To determine The principal stresses at point A . The maximum in-plane shear stress developed at point A . The orientation of the plane for maximum in-plane shear stress. To sketch: The results on an element located at point A . Explanation Given information: Diameter of the shaft is 2 in. Calculation: Let N , V , and M be the normal force, shear force, and moment at the left cut segment of the shaft. Sketch the free body diagram of the left cut segment of the shaft as shown in Figure 1. Apply equations of equilibrium. Sum of forces in the x -direction is equal to 0. Σ F x = 0 N − 3 , 000 lb = 0 N = 3 , 000 lb Sum of forces in the y -direction is equal to 0. Σ F y = 0 − V + 75 lb = 0 V = 75 lb Sum of moments at C is equal to 0. Σ M C = 0 ( 75 lb ) ( 24 in . ) − M = 0 M = 1 , 800 lb ⋅ in . Find the cross sectional area ( A ) of the 1 in. radius of the shaft: A = π ( 1 in . ) 2 = 3.1416 in 2 Find the moment of inertia ( I ) of the 1 in. radius of the shaft: I = π 4 ( 1 in . ) 4 = 0.7854 mm 4 Find the first moment of area at point A ( Q A ) : The value of y ¯ ' is 0. Therefore, Q A = 0 Refer the cross section, The centroid distance of the section ( y ) to point A is 1 in . Find the normal stress at point A ( σ A ) using the relation: σ A = N A − M y I = 3 , 000 lb 3.1416 in 2 − ( 1 , 800 lb ⋅ in . ) ( 1 in . ) 0.7857 in 4 = − 1.336 × 10 3 psi = − 1.34 ksi There is no shear stress acting on the element. Therefore the value of the shear stress at point A ( τ A ) is 0. The value of normal stresses are σ x = − 1.34 ksi and σ y = 0 . The value of shear stress is τ x y = 0 . Sketch the state of stress at point A on the element as shown in Figure 2. Find the in-plane principal stresses ( σ 1 ) and ( σ 2 ) using the relation: The principal stress σ 1 = σ y = 0 . The principal stress σ 2 = σ x = − 1.34 ksi . Therefore, the principal stresses at point A ( σ 1 ) and ( σ 2 ) are 0 _ and − 1.34 ksi _ . Find the value of maximum in-plane shear stress ( τ max in-plane ) using the relation: τ max in-plane = ( σ x − σ y 2 ) 2 + τ x y 2 (1) Substitute − 1

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 9.3, Problem 9.34P

To determine

The principal stresses at point A.

The maximum in-plane shear stress developed at point A.

The orientation of the plane for maximum in-plane shear stress.

To sketch:

The results on an element located at point A.

Expert Solution & Answer

Want to see the full answer?

Check out a sample textbook solution

See solution

Chapter 9.3, Problem 9.33P

Chapter 9.3, Problem 9.35P

Students have asked these similar questions

Q6] (20 Marks) Select the most suitable choice for the following statements: modo digi -1A 10 af5 1 -The copper-based alloy which is responded to age hardening is a) copper-nickel b) aluminum bronze c) copper - beryllium d) brass besincaluy 2- Highly elastic polymers may experience elongations to greater than.... b) 500% bromsia-P c) 1000%. d) 1200% 15m or -2 a)100% 3- The cooling rate of quenching the steel in saltwater will be ......the cooling rate of quenching ir c) faster than sold) none of them a) slower than 4- Adding of a) Cr b) the same as ...... Will lead to stabilize the b) Mo 10 austenite in steel. c) Nimble avolls 1d) Sized loloin nl 5- The adjacent linear chains of crosslinked polymers are joined one to another at various positic DIR... by.........bonds c) covalent noisqo gd) ionic lg 120M 6- For the ceramic with coordination number 6 the cation to anion radius ratio will be a) Van der Waals a) 0.155-0.225 a) linear b) hydrogen (b) 0.225-0.414 c) 0.414 0.732 ..polymers.…

Examine Notes: Attempt Six Questions Only. rever necessa , Q1] (20 Marks) Answer with true (T) or false (F), corrects the wrong phrases, and gives sho reasons for correct and corrected statements: 1- High chromium irons are basically grey cast irons alloyed with 12 to 30 % Cr. yous board-19qgo orT-1 2- The drawbacks of Al- Li alloys are their high young modulus and high density.&M 0) (0 3- Vulcanized rubbers are classified under thermoplastic polymers. 4- Diamond is a stable carbon polymorph at room temperature and atmospheric pressure. ( 5- The metallic ions of ceramic are called anions, and they are positively charged. yldgiH-S 69001(6

H.W 5.4 Calculate the load that will make point A move to the left by 6mm, E-228GPa. The diameters of the rods are as shown in fig. below. 2P- PA 50mm B 200mm 2P 0.9m 1.3m

Chapter 9 Solutions

Mechanics of Materials

Ch. 9.3 - In each case, the state of stress x, y, xy...Ch. 9.3 - Given the state of stress shown on the element,...Ch. 9.3 - Determine the normal stress and shear stress...Ch. 9.3 - Determine the equivalent state of stress on an...Ch. 9.3 - Also, find the corresponding orientation of the...Ch. 9.3 - Determine the equivalent state of stress on an...Ch. 9.3 - Determine the maximum principal stress at point B.Ch. 9.3 - Determine the principal stress at point C.Ch. 9.3 - Prove that the sum of the normal stresses x + y =...Ch. 9.3 - 9-2. The state of stress at a point in a member 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.

The principal stresses at point A . The maximum in-plane shear stress developed at point A . The orientation of the plane for maximum in-plane shear stress. To sketch: The results on an element located at point A .

Concept explainers

Videos

The principal stresses at point A. The maximum in-plane shear stress developed at point A. The orientation of the plane for maximum in-plane shear stress. To sketch: The results on an element located at point A.

Want to see the full answer?

Chapter 9 Solutions

The principal stresses at point A.

The maximum in-plane shear stress developed at point A.

The orientation of the plane for maximum in-plane shear stress.

To sketch:

The results on an element located at point A.