ACCORDING TO THE MAXIMUM PRINCIPAL STRESS THEORY, WHEN A SHAFT IS SUBJECTED TO A BENDING MOMENT OF 2 KN-M AND TORQUE OF 2 KN-M. IF THE ALLOWABLE STRESS IN AXIAL TENSION IS 220 KN/M², THEN FIND THE DIAMETER OF THE SHAFT.

ACCORDING TO THE MAXIMUM PRINCIPAL STRESS THEORY, WHEN A SHAFT IS SUBJECTED TO A BENDING MOMENT OF 2 KN-M AND TORQUE OF 2 KN-M. IF THE ALLOWABLE STRESS IN AXIAL TENSION IS 220 KN/M², THEN FIND THE DIAMETER OF THE SHAFT.

Structural Analysis

6th Edition

ISBN:9781337630931

Author:KASSIMALI, Aslam.

Publisher:KASSIMALI, Aslam.

Chapter2: Loads On Structures

Section: Chapter Questions

Problem 1P

See similar textbooks

Related questions

Q: A couple of magnitude M is applied to a square bar of side a. For each of the orientations shown,…

Q: 1 The vertical stress at point P, due to the point load Q on the ground surface as shown in figure…

Q: The cross-section of a shaft 90 mm diameter is subjected to a bending moment of 6 kNm and a twisting…

Q: A thin rectangular plate having the dimensions of a -20 mm and b-12 mm is acted upon by a stress…

A: A rectangular plate given asThe resulting strain Need to findChange in length in diagonal AC and QB

Q: A cylindrical shaft of 75mm diameter is subjected to a maximum bending moment of 2500 Nm and a…

Q: The solid circular rod has a cross-sectional area of 460 mm². It is subjected to a uniform axial…

Q: The stresses shown act at a point on the free surface of a machine component. Normal and shear…

A: Using transformation equation

Q: A two-dimensional stress system has like stresses q=100 N/mm²ando,= 200 N/mm² in two mutually…

A: Given data- Stresses-σx=100 N/mm2σy=200 N/mm2To Find-The coordinates of the centre of the Mohr's…

Q: If the principal stresses in a plane stress problem, are 0₁ = 100 MPa, o₂ = 40 MPa, the magnitude of…

A: Given- Principal stresses- σ1=100 MPa,σ2=40 MPa, It is needed to find the magnitude of maximum shear…

Q: A steel shaft AB of length 1.5 m and diameter 25 mm () carries a distributed torque of intensity t =…

Q: · The forces as given below are applied as shown in the figure. F1=10 kN, Fz=5 kN and F3=3 kN. The…

A: twisting moment - in twisting direction of load and eccentricity both are in transverse direction.…

Q: 19. Compute the stress when the temperature drops to -20°C. The cross- sectional area of the rod is…

Q: ee surface of a machine component. Normal and shear stress magnitudes acting on horizontal and…

A: Radius of the circleR=(sx-sy2)2+sxy2R=(23.6-2.82)2+72R=12.54 ksisAVG=sx+sy2=23.6+2.82sAVG=13.2ksi

Q: 12 МPа 6 MPa 8 MPa

Q: 50 cm-long circular rod has a diameter of 10 mm and is acted pon by a 20 kN load acting to its…

A: Length of rod = 50 cm Diameter of rod = 10 mm Load = 20 KN E = 200 Gpa

Q: [20 10 01 [o] 10 100 0 0 5] a. Determine the stress matrix for this stress state if the axes are…

A: Data given Find: (a) Determine the stress matrix for the given stress state if the axes are…

Q: ) For the given stress condition find: 1. Principal stresses and their orientation. 2. The stresses…

Q: Problem-3 b A hollow shaft of 20 mm diameter and 16 mm inside diameter is subjected to a torque of…

A: Given- Outside diameter D=20 mm, Inside diameter d=16 mm, Torque T=40 N-m, It is needed to find the…

Q: 6) Across Plane #1, the state of stress is measured as: On 1.1 kb and os= 0.4 kb Across Plane #2…

Q: The following shaft has a diameter of 60 mm, and it is made of Aluminum (G = 29 GPa). The shear…

A: Given, d=60 mmG=29 GPaτ= 200 MPaFOS=2.5ϕ=3 degree

Q: At a point, the major principal stress is 5000 psf compression and the minor principai stress is…

A: Major principal stress = 5000 psf (compression) Minor principal stress = 2000 psf (compression) It…

Q: Problem 1 of 1: Top side of a triangular section long body (t = 0.8 m) is subjected to a uniform…

A: modulus of elasticity E=195 GPa poission ratio ν=0.32 and volume of triangular body V=area…

Q: Q3(b) If the yield stress in the tensile test is equal to oy, sketch the yield (or failure) surface…

Q: The stresses shown act at a point in a stressed body. The stress magnitudes are Sx = 50 MPa and S, =…

A: Determine the normal and shear stresses at this point on the inclined plane shown.

Q: A concentrated vertical load of 3000 kN is applied on a horizontal ground surface. Points P and Q…

A: given- load=3000 kNdepths=1 m & 2 m

Q: Current Attempt in Progress Consider a point in a structural member that is subjected to plane…

Q: The solid circular rod has a cross-sectional area of 470 mm². It is subjected to a uniform axial…

Q: The rigid beam BC shown in Figure is supported by rods (1) and (2) that have cross-sectional areas…

Q: In a circular shaft under torque, maximum torsional stress occurs at * .shaft center True O False

A: In a circular shaft under torque, maximum torisional stress occurs at shaft center. A) true B) false

Q: A solid circular torsional member OPQ is subjected to torsional moments as shown in the figure. The…

Q: Using Mohr's Circle Determine: a.) The principal planes and directions b.) The principal Stresses…

Q: iameter hole in the support plate. When a load P is applied to the rod, the rod head rests on the…

Q: The solid circular rod has a cross kN/m. Two concentrated 0.5 m and b=0.9m. x a B loads : a W b

Q: A horizontal load P is applied to an assembly consisting of two inclined bars, as shown in the…

A: This problem is related to Strength of Material . In this question we calculate angle and Maximum…

Concept explainers

Graphical Methods

The graphical method is a method of solving equations involving more than one variable using the Cartesian coordinate system. In civil engineering, the use of the graphical method is limited to the problems involving two variables. The graphical method has a great impact in solving the questions regarding engineering mechanics.

Question

ACCORDING TO THE MAXIMUM PRINCIPAL STRESS THEORY,
WHEN A SHAFT IS SUBJECTED TO A BENDING MOMENT OF 2
KN-M AND TORQUE OF 2 KN-M. IF THE ALLOWABLE STRESS
IN AXIAL TENSION IS 220 KN/M², THEN FIND THE DIAMETER
OF THE SHAFT.

Expert Solution

This question has been solved!

Explore an expertly crafted, step-by-step solution for a thorough understanding of key concepts.

SEE SOLUTION Check out a sample Q&A here

Step 1

VIEW

Step 2

VIEW

Step by step

Solved in 2 steps

SEE SOLUTION Check out a sample Q&A here

Knowledge Booster

Learn more about

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

Similar questions

I need the answer as soon as possible
Solve
Consider a point in a structural member that is subjected to plane stress. Normal and shear stress magnitudes acting on horizontal and vertical planes at the point are Sy-17 MPa, S, - 60 MPa, and Swy- 29 MPa. (a) Draw Mohr's circle for this state of stress. (b) Determine the principal stresses (ơp > op2) and the maximum in-plane shear stress Tmax acting at the point. (c) Find the smallest rotation angle 0, (counterclockwise is positive, clockwise is negative) that will rotate to principal directions. Then show these stresses in an appropriate sketch (e.g, see Figure 12.15 or Figure 12.16) Say Answers: MPa. Opl = MPa. MPa. Tmax = 0, =
A steel bar 50 mm diameter and 1 m long just fits within a rigid hole carries a compressive load of 300 kN. Consider E= 200 GPa for steel and Poissons ratio υ = 1/4. Determine: (a.)Stress developed along lateral direction in MPa. (b.) Stress developed along lateral direction in MPa if the rigid hole has a diameter of 51 mm.
Given the state of stress shown, determine the normal stress (in MPa) acting on the inclined plane a-a if S = 93.19 MPa, N = 42.03 MPa, and A = 36.4°. Express final answer in four decimal places.
The stresses shown act at a point in a stressed body. The stress magnitudes are Sx = 85 MPa and Sy = 215 MPa, acting in the directions indicated in the figure. Using the equilibrium equation approach, determine the normal and shear stresses at this point on the inclined plane shown. Assume B = 50°. Sx
A 2 m long shaft is transmitting 90 kW at 150 RPM. The allowable shear stress in the material is 60 MPa. If the shaft is not allowed to twist more than 1°, then the diameter of the shaft required will be (Take G = 85 GPa)