20, w = 100plf)

M= wL2

/8 = 100 x 202

/8 M = 5000#, ok

7/29/2019 13 Bending

11/11Bending Copyright Prof Schierle 2011 11

Bending m em bers a re com m onHence: Ben ing i s im por ta n t