The stress concentration occurs whenever there is an abrupt change in the cross-section of a component or there is any discontinuity in the material. The figure given below shows a flat plate with a hole of diameter d. The plate is fixed at one end and the other end is subjected to a tensile load of P = 41 kN due to which there is a change in length of 0.4 mm. The thickness of the plate is 11 mm. The maximum stress developed in the flat plate is 235 MPa. Take Young's modulus(E) = 210 GPa and theoretical stress concentration factor =2, Calculate the following values: i) Width of the plate (W2) in mm ii) Nominal Stress in MPa iii) Diameter of the hole (d) in mm Hole with stress concentration factor 2 35 mm W2 35 mm P 300 mm 350 mm- 250 mm

The stress concentration occurs whenever there is an abrupt change in the cross-section of a component or there is any discontinuity in the material. The figure given below shows a flat plate with a hole of diameter d. The plate is fixed at one end and the other end is subjected to a tensile load of P = 41 kN due to which there is a change in length of 0.4 mm. The thickness of the plate is 11 mm. The maximum stress developed in the flat plate is 235 MPa. Take Young's modulus(E) = 210 GPa and theoretical stress concentration factor =2, Calculate the following values: i) Width of the plate (W2) in mm ii) Nominal Stress in MPa iii) Diameter of the hole (d) in mm Hole with stress concentration factor 2 35 mm W2 35 mm P 300 mm 350 mm- 250 mm

Steel Design (Activate Learning with these NEW titles from Engineering!)

6th Edition

ISBN:9781337094740

Author:Segui, William T.

Publisher:Segui, William T.

Chapter1: Introduction

Section: Chapter Questions

Problem 1.5.5P: The results of a tensile test are shown in Table 1.5.2. The test was performed on a metal specimen...

See similar textbooks

Related questions

Q: The stiffened top skin of a wing structure is modelled as a panel has simply supported edges as…

A: step 1-

Q: Two Triangular wedges are glued together as shown in the following figure. The stress acting normal…

A: Stresses acting on the planes-To Find-The value of the normal stress

Q: An aluminum rod with a semicircular cross section of radius r = 12 mm (Fig. a) is bent int he shape…

A: First we summarise the given data : r = 12 mm mean radius of circular arc = 2.5 m E = 70 GPa To…

Q: An aluminum rod is rigidly attached between a steel rod and a bronze rod as shown in Fig. P-108.…

A: To find out Maximum value of P so that stresses should not be cossing their limits.

Q: An extruded polymer beam is subjected to a bending moment M. The length of the beam is L = 560 mm.…

Q: Q.3: A steel rod with a cross sectional area of 150 mm² is stretched between two fixed points. The…

A: A ) Given data : Cross sectional area of steel rod = 150 mm2 Tensile load at 20oC is = 5000 N α =…

Q: A high-density polyethylene [E= 104 ksi and a 74 x 10-6/°F] block (1) is positioned in a fixture.…

Q: The state of stress at a point in a structural member may be represented by a two-dimensional stress…

Q: The stresses at a point in a bar are o1 = 28 N/mm2 (tensile) and o2=12N/mm2 (compressive) as shown…

A: In the question angle made by inclined surface with major principle plane is -35 because we have…

Q: The state of stress at a certain point in a stressed body is as shown in the figure. Normal stress…

A: We have to determine the radius of the Mohr circle

Concept explainers

Material Properties

Construction in civil engineering demands high strength, durability, and longevity. These targets can be achieved by better design and analysis, along with the correct choice of materials. Different materials have different properties and hence, they respond and performs differently under the various condition of external loadings. It is therefore a critical task of an engineer to select the right material based on the property that the material is supposed to function in real-time. When speaking about the property, there are different categories of material properties like chemical property, physical property, mechanical property, electrical property, magnetic property, thermal property, etc. These material properties form the basis of material selection in various applications.

Question

The stress concentration occurs whenever there is an abrupt change in the cross-section of a component or there is any discontinuity in
the material. The figure given below shows a flat plate with a hole of diameter d. The plate is fixed at one end and the other end is
subjected to a tensile load of P = 41 kN due to which there is a change in length of 0.4 mm. The thickness of the plate is 11 mm. The
maximum stress developed in the flat plate is 235 MPa. Take Young's modulus(E) = 210 GPa and theoretical stress concentration factor
=2,
Calculate the following values:
i) Width of the plate (W2) in mm
ii) Nominal Stress in MPa
iii) Diameter of the hole (d) in mm
Hole with stress
concentration factor 2
35 mm
W2
35 mm
P
300 mm
350 mm-
250 mm

Expert Solution

This question has been solved!

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

This is a popular solution!

SEE SOLUTION Check out a sample Q&A here

Step 1

VIEW

Step 2

VIEW

Step 3

VIEW

Trending now

This is a popular solution!

Step by step

Solved in 3 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

The results of a tensile test are shown in Table 1.5.2. The test was performed on a metal specimen with a circular cross section. The diameter was 3 8 inch and the gage length (The length over which the elongation is measured) was 2 inches. a. Use the data in Table 1.5.2 to produce a table of stress and strain values. b. Plot the stress-strain data and draw a best-fit curve. c. Compute the, modulus of elasticity from the initial slope of the curve. d. Estimate the yield stress.
The data in Table 1.5.3 were obtained from a tensile test of a metal specimen with a rectangular cross section of 0.2011in.2 in area and a gage length (the length over which the elongation is measured) of 2.000 inches. The specimen was not loaded to failure. a. Generate a table of stress and strain values. b. Plot these values and draw a best-fit line to obtain a stress-strain curve. c. Determine the modulus of elasticity from the slope of the linear portion of the curve. d. Estimate the value of the proportional limit. e. Use the 0.2 offset method to determine the yield stress.
Compare the engineering and true secant elastic moduli for the natural rubber in Example Problem 6.2 at an engineering strain of 6.0. Assume that the deformation is all elastic.
Solve the problem and give what is asked in this problem. box the final answer clearly and readable.
Determine Value of x if the Normal stress at rod is equal to the stress at rectangular pipe A =400 mm2 F = 8kN Rod A =650 mm2 fectangular Pipe
As shown, an aluminium alloy construction BCD with a circular cross section is fixed at end B and affected by a force of 150 N at the free end D. The diameter of the cross-section a-a is 20 mm. The yield strength of the material is 80 MPa: a) Determine the stresses at point A of the a-a cross-section. As indicated in the picture, draw the stress element in Cartesian coordinates and specify the stress values.(b) Calculate the factor of safety, n for Tresca, and the von Mises yield criterion to see if the structure would yield based on the stresses at point A.(c) In the major stress area, draw the yield loci of both criteria and indicate the operational stress state & why is the Rankine failure criterion inappropriate for aluminium alloys?
Two cylindrical rods AB and BC are welded together at B and loaded as shown. Knowing that the average normal stress must not exceed 175 MPa in rod AB and 150 MPa in rod BC, determine the smallest allowable values of d1 and d2. 300 mm 250 mm A B C dy 40 kN -dr 30 kN
A bar obtained by bonding together pieces of steel (Es = 29 x 106 psi) and brass (E = 15 x 106 psi) has the cross section shown (Fig. a). Determine the maximum stress in the steel and in the brass when the bar is in pure bending with a bending moment M = 40 kip-in. 0.4 in. 3 in. Brass 0.75 in. Steel Brass 0.4 in. (a) 1) Composite bar.
I5 In a tensile test experiment of carbon steel. a standard specimen (D= 0.505 in. Gauge length=2.0 in & total length 8 in) had a 0.2% offset yield strength of 80.000 psi and engineering strain of 0.010 at the yield strength point Calculate: The load (force) at this point. The instantaneous area of the specimen at this point. c. The true stress at this point. d. The true strain at this point. e. The total length of the specimen. if the load is released at this point.
30. An aluminum alloy rod has a circular cross section with a diameter of 8mm. The rod is subjected to a tensile load of 5kN. Assume that the material is in the elastic region and E=69 GPa If Poisson's Ratio is 0.33, what will be the lateral strain? E=68 v=-e(lateral) e(axial)
A symmetrical framework consisting of three pin-connected bars is loaded by a forceP (see figure). The angle between the inclined bars and the horizontal is a = 52°. Theaxial strain in the middle bar is measured as 0.036.Determine the tensile stress in the outer bars if they are constructed of a copper alloyhaving the following stress-strain relationship:
6