Explanation Calculate the normal stress ( σ A ) acting at point A using the relation: σ A = M y z I y (1) Here, M y is the moment in y -direction, z is the distance in z -direction from centroid to point A , and I y is the moment of inertia. Sketch the internal forces and moment in free body diagram at point A as shown in Figure 1. Apply Equilibrium equations to find the value of moment at point A . Sum of moments in y -direction is equal to 0. Σ M y = 0 M y − ( 20 × 10 ) = 0 M y = 200 lb ⋅ in Sum of moments in x -direction is equal to 0. Σ M x = 0 T x + ( 20 × 12 ) = 0 T x = 240 lb ⋅ in Sum of forces in z -direction is equal to 0. Σ V z = 0 V z − 20 = 0 V z = 20 lb Find the moment of inertia of the section ( I ) : Outer radius of the pipe is 1.5 in. and the inner radius of the pipe is 1.375 in. I = π 4 [ ( 1.5 in ) 4 − ( 1.375 in ) 4 ] = 1.1687 in 4 Find the polar moment of inertia of the section ( J ) : J = π 2 [ ( 1.5 in ) 4 − ( 1.375 in ) 4 ] = 2.3374 in 4 Sketch the cross section at point A as shown in Figure 2. Find the first moment of area at point A ( Q A ) z using the relation: ( Q A ) z = Σ y ¯ ' A ' (2) Here, y ¯ ' is the centroid distance to point A and A ' is the area from centroid to point A . Refer to Figure 2. ( Q A ) z = 4 × 1.5 in . 3 π [ 1 2 × π × ( 1.5 in . ) 2 ] − 4 × 1.375 in . 3 π [ 1 2 × π × ( 1.375 in . ) 2 ] = 0.51693 in 3 . Substitute 200 lb ⋅ in for M y , 0 for z , and 1.1687 in 4 for I y in Equation (1). σ A = − ( 200 lb ⋅ in ) ( 0 ) 1

9th Edition

ISBN: 9780133254426

Author: Russell C. Hibbeler

Publisher: Prentice Hall

See similar textbooks

Concept explainers

Pressure Vessels

The term pressure vessel refers to a closed container that contains gas or liquid at a different pressure than environment pressure. The pressure vessels have a wide scale of applications in our day-to-day lives and other industrial places. Some common e…

Videos

Question

Chapter 9, Problem 9.91RP

To determine

The principal stresses at point A.

Expert Solution & Answer

Want to see the full answer?

Check out a sample textbook solution

See solution

Chapter 9, Problem 9.90RP

Chapter 9, Problem 9.92RP

Students have asked these similar questions

First monthly exam Gas dynamics Third stage Q1/Water at 15° C flow through a 300 mm diameter riveted steel pipe, E-3 mm with a head loss of 6 m in 300 m length. Determine the flow rate in pipe. Use moody chart. Q2/ Assume a car's exhaust system can be approximated as 14 ft long and 0.125 ft-diameter cast-iron pipe ( = 0.00085 ft) with the equivalent of (6) regular 90° flanged elbows (KL = 0.3) and a muffler. The muffler acts as a resistor with a loss coefficient of KL= 8.5. Determine the pressure at the beginning of the exhaust system (pl) if the flowrate is 0.10 cfs, and the exhaust has the same properties as air.(p = 1.74 × 10-3 slug/ft³, u= 4.7 x 10-7 lb.s/ft²) Use moody chart (1) MIDAS Kel=0.3 Q3/Liquid ammonia at -20°C is flowing through a 30 m long section of a 5 mm diameter copper tube(e = 1.5 × 10-6 m) at a rate of 0.15 kg/s. Determine the pressure drop and the head losses. .μ= 2.36 × 10-4 kg/m.s)p = 665.1 kg/m³

2/Y Y+1 2Cp Q1/ Show that Cda Az x P1 mactual Cdf Af R/T₁ 2pf(P1-P2-zxgxpf) Q2/ A simple jet carburetor has to supply 5 Kg of air per minute. The air is at a pressure of 1.013 bar and a temperature of 27 °C. Calculate the throat diameter of the choke for air flow velocity of 90 m/sec. Take velocity coefficient to be 0.8. Assume isentropic flow and the flow to be compressible. Quiz/ Determine the air-fuel ratio supplied at 5000 m altitude by a carburetor which is adjusted to give an air-fuel ratio of 14:1 at sea level where air temperature is 27 °C and pressure is 1.013 bar. The temperature of air decreases with altitude as given by the expression The air pressure decreases with altitude as per relation h = 19200 log10 (1.013), where P is in bar. State any assumptions made. t = ts P 0.0065h

36 2) Use the method of MEMBERS to determine the true magnitude and direction of the forces in members1 and 2 of the frame shown below in Fig 3.2. 300lbs/ft member-1 member-2 30° Fig 3.2. https://brightspace.cuny.edu/d21/le/content/433117/viewContent/29873977/View

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 .

Concept explainers

Videos

The principal stresses at point A.

Want to see the full answer?

Chapter 9 Solutions