To determine The average normal stress developed in the bolt shank and sleeve ( σ b and σ s ) . Explanation Given information: Initially the A-36 bolt shank fits snugly against the rigid caps E and F on the 6061-T6 aluminum sleeve. The modulus of elasticity of the steel ( E st ) is 200 × 10 3 N/mm 2 . The modulus of elasticity of the aluminum ( E al ) is 68.9 × 10 3 N/mm 2 . The diameter of the bolt shank ( d b ) is 60 mm . The outer diameter of the sleeve ( D s ) is 250 mm . The thickness of the sleeve ( t ) is 15 mm . The length of the bolt shank ( L b ) is 500 mm . The length of the sleeve ( L s ) is 450 mm . Calculation: Show the free-body diagram of the cut part of the assembly as in Figure 1. Refer Figure 1. Here, F b is force of the bolt shank and F s is force of the sleeve. Calculate the vertical forces by applying the equation of equilibrium: Sum of vertical forces is equal to 0. ∑ F y = 0 F s − F b = 0 (1) When the nut is tightened 3 4 of a turn, the unconstrained bolt will be shortened by deformation due to bolt shank ( δ b ) . Therefore, δ b = 3 4 × 1 = 0.75 mm Calculate the area of the bolt shank ( A b ) : A b = π 4 d b 2 (2) Substitute 60 mm for d b in Equation (2). A b = π 4 × 60 2 = 2827.43 mm 2 Calculate the inner diameter area of the sleeve ( d s ) : d s = D s − 2 t (3) Substitute 250 mm for D s and 15 mm for t in Equation (3). d s = 250 − ( 2 × 15 ) = 220 mm Calculate the area of the sleeve ( A s ) : A s = π 4 ( D s 2 − d s 2 ) (4) Substitute 250 mm for D s and 220 mm for d s in Equation (4). A s = π 4 × ( 250 2 − 220 2 ) = 11074.11 mm 2 Show the initial and final position of the assembly as in Figure 2. Refer Figure 2. Here, δ b is deformation due to bolt shank, δ F b is deformation due to F b , and δ F s is deformation due to F s . Calculate the forces of the bolt shank and sleeve ( F b and F s ) using the deformation: δ b − δ F b = δ F b δ b − F b L b A b E st = F s L s A s E al (5) Substitute 0.75 mm for δ b , 500 mm for L b , 450 mm for L s , 2827

9th Edition

ISBN: 9780133254426

Author: Russell C. Hibbeler

Publisher: Prentice Hall

See similar textbooks

Concept explainers

Combined Loading

Any functioning component, such as machine parts, structural members, pressure vessels, and so on, are all under the influence of more than one force. When anybody is under the influence of more than one force such as shear forces, bending moments, axial…

Videos

Question

Chapter 4.5, Problem 4.65P

To determine

The average normal stress developed in the bolt shank and sleeve (σb and σs) .

Expert Solution & Answer

Want to see the full answer?

Check out a sample textbook solution

See solution

Chapter 4.5, Problem 4.64P

Chapter 4.5, Problem 4.66P

Students have asked these similar questions

Example-1: l D A uniform rotor of length 0.6 m and diameter 0.4 m is made of steel (density 7810 kg/m³) is supported by identical short bearings of stiffness 1 MN/m in the horizontal and vertical directions. If the distance between the bearings is 0.7 m, determine the natural frequencies and plot whirl speed map. Solution: B

find the laplace transform for the flowing function 2(1-e) Ans. F(s)=- S 12) k 0 Ans. F(s)= k s(1+e) 0 a 2a 3a 4a 13) 2+ Ans. F(s)= 1 s(1+e") 3 14) f(t)=1, 0

Find the solution of the following Differential Equations Using Laplace Transforms 1) 4y+2y=0. y(0)=2. y'(0)=0. 2) y+w²y=0, (0)=A, y'(0)=B. 3) +2y-8y 0. y(0)=1. y'(0)-8. 4)-2-3y=0, y(0)=1. y'(0)=7. 5) y-ky'=0, y(0)=2, y'(0)=k. 6) y+ky'-2k²y=0, y(0)=2, y'(0) = 2k. 7) '+4y=0, y(0)=2.8 8) y+y=17 sin(21), y(0)=-1. 9) y-y-6y=0, y(0)=6, y'(0)=13. 10) y=0. y(0)=4, y' (0)=0. 11) -4y+4y-0, y(0)=2.1. y'(0)=3.9 12) y+2y'+2y=0, y(0)=1, y'(0)=-3. 13) +7y+12y=21e". y(0)=3.5. y'(0)=-10. 14) "+9y=10e". y(0)=0, y'(0)=0. 15) +3y+2.25y=91' +64. y(0)=1. y'(0) = 31.5 16) -6y+5y-29 cos(2t). y(0)=3.2, y'(0)=6.2 17) y+2y+2y=0, y(0)=0. y'(0)=1. 18) y+2y+17y=0, y(0)=0. y'(0)=12. 19) y"-4y+5y=0, y(0)=1, y'(0)=2. 20) 9y-6y+y=0, (0)-3, y'(0)=1. 21) -2y+10y=0, y(0)=3, y'(0)=3. 22) 4y-4y+37y=0, y(0)=3. y'(0)=1.5 23) 4y-8y+5y=0, y(0)=0, y'(0)=1. 24) ++1.25y-0, y(0)=1, y'(0)=-0.5 25) y 2 cos(r). y(0)=2. y'(0) = 0. 26) -4y+3y-0, y(0)=3, y(0) 7. 27) y+2y+y=e y(0)=0. y'(0)=0. 28) y+2y-3y=10sinh(27), y(0)=0. y'(0)=4. 29)…

Chapter 4 Solutions

Mechanics of Materials

Ch. 4.2 - In each case, determine the internal normal force...Ch. 4.2 - Determine the internal normal force between...Ch. 4.2 - The post weighs 8kN/m. Determine the internal...Ch. 4.2 - The rod is subjected to an external axial force of...Ch. 4.2 - The rigid beam supports the load of 60 kN....Ch. 4.2 - The 20-mm-diameter A-36 steel rod is subjected to...Ch. 4.2 - Segments AB and CD of the assembly are solid...Ch. 4.2 - The 30-mm-diameter A992 steel rod is subjected to...Ch. 4.2 - If the 20-mm-diameter rod is made of A-36 steel...Ch. 4.2 - The 20-mm-diameter 2014-T6 aluminum rod is...

Knowledge Booster

Learn more about

Need a deep-dive on the concept behind this application? Look no further. Learn more about this topic, mechanical-engineering and related others by exploring similar questions and additional content below.