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BME 315 – Biomechanics Chapter 4. Mechanical Properties of the Body Professor: Darryl Thelen University of Wisconsin-Madison Fall 2009.

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Presentation on theme: "BME 315 – Biomechanics Chapter 4. Mechanical Properties of the Body Professor: Darryl Thelen University of Wisconsin-Madison Fall 2009."— Presentation transcript:

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2 BME 315 – Biomechanics Chapter 4. Mechanical Properties of the Body Professor: Darryl Thelen University of Wisconsin-Madison Fall 2009

3 Shear Forces and Bending Moments  Define: V – shear force acting on a beam cross-section Mz – bending moment acting on a beam cross- section  Considering a differential element, find that

4 Shear Forces and Bending Moments Steps: 1.Draw a free body diagram of entire beam, determine reactions at supports 2.Construct FBD for each part of interest, making cuts between any changes in load, geometry or material properties 3.Construct your shear and bending moment diagrams directly below the beam drawing 4.Check your solution by confirming that V(x)=dM z /dx

5 Example  Draw the bending moment and shear force diagrams

6 Pure Bending Beam deforms with a constant curvature , as a result  plane sections remain plane  a neutral axis NA exists that does not change length  above the NA, elements are in compression  below the NA, elements are in tension  where will largest tensile strain be?  where will largest compressive strain be?  what will a rectangular cross-section look like in deformed configuration? 

7 Stresses Due to Shear Forces and Bending Moments  To balance V and M z at a cross-section, need Normal stress  xx Shear stress  xy Flexure formula (for normal stress)  M z (x) – internal bending moment at a cross-section x  y – distance from the neutral axis to point of interest  I z – area moment of inertia about the z axis

8 ©2001 Brooks/Cole, a division of Thomson Learning, Inc. Thomson Learning ™ is a trademark used herein under license.

9 Area Moments of Inertia Beam with a rectangular cross-section  I = bh 3 /12 b h y Beam with a solid circular cross-section  I = D 4 /64 = r 4 /4 D y Beam with a hollow cylindrical cross-section  I = (D o 4 -D i 4 )/64 = (r o 4 -r i 4 )/4

10 Example  Beam has a rectangular cross-section, b x h  Determine maximum  xx

11 Example 2  A cantilever beam is subjected to a distributed load q o (force/unit length)  Draw the shear and bending moment diagrams

12 Example 2  Beam has a solid circular cross-section of radius r  Determine the location and magnitude of the maximum tensile and compressive stresses

13 ©2001 Brooks/Cole, a division of Thomson Learning, Inc. Thomson Learning ™ is a trademark used herein under license. Beam Deflections

14 Deformation of a Beam Under Transverse Loading v(x) – vertical deflection of a point on the neutral axis of a beam General relationship between bending moment and beam deflection at a point x Thus the internal bending moment provides a quantitative indicator of how curvature, d 2 v/dx 2, varies along a beam ©2001 Brooks/Cole, a division of Thomson Learning, Inc. Thomson Learning ™ is a trademark used herein under license.

15 Equation of the Elastic Curve: Boundary Conditions Constants can be determined from boundary conditions Known values of v or dv/dx at a specific location Three cases for statically determinant beams, –Simply supported beam –Overhanging beam –Cantilever beam More complicated loadings require multiple integrals for each section of the beam, and enforcing continuity of displacement and slope. Integrating the beam equation, one gets v v A =0 v B =0 v A =0 v B =0 v A =0 dv A /dx=0

16 Beam Boundary Conditions Photo credits: Wendy Crone Lot 17 parking deck Footbridge over campus drive Support bolt in ECB Identify the supports that are idealized as roller, pin or fixed.

17 L Example Find an expression for the elastic curve v(x) Given P=100 lbs, L=4’, E=30,000 ksi, I=350 in 4, find the max deflection v max

18 Using tables.....

19 Method of Superposition  For linearly elastic materials, the deformations of beams subjected to combinations of loadings may be obtained as the linear combination of the deformations from the individual loadings  Thus, for complex loadings, Break the problem down into combinations of simpler cases Solve the simpler cases Add the total solution to get the net solution = +

20 Example  Find an expression for the elastic curve due to the combined loading where L=8 m, EI=constant, w = 20 kN/m, P = 150 kN  At what point x is their maximum deflection? 22

21 Example  Find an expression for the elastic curve due to the combined loading where L AD =2 m, L=8 m, EI=constant  At what point x is their maximum deflection?


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