Explanation Given information: The load at x − x cross-section is 300 kN . Diameter of aluminium cylinder is d = 200 mm . Angle of inclination of a − a section with x − x section is θ = 30 ° . Free body diagram of the cylinder at x − x section: Compressive stress in plate is given by using the formula: σ C = − P A c y l i n d e r = − 300 kN × 1000 N kN π 4 d x − x 2 = − 4 × 300000 N π × 200 2 = − 9.55 MPa Calculation: Free body diagram of the cylinder at a − a section: For the normal stress at a − a section: σ n a − a = σ C cos 2 θ σ n a − a = − 9

11th Edition

ISBN: 9780137605521

Author: HIBBELER

Publisher: RENT PEARS

See similar textbooks

Concept explainers

Axial Load

When a bar of the uniform cross-section is acted by a load, such that the load acts along the longitudinal axis of the bar, such kinds of loadings are known as axial loadings. In general terms, a load is an external force acting on a component. An axial …

Videos

Question

Chapter 1.7, Problem 81P

To determine

The normal and shear stress at a−a section.

Expert Solution & Answer

Want to see the full answer?

Check out a sample textbook solution

See solution

Chapter 1.7, Problem 80P

Chapter 1.7, Problem 82P

Students have asked these similar questions

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)…

Auto Controls A union feedback control system has the following open loop transfer function where k>0 is a variable proportional gain i. for K = 1 , derive the exact magnitude and phase expressions of G(jw). ii) for K = 1 , identify the gaincross-over frequency (Wgc) [where IG(jo))| 1] and phase cross-overfrequency [where <G(jw) = - 180]. You can use MATLAB command "margin" to obtain there quantities. iii) Calculate gain margin (in dB) and phase margin (in degrees) ·State whether the closed-loop is stable for K = 1 and briefly justify your answer based on the margin . (Gain marginPhase margin) iv. what happens to the gain margin and Phase margin when you increase the value of K?you You can use for loop in MATLAB to check that.Helpful matlab commands : if, bode, margin, rlocus NO COPIED SOLUTIONS

Chapter 1 Solutions

MECHANICS OF MATERIALS

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.

The normal and shear stress at a − a section.

Solution Summary: The author explains the normal and shear stress at a-a section.

Concept explainers

Videos

The normal and shear stress at a−a section.

Want to see the full answer?

Chapter 1 Solutions