The initial compressive force of a steel column can be determined by Euler's buckling formula. The column has the following properties:A = 9484 mm2Fy = 345 MPaIx = 164 x 106 mm4ly = 23 x 106 mm4E = 200 GPaProportional limit, f = 290 MPa%3D%3DThe x-axis has an unbraced length of 10 m which is pinned athe top and fixed at the bottom with an k = 0.70. The y-axis has unbraced length of 5 m due to the bracing at the mid-height.Determine the factor of safety for the load shown with respect to buckling in the plane of the figure.Select the correct response:1.241.181.414.373

Area(A)=9484 mm2, Yield strength(Fy)=345 Mpa, Ix=164×106 mm4Iy=23×106 mm4 Modulus of…

The initial compressive force of a steel column can be determined by Euler's buckling formula. The column has the following properties: A = 9484 mm2 Fy = 345 MPa E = 200 GPa x = 164 x 10° mm4 ly = 23 x 106 mm4 The x-axis has an unbraced length of 10 m which is pinned ahihe top and fixed at the bottom with an k = 0.70. The y-axis has unbraced length of 5 m due to the bracing at the mid-height Proportional limit, f = 290 MPa Determine the factor of safety for the load shown with respect to buckling in the plane of the figure. Select the correct response: 124 1.18 1.41 O 4.373

The initial compressive force of a steel column can be determined by Euler's buckling formula. The column has the following properties: A = 9484 mm2 Fy = 345 MPa E = 200 GPa x = 164 x 10° mm4 ly = 23 x 106 mm4 The x-axis has an unbraced length of 10 m which is pinned ahihe top and fixed at the bottom with an k = 0.70. The y-axis has unbraced length of 5 m due to the bracing at the mid-height Proportional limit, f = 290 MPa Determine the factor of safety for the load shown with respect to buckling in the plane of the figure. Select the correct response: 124 1.18 1.41 O 4.373

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. Compute the internal forces in bars BC and CD. Indicate if they are in compression or tension.…

A: A truss member is given belowNeed to find a. compute the internal forces in bars BC and CD. Indicate…

Q: Determine the internal forces at point J for the structure shown, where member BD can only carry…

A: Given:- a=160 mmP=450 N To find:- Internal force at point J

Q: The initial compressive force of a steel column can be determined by Euler's buckling formula. The…

Q: 9. The rigid bar ABC shown in the figure is hinged at A and supported by a steel rod at B. Determine…

A: Given Length = 4 ft Area = 0.5 sq.in Elastic modulus =29(10)6 psi Required = value of P satisfying…

Q: Please help with this practice question. Thanks Question 1 ) Determine the y-component of force at…

A: We are required to find the Y component of force at pin D

Q: For certain values of w and P, the maximum positive bending moment in the beam is +7044 lb-ft, and…

A: Here, M = 7044 lb.ft Iz = 18.3990 in4 For positive bending moment,…

Q: The rigid bar BDE is supported by two links AB and CD. Link AB is made of aluminum (E = 70 GPa) and…

Q: As shown in the figure, aluminum, E= 70GPa links (1) and (2) support rigid beam ABC. Link (1) has a…

Q: X is L/2 For the frame shown above is constructed out of circular tubes for both the beam and…

Q: A pin-connected truss is loaded and supported as shown. Each aluminum member has a cross-sectional…

A: Given data- Cross sectional area of Truss member = 3.6 in2 Maximum normal stress in each member = 50…

Q: A pin-connected truss is loaded and supported as shown. Each aluminum member has a cross- sectional…

Q: Determine the the vertical component of the displacement at point"d" for the shown truss. The areas…

A: Area of cross section given in mm2 along each member. E = 200 GPa Determine the vertical…

Q: The principal stresses are incorrect but the maximum shear stress is correct. What are the two…

A: Compressive stress is the force that deforms the material in a way that reduces the volume of the…

Q: 8 kN 4 m 15 kN 5 kN 2 m 2 kN 2m 2m A B с D E The aluminum rod shown has a diameter of 20 mm and is…

A: Given: diameter, D = 20 mm E = 70 GPa

Q: 21 P3. 2) Determine the displacement of each member or E = 200 GPa %3D |P 650 kN %3D Diameter 1 =…

Q: P5 The pin-connected structure shown in the figure consists of a rigid beam ABCD and two supporting…

Q: The two vertical steel [E = 200 GPa] rods that support rigid bar ABCD are initially free of stress.…

Concept explainers

Types and properties of steel

Steel is an alloy of iron. It is formed by heating iron and combining carbon (0.1-1.7%) with it and minor amounts of other elements such sulfur, chromium, manganese, nickel, etc. Thousands of steel grades have been published, registered, or regulated around the world, each with its chemical composition, and special numbering systems have been devised in various countries to classify the vast number of alloys.

Question

The initial compressive force of a steel column can be determined by Euler's buckling formula. The column has the following properties:
A = 9484 mm2
Fy = 345 MPa
Ix = 164 x 106 mm4
ly = 23 x 106 mm4
E = 200 GPa
Proportional limit, f = 290 MPa
%3D
%3D
The x-axis has an unbraced length of 10 m which is pinned athe top and fixed at the bottom with an k = 0.70. The y-axis has unbraced length of 5 m due to the bracing at the mid-height.
Determine the factor of safety for the load shown with respect to buckling in the plane of the figure.
Select the correct response:
1.24
1.18
1.41
4.373

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

Given data

VIEW

Calculation

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

The initial compressive force of a steel column can be determined by Euler's buckling formula. The column has the following properties: A = 9484 mm² Fy=345 MPa x = 164 x 106 mm4 ly = 23 x 106 mm4 Proportional limit, fs = 290 MPa The x-axis has an unbraced length of 10 m which is pinned at the top and fixed at the bottom with an k = 0.70. The y-axis has unbraced length of 5 m due to the bracing at the mid-height. Compute the initial compressive load of the column. Select the correct response: 6606.59 KN 3706.14 KN E = 200 GPa 1816.01 kN 926.53 kN
The two vertical steel [E = 200 GPa] rods that support rigid bar ABCD are initially free of stress. Rod (1) has an area of A1 = 410 mm2 and a length of L1 = 3.2 m. Rod (2) has an area of A2 = 355 mm2 and a length of L2 = 1.3 m. Assume dimensions of a = 3.4 m, b = 1.2 m, and c = 1.5 m. After a load of P = 50 kN is applied to the rigid bar at D, determine: (a) the normal stresses in rods (1) and (2) (b) the magnitude of the downward deflection of the rigid bar at D. y,v (2) L2 a B C D x,u Rigid bar |(1) Answer: (a) o1 = i MPa 02 = i MPa (b) VD = i mm
A pin-jointed truss ABC shown below consists of two links AB and BC. These links are 6 mm thick and 20 mm wide as shown. There are support brackets at A and C. All pins are 12 mm in diameter and a vertical force, P = 9 kN is applied at pin B as shown. The angle 0 =38 deg. and the angle ẞ = 0 + 20 deg. The normal stress in link AB is (The result is accurate to one decimal place, in the unit MPa. Please write down only the value of your final result) · 12mm 9 20mm -1.0m⋅ B
The members of the truss shown have equal cross-sectional areas A = 400 mm2. The suspended mass is m = 3400 kg. What is the normal stress in the member DE? Answer in MPa and use two decimals places.