Stress Analysis Lecture 3 ME 276 Spring Dr./ Ahmed Mohamed Nagib Elmekawy

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1 Stress Analysis Lecture 3 ME 276 Spring Dr./ Ahmed Mohamed Nagib Elmekawy

2 Axial Stress 2

3 Beam under the action of two tensile forces 3

4 Beam under the action of two tensile forces 4

5 Shear Stress Average Shear Stress V A V A is the average shear stress at the section is internal resultant shear force at the section determined from the equations of equilibrium is the area at the section

6 Stress Single Shear Double Shear P A F A F 2A

7 Stress Example 1 Determine the average shear stress in the 20-mm-diameter pin at A and the 30-mm diameter pin at B that support the beam in the attached figure. kn A A F kn A A F kn F F M y y y x x x B B A *

8 Stress The resultant force acting on pin A is kn 2 2 FA Ax Ay 36 FA 21.36*1000 A 34 MPa 2A 2 A 2* 20 4 double shear F A 21.5*1000 B B B 4 30 MPa single shear

9 Stress Allowable Stress An engineer on charge of the design of a structural or mechanical element must restrict the stress in the material to a level that will be safe. So it becomes necessary to perform the calculations using a safe or allowable stress. To ensure safety, it is necessary to choose an allowable stress that restrict the applied load to one that is less than the load the member can fully support. One method of specifying the allowable load for the design or analysis of a member is to use a number called the factor of safety. f. s. fail allow f. s. fail allow

10 Stress Example 1 The control arm is subjected to the loading shown in the figure. Determine to the nearest ¼ in. the required diameter of the steel pin at C if the allowable shear stress for the steel is allow = 8 ksi.

11 Stress kip F F M AB AB C 3 0 * *3 *8 0 kip C C F x x x kip C C F y y y kip C C F y x C

12 Stress C FC 2A allow A 8 A in 2 2 A d 4 d in d 4A 4* in Use a pin having a diameter of d in

13 Torsion Stress 13

14 Torsion Stress 14

15 Torsion Stress 15

16 Torsion Stress 16

17 Torsional Deformation of a Circular Shaft

18 Angle of Twist 18

19 The Torsion Formula If the material is linear-elastic, then Hook s law applies, = G, and consequently a linear variation in shear strain, leads to a linear variation in shear stress along any radial line on the cross section. T J and max Tc J Where T J c max the resultant internal torque acting at the cross section. the polar moment of inertia of the cross-sectional area. the outer radius of the shaft. the maximum shear stress in the shaft, which occurs at the outer surface. the radius

20 The Torsion Formula J 2 d 32 4 c 4 4 J c o ci d 4 o d 4 i

21 Shearing Strain 21

22 Torque stress summary Shear Stress τ = Gγ= Tr = 16T J π d 3 τ : Shear Stress r : shaft radius G: Modulus of rigidity d : Shaft diameter γ: Shear Strain J : Polar moment of inertia For a hollow circular shaft of inner radius c 1 and outer radius c 2,the polar moment of inertia is J = π 32 d o 4 d i 4 : Angle of Twist 22

23 Torsion of a Shaft with Circular Cross-Section Finite Element Analysis 23

24 Torsion of a Shaft with Circular Cross-Section Finite Element Analysis 24

25 Torsion of a Beam with the Square Cross-Section Finite Element Analysis 25

26 Torsion of a Beam with the Square Cross-Section Finite Element Analysis 26

27 The Torsion Formula Example 1 The shaft shown in the attached figure is supported by two bearings and is subjected to three torques. Determine the shear stress developed at points A and B located at section a-a of the shaft. The shaft diameter is 75 mm kn.m 3 kn.m 1.25 kn.m Torque diagram

28 The Torsion Formula A Tr J 1.25*1000*1000*750 / MPa B Tr J 1.25*1000*1000*150 / MPa

29 The Torsion Formula Example 2 The pipe shown in the attached figure has an inner diameter of 80 mm and an outer diameter of 100 mm. If its end is tightened against the support at A using a torque wrench at B, determine the shear stress developed in the material at the inner and outer walls along the central portion of the pipe when the 80-N forces are applied to the wrench.

30 The Torsion Formula M y 0; 80*0.3 80*0.2 T T 40 N.m 0 o Tc o J * MPa

31 The Power Transmission i Tc i J 40000* The Power Transmission MPa 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 a torque that depends on the power generated by the machine and the angular speed of the shaft. P T

32 The Power Transmission Example 1 A solid steel shaft AB shown in the figure is to be used to transmit 3750 W from the motor M to which it is attached. If the shaft rotates at = 175 rpm and the steel has an allowable shear stress of allow = 100 MPa determine the required diameter of the shaft to the nearest mm. P T T max 3750* N.mm 2 * d T Tc 2 16T allow J 4 d d 32 3

33 The Angle of Twist d Angle of Twist 100 d mm d 3 22 mm Occasionally the design of a shaft depends on restricting the amount of rotation or twist that may occur when the shaft is subjected to a torque. Furthermore, being able to compute the angle of twist for a shaft is important when analyzing the reactions on statically indeterminate shafts.

34 The Angle of Twist d dx; d T J x x G dx G and L 0 T J T J x x G dx Constant Torque and Cross-Sectional Area TL JG The similarities between the above equations and those for an axially loaded members should be noted.

35 The Angle of Twist The equation of angle of twist is often used to determine the shear modulus of elasticity G of a material. To do so, a specimen of known length and diameter is placed in a testing machine like shown in the attached figure. The applied torque T and angle of twist are then measured along the length L. Multiple Torques TL JG

36 The Angle of Twist Sign Convention A/ D TL JG 80000L 70000L JG AB JG BC JG L CD

37 The Angle of Twist Example 1 The gears attached to the fixed-end steel shaft are subjected to the torques shown in the figure. If the shear modulus of elasticity is 80 GPa and the shaft has a diameter of 14 mm, determine the displacement of tooth P on gear A. The shaft turns freely within the bearing at B.

38 The Angle of Twist A TL JG rad * * * Since the answer is negative, by the right hand rule the thumb is directed toward the end E of the shaft and therefore gear A will rotate as shown in the attached figure. The displacement of tooth P on gear A is s P r A * mm

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