Section 15: Introduction to Stress and Bending 15-1 Bending - - PowerPoint PPT Presentation

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Section 15: Introduction to Stress and Bending 15-1 Bending - - PowerPoint PPT Presentation

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Section 15: Introduction to Stress and Bending 15-1 Bending Bending Long bones: beams Long bones: beams T Compressive stress: inner portion p Tensile stress: outer C p portion axis i Max stresses near the edges,

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SLIDE 1

Section 15: Introduction to Stress and Bending

15-1

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SLIDE 2

Bending Bending

  • Long bones: beams

Long bones: beams

  • Compressive stress:

inner portion

T

p

  • Tensile stress: outer

portion

C i

p

  • Max stresses near the

edges, less near the

axis axis

neutral axis

(M )/I

y axis

15-2 From: Noffal

σx=(Mb·y)/I

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SLIDE 3

Bending Moments Bending Moments

  • Shear stresses max

Shear stresses max at neutral axis and zero at the surface

Q

  • τ = (Q·V)/(I ·b)
  • Q= area moment

Q y

  • V= vertical shear

force

h y b

15-3 From: Noffal

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SLIDE 4

Behavior of Bone Under Bending Behavior of Bone Under Bending

  • Bending subjects bone to a combination of

t i d i (t i tension and compression (tension on one side of neutral axis, compression on the other side, and no stress or strain along the neutral side, and no stress or strain along the neutral axis)

  • Magnitude of stresses is proportional to the

Magnitude of stresses is proportional to the distance from the neutral axis (see figure)

  • Long bone subject to increased risk of

Long bone subject to increased risk of bending fractures

15-4 From: Brown

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SLIDE 5

Bending Bending

  • Cantilever bending

Cantilever bending

  • Compressive force

acting off-center from g long axis

15-5 From: Noffal

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SLIDE 6

Bending Loading

15-6 From: Brown

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SLIDE 7

Muscle Activity Changing Stress Di t ib ti Distribution

15-7 From: Brown

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SLIDE 8

Various Types of Beam Loading and Support a ous ypes o ea

  • ad g a d Suppo t
  • Beam - structural member designed to support

loads applied at various points along its length.

  • Beam can be subjected to concentrated loads or

distributed loads or combination of both.

  • Beam design is two-step process:

1) d t i h i f d b di 1) determine shearing forces and bending moments produced by applied loads 2) select cross-section best suited to resist shearing forces and bending moments

15-8 From: Rabiei

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SLIDE 9

6.1 SHEAR AND MOMENT DIAGRAMS DIAGRAMS

  • In order to design a beam, it is necessary to

determine the maximum shear and moment in the beam

  • Express V and M as functions of arbitrary position

x along axis.

  • These functions can be represented by graphs

called shear and moment diagrams

  • Engineers need to know the variation of shear and

t l th b t k h t moment along the beam to know where to reinforce it

15-9 From: Wang

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SLIDE 10

Diagrams

15-10 From: Hornsey

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SLIDE 11

6.1 SHEAR AND MOMENT DIAGRAMS DIAGRAMS

  • Shear and bending-moment functions must be

determined for each region of the beam between g any two discontinuities of loading

15-11 From: Wang

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SLIDE 12

6.1 SHEAR AND MOMENT DIAGRAMS DIAGRAMS

Beam sign convention

  • Although choice of sign convention is arbitrary, in

Although choice of sign convention is arbitrary, in this course, we adopt the one often used by engineers:

15-12 From: Wang

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SLIDE 13

6.1 SHEAR AND MOMENT DIAGRAMS DIAGRAMS

Procedure for analysis Support reactions

  • Determine all reactive forces and couple moments

acting on beam

  • Resolve all forces into components acting

Resolve all forces into components acting perpendicular and parallel to beam’s axis Shear and moment functions

  • Specify separate coordinates x having an origin at

beam’s left end, and extending to regions of beam between concentrated forces and/or couple p moments, or where there is no discontinuity of distributed loading

15-13 From: Wang

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SLIDE 14

6.1 SHEAR AND MOMENT DIAGRAMS DIAGRAMS

Procedure for analysis Shear and moment functions

  • Section beam perpendicular to its axis at each

distance x

  • Draw free-body diagram of one segment

Draw free body diagram of one segment

  • Make sure V and M are shown acting in positive

sense, according to sign convention

  • Sum forces perpendicular to beam’s axis to get

shear

  • Sum moments about the sectioned end of segment

Sum moments about the sectioned end of segment to get moment

15-14 From: Wang

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SLIDE 15

6.1 SHEAR AND MOMENT DIAGRAMS DIAGRAMS

Procedure for analysis Shear and moment diagrams

  • Plot shear diagram (V vs. x) and moment diagram

(M vs. x)

  • If numerical values are positive values are plotted

If numerical values are positive, values are plotted above axis, otherwise, negative values are plotted below axis It is con enient to sho the shear and moment

  • It is convenient to show the shear and moment

diagrams directly below the free-body diagram

15-15 From: Wang

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SLIDE 16

EXAMPLE 6 6 EXAMPLE 6.6

Draw the shear and moment diagrams for beam shown below.

15-16 From: Wang

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SLIDE 17

EXAMPLE 6 6 (SOLN) EXAMPLE 6.6 (SOLN)

Support reactions: Shown in free-body diagram. Shear and moment functions Shear and moment functions Since there is a discontinuity of distributed load and a concentrated load at beam’s center, two , regions of x must be considered. 0 ≤ x1 ≤ 5 m, ≤

1 ≤

, +↑ Σ Fy = 0; ... V = 5.75 N + Σ M = 0; ... M = (5.75x1 + 80) kN·m

15-17 From: Wang

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SLIDE 18

EXAMPLE 6 6 (SOLN) EXAMPLE 6.6 (SOLN)

Shear and moment functions 5 m ≤ x ≤ 10 m 5 m ≤ x2 ≤ 10 m, +↑ Σ Fy = 0; ... V = (15.75 − 5x2) kN ↑

y

; (

2)

+ Σ M = 0; ... M = (−5.75x2

2 + 15.75x2 +92.5) kN·m

; (

2 2

) Check results by applying w = dV/dx and V = dM/dx. Check results by applying w dV/dx and V dM/dx.

15-18 From: Wang

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SLIDE 19

EXAMPLE 6 6 (SOLN) EXAMPLE 6.6 (SOLN)

Shear and moment diagrams

15-19 From: Wang

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SLIDE 20

6.2 GRAPHICAL METHOD FOR CONSTRUCTING SHEAR AND MOMENT CONSTRUCTING SHEAR AND MOMENT DIAGRAMS

Regions of concentrated force and moment

15-20 From: Wang

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SLIDE 21

6.2 GRAPHICAL METHOD FOR CONSTRUCTING SHEAR AND MOMENT CONSTRUCTING SHEAR AND MOMENT DIAGRAMS

Regions of concentrated force and moment

15-21 From: Wang

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SLIDE 22

Sample Problem 7 2 Sample Problem 7.2

SOLUTION:

  • Taking entire beam as a free-body,

Taking entire beam as a free body, calculate reactions at B and D.

  • Find equivalent internal force-couple

t f f b di f d b Draw the shear and bending moment systems for free-bodies formed by cutting beam on either side of load application points. g diagrams for the beam and loading shown.

  • Plot results.

15-22 From: Rabiei

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SLIDE 23

Sample Problem 7.2

SOLUTION: SOLUTION:

  • Taking entire beam as a free-body, calculate

reactions at B and D.

  • Find equivalent internal force-couple systems at

sections on either side of load application points.

∑

= :

y

F kN 20

1 =

− − V kN 20

1

− = V

∑

y 1 1

:

2 =

∑ M

( )( )

m kN 20

1 =

+ M

1 =

M Similarly m kN 50 kN 26 m kN 50 kN 26

4 4 3 3

⋅ − = = ⋅ − = = M V M V Similarly, m kN 50 kN 26 m kN 50 kN 26

6 6 5 5 4 4

⋅ − = = ⋅ − = = M V M V

15-23 From: Rabiei

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SLIDE 24

Sample Problem 7.2

  • Plot results.

Note that shear is of constant value b d l d d between concentrated loads and bending moment varies linearly.

15-24 From: Rabiei