Lecture7 - LECTURE Third Edition SHAFTS: STATICALLY...

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Unformatted text preview: LECTURE Third Edition SHAFTS: STATICALLY INDETERMINATE SHAFTS • A. J. Clark School of Engineering •Department of Civil and Environmental Engineering 7 Chapter 3.6 by Dr. Ibrahim A. Assakkaf SPRING 2003 ENES 220 – Mechanics of Materials Department of Civil and Environmental Engineering University of Maryland, College Park LECTURE 7. SHAFTS: STATICALLY INDETERMINATE SHAFTS (3.6) Stresses in Oblique Planes Slide No. 1 ENES 220 ©Assakkaf Up to this point, the stresses in a shaft has been limited to shearing stresses. This due to the fact that the selection of the element under study was oriented in such a way that its faces were either perpendicular or parallel to the axis of the shaft (see Fig. 15) 1 LECTURE 7. SHAFTS: STATICALLY INDETERMINATE SHAFTS (3.6) Stresses in Oblique Planes Slide No. 2 ENES 220 ©Assakkaf Fig. 15 τ τ Axis of shaft LECTURE 7. SHAFTS: STATICALLY INDETERMINATE SHAFTS (3.6) Stresses in Oblique Planes Slide No. 3 ENES 220 ©Assakkaf From our discussion of the torsional loading on a shaft, we know this loading produces shearing stresses τ in the faces perpendicular to the axis of the shaft. But due to equilibrium requirement, there are equal stresses on the faces formed by the two planes containing the axis of the shaft. 2 Slide No. 4 LECTURE 7. SHAFTS: STATICALLY INDETERMINATE SHAFTS (3.6) Stresses in Oblique Planes ENES 220 ©Assakkaf Other Stresses Induced By Torsion – It is necessary to make sure that whether the transverse plane is a plane of maximum shearing stress and whether there are other significant stresses induced by torsion. – Consider the following shaft (Fig. 16), which is subjected to a torque T. Slide No. 5 LECTURE 7. SHAFTS: STATICALLY INDETERMINATE SHAFTS (3.6) Stresses in Oblique Planes ENES 220 ©Assakkaf Other Stresses Induced By Torsion Fig. 16 t y y (c) n τx y dA cos α τn t dA α α A σn dA x y τ xy τyx dA sin α x τ yx α τ yx (a) τ xy (b) x 3 LECTURE 7. SHAFTS: STATICALLY INDETERMINATE SHAFTS (3.6) Stresses in Oblique Planes Slide No. 6 ENES 220 ©Assakkaf Other Stresses Induced By Torsion – The stresses at point A in the shaft of Fig. 16a is analyzed. – A differential element taken from the shaft at point A and the stresses acting on transverse and longitudinal planes are shown in Fig. 16b. – The shearing stress τxy can be determined from Tc τ xy = J LECTURE 7. SHAFTS: STATICALLY INDETERMINATE SHAFTS (3.6) Stresses in Oblique Planes Slide No. 7 ENES 220 ©Assakkaf Other Stresses Induced By Torsion – Let assume that differential element of Fig. 16b has length dx, height dy, and thickness dz. – If a shearing force Vx = τxy dx dy is applied to the top surface of the element, the equation of equilibrium ∑Fx = 0 then will require application of an opposite shear force V’x at the bottom of the element. 4 Slide No. 8 LECTURE 7. SHAFTS: STATICALLY INDETERMINATE SHAFTS (3.6) Stresses in Oblique Planes ENES 220 ©Assakkaf Other Stresses Induced By Torsion Vx = τ yx dx dz y Vx V y = τ xy dy dz Vy Vy α ∆z z ∆y ∆x x Vx Fig. 17 LECTURE 7. SHAFTS: STATICALLY INDETERMINATE SHAFTS (3.6) Stresses in Oblique Planes Slide No. 9 ENES 220 ©Assakkaf Other Stresses Induced By Torsion – If ∑Fx = 0 then requires application of an opposite shear force V’x at the bottom of the element, then it will the element subjected to a clockwise couple. – This clockwise couple must be balanced by counterclockwise couple composed of Vx applied to the vertical faces of the element. 5 Slide No. 10 LECTURE 7. SHAFTS: STATICALLY INDETERMINATE SHAFTS (3.6) Stresses in Oblique Planes ENES 220 ©Assakkaf Other Stresses Induced By Torsion – The application of the equilibrium moment equation ∑Mz = 0 gives τ yx (dx dz ) dy = τ xy (dy dz ) dx – From which the important result τ yx = τ xy (27) LECTURE 7. SHAFTS: STATICALLY INDETERMINATE SHAFTS (3.6) Stresses in Oblique Planes Slide No. 11 ENES 220 ©Assakkaf Other Stresses Induced By Torsion – If the equations of equilibrium are applied to the free-body diagram of Fig. 16c (which is a wedge-shaped part of the differential element of Fig. 16b with dA being the area of the inclined face), the following results are obtained + ∑F =0 (28) t τ nt dA − τ xy (dA cos α ) cos α + τ yx (dA sin α )sin α = 0 6 Slide No. 12 LECTURE 7. SHAFTS: STATICALLY INDETERMINATE SHAFTS (3.6) Stresses in Oblique Planes ENES 220 ©Assakkaf Other Stresses Induced By Torsion Fig. 16 y y t n τn t dA τx y dA cos α A (c) α α x y σn dA τ xy τyx dA sin α x τ yx (a) τ xy α τ yx (b) x Slide No. 13 LECTURE 7. SHAFTS: STATICALLY INDETERMINATE SHAFTS (3.6) Stresses in Oblique Planes ENES 220 ©Assakkaf Other Stresses Induced By Torsion t ∑F t =0 τ nt dA − τ xy (dA cos α ) cos α + τ yx (dA sin α )sin α = 0 From which τ nt = τ xy (cos 2 α − sin 2 α ) = τ xy cos 2α (29) n τn t dA τx y dA cos α + y α α σn dA τyx dA sin α Fig. 16c 7 Slide No. 14 LECTURE 7. SHAFTS: STATICALLY INDETERMINATE SHAFTS (3.6) Stresses in Oblique Planes ENES 220 ©Assakkaf Other Stresses Induced By Torsion – Likewise, if we take summation of forces in the n direction (see Fig. 16c), then the results would be + ∑F n =0 σ n dA − τ xy (dA cos α )sin α − τ yx (dA sin α ) cos α = 0 (30) Slide No. 15 LECTURE 7. SHAFTS: STATICALLY INDETERMINATE SHAFTS (3.6) ENES 220 ©Assakkaf Stresses in Oblique Planes Other Stresses Induced By Torsion t ∑F t =0 σ n dA − τ xy (dA cos α )sin α − τ yx (dA sin α ) cos α = 0 From which σ n = 2τ xy sin α cos α = τ xy sin 2α (31) n τn t dA τx y dA cos α + y α α σn dA τyx dA sin α Fig. 16c 8 Slide No. 16 LECTURE 7. SHAFTS: STATICALLY INDETERMINATE SHAFTS (3.6) Stresses in Oblique Planes ENES 220 ©Assakkaf Maximum Normal Stress due to Torsion on Circular Shaft The maximum compressive normal stress σmax can be computed from σ max = τ max = Tmax c J (32) LECTURE 7. SHAFTS: STATICALLY INDETERMINATE SHAFTS (3.6) Stresses in Oblique Planes Slide No. 17 ENES 220 ©Assakkaf Example 4 A cylindrical tube is fabricated by butt-welding a 6 mm-thick steel plate along a spiral seam as shown. If the maximum compressive stress in the tube must be limited to 80 MPa, determine (a) the maximum torque T that can be applied and (b) the factor of safety with respect to the failure by fracture for the weld, when a torque of 12 kN.m is applied, if the ultimate strengths of the weld metal are 205 MPa in shear and 345 MPa in tension. 9 Slide No. 18 LECTURE 7. SHAFTS: STATICALLY INDETERMINATE SHAFTS (3.6) Stresses in Oblique Planes ENES 220 ©Assakkaf Example 4 (cont’d) T T Weld 300 60 150 mm 30 Fig. 18 Slide No. 19 LECTURE 7. SHAFTS: STATICALLY INDETERMINATE SHAFTS (3.6) Stresses in Oblique Planes ENES 220 ©Assakkaf Example 4 (cont’d) (a) The polar moment of area for the cylindrical tube can be determined from Eq.14 as J= π (r 2 4 o ) − ri 4 = 4 4 τ 150 150 − 6 − 2 2 2 6 4 = 14.096 × 10 mm The maximum torque can be computed from Eq. 32 as σ max = σ J 80 × 106 (14.096 × 10 −6 ) Tmax c ⇒ Tmax = max = 75 ×10 −3 J c 3 = 15.036 × 10 N ⋅ m = 15.036 kN ⋅ m 10 LECTURE 7. SHAFTS: STATICALLY INDETERMINATE SHAFTS (3.6) Stresses in Oblique Planes Slide No. 20 ENES 220 ©Assakkaf Example 4 (cont’d) (b)The normal stress σn and shear stress τnt on the weld surface are given by Eqs. 30 and 29 as Tc 12 ×103 (75 × 10 −3 ) sin 2α = sin 2(60 0 ) = 55.29 MPa (T) J 14.096 × 10 −6 Tc 12 × 103 (75 ×10 −3 ) τ nt = τ xy cos 2α = cos 2α = cos 2(60 0 ) = −31.92 MPa J 14.096 ×10 −6 σ n = τ xy sin 2α = LECTURE 7. SHAFTS: STATICALLY INDETERMINATE SHAFTS (3.6) Stresses in Oblique Planes Slide No. 21 ENES 220 ©Assakkaf Example 4 (cont’d) The factors of safety with respect to failure by fracture for the weld are 345 σ ult = = 6.24 σ n 55.29 τ 205 FSτ = ult = = 6.42 τ tn 31.92 FSσ = 11 LECTURE 7. SHAFTS: STATICALLY INDETERMINATE SHAFTS (3.6) Statically Indeterminate Shafts Slide No. 22 ENES 220 ©Assakkaf Up to this point, all problems discussed are statically determinate, that is, only the equations of equilibrium were required to determine the torque T at any section of the shaft. It is often for torsionally loaded members to be statically indeterminate in real engineering applications. LECTURE 7. SHAFTS: STATICALLY INDETERMINATE SHAFTS (3.6) Statically Indeterminate Shafts Slide No. 23 ENES 220 ©Assakkaf When this occurs, distortion equations involving angle of twist θ must written until the total number of equations agrees with the number of unknowns to be determined. A simplified angle of twist diagram will often be of great assistance in obtaining the correct equations. 12 Slide No. 24 LECTURE 7. SHAFTS: STATICALLY INDETERMINATE SHAFTS (3.6) Statically Indeterminate Shafts ENES 220 ©Assakkaf Example 5 A steel shaft and aluminum tube are connected to a fixed support and to a rigid disk as shown in the figure. Knowing that the initial stresses are zero, determine the minimum torque T0 that may be applied to the disk if the allowable stresses are 120 MPa in the steel shaft and 70 MPa in the aluminum tube. Use G = 80 GPa for steel and G = 27 GPa for aluminum. Slide No. 25 LECTURE 7. SHAFTS: STATICALLY INDETERMINATE SHAFTS (3.6) Statically Indeterminate Shafts ENES 220 ©Assakkaf Example 5 (cont’d) Aluminum 8 mm 76 mm Rigid disk 50 mm Steel 500 mm 13 Slide No. 26 LECTURE 7. SHAFTS: STATICALLY INDETERMINATE SHAFTS (3.6) Statically Indeterminate Shafts ENES 220 ©Assakkaf Example 5 (cont’d) – Free-body diagram for the rigid disk Tal T0 From statics, T0 = Tal + Tst T st – Deformation (39) Tal Lal Tst Lst = J alGal J st Gst θ al = θ st ⇒ (40) Slide No. 27 LECTURE 7. SHAFTS: STATICALLY INDETERMINATE SHAFTS (3.6) Statically Indeterminate Shafts ENES 220 ©Assakkaf Example 5 (cont’d) – Properties of the aluminum tube Gal = 27 GPa 38 mm ri = 30 mm = 0.030 m ro = 38 mm = 0.038 m 30 mm 76 mm J al = [(0.038) − (0.030) ] = 2 π 4 4 = 2.003 × 10 −6 m 4 14 Slide No. 28 LECTURE 7. SHAFTS: STATICALLY INDETERMINATE SHAFTS (3.6) Statically Indeterminate Shafts ENES 220 ©Assakkaf Example 5 (cont’d) – Properties of the steel tube Gst = 80 GPa 25 mm c = 25 mm = 0.025 m J st = 50 mm [(0.025) ] = 2 π 4 = 0.6136 × 10 −6 m 4 LECTURE 7. SHAFTS: STATICALLY INDETERMINATE SHAFTS (3.6) Statically Indeterminate Shafts Slide No. 29 ENES 220 ©Assakkaf Example 5 (cont’d) Substituting these input values in Eq. 40, gives Tal Lal Tst Lst = J alGal J st Gst Tal (0.5) Tst (0.5) = −6 2.003 × 10 (27) 0.6136 ×10 −6 (80) Tst = 0.908Tal (41) 15 LECTURE 7. SHAFTS: STATICALLY INDETERMINATE SHAFTS (3.6) Statically Indeterminate Shafts Slide No. 30 ENES 220 ©Assakkaf Example 5 (cont’d) Let’s assume that the requirement τst is less or to equal to 120 MPa, therefore Tst = τ st J st cst = 120 × 106 (0.6136 × 10 −6 ) = 2945 N ⋅ m 0.025 From Eq. 39, we have Tst = 0.908Tal 2945 = 0.908Tal ⇒ Tal = 3244 N ⋅ m LECTURE 7. SHAFTS: STATICALLY INDETERMINATE SHAFTS (3.6) Statically Indeterminate Shafts Slide No. 31 ENES 220 ©Assakkaf Example 5 (cont’d) Let’s check the maximum stress τal in aluminum tube corresponding to Tal = 3244 N·m: Tc 3244(0.038) τ al = al al jal = 2.003 × 10 −6 = 61.5 MPa < 70 MPa OK Hence, the max permissible torque T0 is computed from Eq. 39 as T0 = Tal + Tst = 3244 + 2945 = 6189 N ⋅ m = 6.2 kN ⋅ m 16 Slide No. 32 LECTURE 7. SHAFTS: STATICALLY INDETERMINATE SHAFTS (3.6) Statically Indeterminate Shafts ENES 220 ©Assakkaf Example 6 A circular shaft AB consists of a 10-in-long, 7/8 indiameter steel cylinder, in which a 5-in.-long, 5/8-in.diameter cavity has been drilled from end B. The shaft is attached to fixed supports at both ends, and a 90 lb – ft torque is applied at its mid-section. Determine the torque exerted on the shaft by each of the supports. Slide No. 33 LECTURE 7. SHAFTS: STATICALLY INDETERMINATE SHAFTS (3.6) Statically Indeterminate Shafts ENES 220 ©Assakkaf Example 6 • Given the shaft dimensions and the applied torque, we would like to find the torque reactions at A and B. • From a free-body analysis of the shaft, TA + TB = 90 lb ⋅ ft which is not sufficient to find the end torques. The problem is statically indeterminate. • Divide the shaft into two components which must have compatible deformations, φ = φ1 + φ2 = TA L1 TB L2 − =0 J1G J 2G LJ TB = 1 2 TA L2 J1 • Substitute into the original equilibrium equation, LJ TA + 1 2 TA = 90 lb ⋅ ft L2 J1 17 ...
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