Chapter 5

1. Chapter 5 Torsion

2. Introduction • Torsional effects differ according to the shape of the member. Three main classifications can be considered: • Circular members (hollow or solid). • Noncircular members. • Thin-walled hollow shafts. The torsion theory for each of the above classification is to be considered.

3. Torsional Loads on Circular Shafts1.Net Torque Due to Internal Shear Stresses • Net of the internal shearing stresses is an internal torque, equal and opposite to the applied torque, • Unlike the normal stress due to axial loads, the distribution of shearing stresses due to torsional loads can not be assumed uniform. • As a result of applying torsion two effects are observed: • Shearing stresses are set up on cross section perpendicular to the axis of the bar as shown.

4. Shaft Deformations:2. Angular displacements of one end cross section wrt the other are set. • The angle of twist is proportional to the applied torque and to the shaft length as shown • Every cross-section of a circular shaft remains plane and undistorted. • Cross-sections for hollow and solid circular shafts remain plane and undistorted because a circular shaft is axisymmetric. • Cross-sections of noncircular shafts are distorted when subjected to torsion.

5. Shearing Strain • Since the ends of the element remain planar, the shear strain is equal to angle of twist. • It follows that • Shear strain is proportional to twist and radius

6. Stresses in Elastic Range • Multiplying the previous equation by the shear modulus, From Hooke’s Law, The shearing stress varies linearly with the radial position in the section. • The sum of the moments from the internal stress distribution is equal to the torque on the shaft: • The elastic torsion formulas is thus:

7. Normal Stresses • Elements with faces parallel and perpendicular to shaft axis are subjected to shear stresses only. • Consider an element at 45o to the shaft axis, • Element a is in pure shear. • Element c is subjected to a tensile stress on two faces and compressive stress on the other two. • Note that all stresses for elements a and c have the same magnitude

8. Torsional Failure Modes • Ductile materials generally fail in shear. Brittle materials are weaker in tension than shear. • When subjected to torsion, a ductile specimen breaks along a plane of maximum shear, i.e., a plane perpendicular to the shaft axis. • When subjected to torsion, a brittle specimen breaks along planes perpendicular to the direction in which tension is a maximum, i.e., along surfaces at 45o to the shaft axis.

9. Example 1 Given: Shaft BC is hollow with inner and outer diameters 90 mm and 120 mm, respectively. Shafts AB and CD are solid of diameter d. Find: (a) the min. and max. shearing stress in shaft BC, (b) the required diameter d of shafts AB and CD if the allowable shearing stress in these shafts is 65 MPa.

10. Example 2

11. Power Transmission • Shafts and tubes having circular cross sections are often used to transmit power developed by a machine • When used for this purpose, they are subjected to torques that depend on the power generated by the machine and the angular speed of the shaft • Power is defined as the work performed per unit of time • The work transmitted by a rotating shaft equals the torque (T) applied times the shaft`s anglular velocity (ω) • Units: • In the SI system, T (N.m), ω (rad/s) P (watts) [1 w = 1 N.m/s] • In the USCS, P (ft.Ib/s), horsepower (hp) = 550 ft.Ib/s

12. Power Transmission • For machinary, the frequency of the shaft`s rotation (f) is often reported • f is a measure of the number of revolutions or cycles the shaft makes per second and is expressed in herts (1 Hz = 1 cycle/s) • Since 1 cycle = 2л rad, thenω = 2лfand the power equation becomes • The design of a shaft requires finding the design or geometric parameter (J/c)

13. Example 6

14. Angle of Twist in Elastic Range • Recall • In the elastic range apply Hooke’s Law, • Equate previous equations: • If the torsional load or shaft section changes along the length, the angle of rotation is found as the sum of segment rotations

15. Analogy between axial and torsion

16. Example 7

17. 1 Statically Indeterminate Torsionally-Loaded Members • A torsionally loaded shaft may be classified as statically indeterminate if the moment equation of equilibrium, applied about the axis of the shaft, is not adequate to determine the unknown torques acting on the shaft • A free-body diagram of the bar allows us to establish a relationship between the applied torqueT and the reactive torques at the supports TA and TB

18. Statically Indeterminate Torsionally-Loaded Members • The previous equation is not sufficient to detrmine the unknown reactive torques, so the problem is statically indeterminate • In order to establish an additional equation needed for solution, it is necessary to consider the compatibility or kinematic condition • Due to the fixation at either end of the shaft, the angle of twist of one end of the shaft with respect to the other end is zero. • Therfore, the kinematic relationship for the shaft is ØA/B = 0 • This equation can be expressed in terms of the unknown torques using a load-displacement relationship, which depends on the material behavior (e.g. if linear-elastic behavior occurs,Ø = TL/JG can be used)

19. 2 Statically Indeterminate Torsionally-Loaded Members • Realizing that the internal torque in segment AC is +TA and in segment CB the internal torque is −TB • The compatibility equation can be written as • Assuming that JG is constant, we can solve the two equations 1 and 2 for theunknown torques, which gives

20. Example 9

21. Design of Transmission Shafts • Principal transmission shaft performance specifications are: power and speed • Select shaft material and cross-section to meet performance specifications without exceeding allowable shearing stress. • Determine torque applied to shaft at specified power and speed, • Find shaft cross-section given:

22. Torsion of Noncircular Members • Previous formulas are valid for axisymmetric or circular shafts • Planar cross-sections of noncircular shafts do not remain planar and stress and strain distribution do not vary linearly. For uniform rectangular cross-sections: • At large values of a/b, the maximum shear stress and angle of twist for other open sections are the same as a rectangular bar.

23. Thin-Walled Hollow Shafts • Summing forces in the x-direction on AB, • shear stress varies inversely with thickness • Compute the torque from the integral of • Angle of twist:

24. End of torsion Let Learning Continue

25. Analysis and Design of Beams for Bending

26. Introduction Shear and Bending Moment Diagrams Example 1 Relations Among Load, Shear, and Bending Moment Example 2 Design of Prismatic Beams for Bending Example 3 Analysis and Design of Beams for Bending

27. Introduction • Summary: • Beams - structural members supporting transverse loadings which are classified as concentrated or distributed loads • Applied loads result in internal forces consisting of a shear force and a bending couple • Normal stress is often the critical design criteria This requires determination of the location and magnitude of largest bending moment

28. Introduction Classification of Beam Supports

29. Shear and Bending Moment Diagrams • Determination of maximum normal and shearing stresses requires identification of maximum internal shear force and bending couple. • These are determined by passing a section through the beam and applying equilibrium equations on the beam portions. • Sign conventions for shear forces V and V’ and bending couples M and M’

30. Relations Among Load, Shear, and Bending Moment • Relationship between load and shear: • Relationship between shear and bending moment:

31. Common Loading Cases

32. Common Loading Cases

33. Example 1 Given: The timber beam and loading shown Find: draw the shear and bending-moment diagrams and determine the maximum normal stress due to bending.

34. Example 2 Draw the shear and bending moment diagrams for the beam and loading shown.

35. Example 3

36. Example 4

37. Example 5

38. End of beam part 1 Let Learning Continue

39. Pure bending of beams Outline Definition of centroid of an area Pure bending Bending deformation of a straight member The flexure formula Design of prismatic beams for bending Unsymmetric bending

40. Centroidal axis

41. Pure bending of beams Pure bending is the case of moment M≠0 and shear force V=0

42. Bending Deformation of a Straight Member • The member shown is subjected to equal and opposite couples M and M` • Under their action the member will bend, but will remain symmetric with respect to the plane containing the couples • The member bends uniformely since the bending moment is the same at any cross section • The line AB on the upper face intersects the plane of the couples and has a constant curvature • The originally straight line AB is transformed into a circular arc with its center at C • Therefore, we note that CA = CB = CD

43. Bending Deformation of a Straight Member • More insight into the nature of bending is gained from a two-dimensional member with lines drawn on it in a gridwork • When a bending moment is applied, it tends to distort these lines as shown in the figure • It can be seen that the longitudinal lines become curved and the vertical transverse lines remain straight and yet undergo a rotation

44. Bending Deformation of a Straight Member • The behavior of any deformable bar subjected to a bending moment causes the material within the bottom portion of the bar to stretch and the material within the top portion to compress • Consequently, between these two regions there must be a surface called the neutral surface

45. Bending Deformation of a Straight Member • The neutral surface is defined by arc DE and is the surface on which no deformation has taken place. The length of this arc for both the undeformed and deformed member is L = ρƟ • Next, we consider an arc some distance y above the neutral surface • The length of the arc JK can be expressed as L`= (ρ ─ y)Ɵ. The original length JK was equal to L. The deformation of JK is therefore δ = L` ─ L = (ρ ─ y)Ɵ ─ ρƟ = ─ yƟ

46. Bending Deformation of a Straight Member • Knowing the deformation of arcJK is δ = ─ yƟ,the strain in JK can be determined • The minus sign is a result of assuming the bending moment to be concave up • Transverse sections remains plane, so identical deformations occur in all planes parallel to the plane of symmetry • The longitudinal normal strainεxvaries linearly (through the member) with the distance y from the neutral surface

47. Bending Deformation of a Straight Member • The normal strain is a maximum when y is largest. This occures when y = c • The maximum absolute normal strain is εmax = c/ρ • The normal strain at any y location in the beam can be written in terms of the maximum strain • By Poisson`s ratio, there must also be associated strain componenets εy = ─ʋεx and εz = ─ʋεx

48. The Flexure Formula • Here we will develop an equation that relates the longitudinal stress distribution in a beam to the internal resultant bending moment acting on the beam`s cross section • To do this we will assume that the material behaves in a linear-elastic manner so that Hook`s law applies, that is, σ = Eε • A linear variation of normal strain must then be the consequence of a linear variation in normal stress

49. The Flexure Formula • Like the normal strain variation, σ will vary from zero at the member's neutral axis to a maximum value, σmax,a distance c farthest from the neutral axis • The normal strain as a function of y is related to maximum normal strain by εx = ─ (y/c) εmax • Multiplying the strain by the elastic modulus, and noting that σmax = Eεmax, we can write • This equation represents the stress distribution over the cross-sectional area

50. The Flexure Formula • The position of the neutral axis on the cross section can be located by satisfying the condition that the resultant force produced by the stress distribution over the cross-sectional area must be equal to zero • Noting that the force dF = σdA acts on the arbitrary elementdA, we require