SHAFTS: STATICALLY INDETERMINATE SHAFTS

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1 SHAFTS: STATICALLY INDETERMINATE SHAFTS Up to this poit, the resses i a shaft has bee limited to shearig resses. This due to the fact that the selectio of the elemet uder udy was orieted i such a way that its faces were either perpedicular or parlel to the axis of the shaft (see Fig. 15)

2 Fig. 15 τ τ Axis of shaft From our discussio of the torsio loadig o a shaft, we kow this loadig produces shearig resses τ i the faces perpedicular to the axis of the shaft. But due to equilibrium requiremet, there are equ resses o the faces formed by the two plaes cotaiig the axis of the shaft.

3 It is ecessary to make sure that whether the trasverse plae is a plae of maximum shearig ress ad whether there are other sigificat resses iduced by torsio. Cosider the followig shaft (Fig. 16), which is subjected to a torque T. Fig. 16 t y (c) τ t da τ x y da cos σ da x τ yx da si y τ τ yx τ yx y A τ (b) x (a) x

4 The resses at poit A i the shaft of Fig. 16a is ayzed. A differeti elemet take from the shaft at poit A ad the resses actig o trasverse ad logitudi plaes are show i Fig. 16b. The shearig ress τ ca be determied from τ Tc Let assume that differeti elemet of Fig. 16b has legth dx, height dy, ad thickess dz. If a shearig force V x τ dx dy is applied to the top surface of the elemet, the equatio of equilibrium F x 0 the will require applicatio of a opposite shear force V x at the bottom of the elemet.

5 V x τ yx dx dz y V y τ dy dz V x V y z V y x y z V x x Fig. 17 If F x 0 the requires applicatio of a opposite shear force V x at the bottom of the elemet, the it will the elemet subjected to a clockwise couple. This clockwise couple mu be baced by couterclockwise couple composed of V x applied to the vertic faces of the elemet.

6 The applicatio of the equilibrium momet equatio M z 0 gives yx ( dx dz) dy ( dy dz)dx τ τ From which the importat result τ yx τ (27) If the equatios of equilibrium are applied to the free-body diagram of Fig. 16c (which is a wedge-shaped part of the differeti elemet of Fig. 16b with da beig the area of the iclied face), the followig results are obtaied + Ft 0 τ da τ t ( dacos ) cos + τ ( dasi ) si 0 yx (28)

7 Fig. 16 t y (c) τ t da τ x y da cos σ da x τ yx da si y τ τ yx τ yx y A τ (b) x (a) x + Ft 0 τ da τ t From which τ τ t ( dacos ) cos + τ ( dasi ) si ( cos si ) τ cos 2 (29) yx t τ x y da cos y τ t da σ da Fig. 16c τ yx da si

8 Likewise, if we take summatio of forces i the directio (see Fig. 16c), the the results would be + F 0 σ da τ ( dacos ) si τ ( dasi ) cos 0 yx (30) + Ft 0 σ da τ From which y t ( dacos ) si τ ( dasi ) σ 2τ si cos τ si 2 yx cos 0 (31) τ x y da cos τ t da σ da Fig. 16c τ yx da si

9 Maximum Norm Stress due to Torsio o Circular Shaft The maximum compressive orm ress σ max ca be computed from T max c σ max τ max (32) Slide No. 17 Example 4 A cylidric tube is fabricated by butt-weldig a 6 mm-thick eel plate og a spir seam as show. If the maximum compressive ress i the tube mu be limited to 80 MPa, determie (a) the maximum torque T that ca be applied ad (b) the factor of safety with respect to the failure by fracture for the weld, whe a torque of 12 kn.m is applied, if the ultimate regths of the weld met are 205 MPa i shear ad 345 MPa i tesio.

10 Example 4 (cot d) 60 T 30 0 Weld T 30 Fig mm Example 4 (cot d) σ (a) The polar momet of area for the cylidric tube ca be determied from Eq.14 as 4 4 π τ ( ) mm r o r i The maximum torque ca be computed from Eq. 32 as T c 2 6 6) σ max ( c N m kn m max max Tmax

11 Example 4 (cot d) (b)the orm ress σ ad shear ress τ t o the weld surface are give by Eqs. 30 ad 29 as σ τ τ τ t 3 3 Tc (75 10 ) 0 si 2 si 2 si 2(60 ) MPa (T) Tc (75 10 ) 0 cos 2 cos 2 cos 2(60 ) MPa Example 4 (cot d) The factors of safety with respect to failure by fracture for the weld are FS FS σ τ σ ult σ τ ult τ t

12 Staticly Idetermiate Shafts Up to this poit, l problems discussed are aticly determiate, that is, oly the equatios of equilibrium were required to determie the torque T at ay sectio of the shaft. It is ofte for torsioly loaded members to be aticly idetermiate i re egieerig applicatios. Staticly Idetermiate Shafts Whe this occurs, diortio equatios ivolvig agle of twi θ mu writte util the tot umber of equatios agrees with the umber of ukows to be determied. A simplified agle of twi diagram will ofte be of great assiace i obtaiig the correct equatios.

13 Staticly Idetermiate Shafts Example 5 A eel shaft ad umium tube are coected to a fixed support ad to a rigid disk as show i the figure. Kowig that the iiti resses are zero, determie the miimum torque T 0 that may be applied to the disk if the lowable resses are 120 MPa i the eel shaft ad 70 MPa i the umium tube. Use G 80 GPa for eel ad G 27 GPa for umium. Staticly Idetermiate Shafts Example 5 (cot d) 8 mm Alumium Rigid disk 76 mm 50 mm Steel 500 mm

14 Staticly Idetermiate Shafts Example 5 (cot d) Free-body diagram for the rigid disk T 0 T T Deformatio From atics, T T + T 0 (39) θ θ T L G T L G (40) Staticly Idetermiate Shafts Example 5 (cot d) Properties of the umium tube 38 mm 30 mm 76 mm G 27 GPa r 30 mm m i r o 38 mm m 4 4 [( 0.038) ( 0.030) ] π m 4

15 Staticly Idetermiate Shafts Example 5 (cot d) Properties of the eel tube 25 mm 50 mm G c 25 mm m 80 GPa 4 [( 0.025) ] π m 4 Staticly Idetermiate Shafts Example 5 (cot d) Subitutig these iput vues i Eq. 40, gives ( 0.5) T T T L G T T L G (27) T (0.5) (80) (41)

16 Staticly Idetermiate Shafts Example 5 (cot d) Let s assume that the requiremet τ is less or to equ to 120 MPa, therefore T τ c ( From Eq. 39, we have 6 ) 2945 N m T 0.908T T T 3244 N m Staticly Idetermiate Shafts Example 5 (cot d) Let s check the maximum ress τ i umium tube correspodig to T 3244 N m: Tc 3244(0.038) τ 61.5 MPa < j 70 MPa OK Hece, the max permissible torque T 0 is computed from Eq. 39 as T0 T + T N m 6.2 kn m

Staticly Idetermiate Shafts Example 6 A circular shaft AB cosis of a 10-i-log, 7/8 idiameter eel cylider, i which a 5-i.-log, 5/8-i.- diameter cavity has bee drilled from ed B.

Staticly Idetermiate Shafts Example 6 Give the shaft dimesios ad the applied torque, we would like to fid the torque reactios at A ad B.

17 Staticly Idetermiate Shafts Example 6 A circular shaft AB cosis of a 10-i-log, 7/8 idiameter eel cylider, i which a 5-i.-log, 5/8-i.- diameter cavity has bee drilled from ed B. The shaft is attached to fixed supports at both eds, ad a 90 lb ft torque is applied at its mid-sectio. Determie the torque exerted o the shaft by each of the supports. Staticly Idetermiate Shafts Example 6 Give the shaft dimesios ad the applied torque, we would like to fid the torque reactios at A ad B. From a free-body aysis of the shaft, T A + T B 90lb ft which is ot sufficiet to fid the ed torques. The problem is aticly idetermiate. Divide the shaft ito two compoets which mu have compatible deformatios, TAL TBL L φ φ 1 2 T φ2 0 B TA 1G 2G L21 Subitute ito the origi equilibrium equatio, L + 1 T 2 A TA 90lb ft L21

SHAFTS: STATICALLY INDETERMINATE SHAFTS

LECTURE SHAFTS: STATICALLY INDETERMINATE SHAFTS Third Editio A.. Clark School of Egieerig Departmet of Civil ad Eviromet Egieerig 7 Chapter 3.6 by Dr. Ibrahim A. Assakkaf SPRING 2003 ENES 220 Mechaics