DR LARISA MALYSHEVA QB1 DIRECT STRESS Two round solid bars of different materials are connected (in parallel) and have a common tensile force, P, of 190 kN is applied,ras shown in Figure QB1. For each bar, determine the: a) b) Stress. Strain and Strain in the diameter. Use the following information. Clearly show all your derivations. The original length of each bar = 1.7m. Steel Bar: 32mm Dia. Poisson's Ratio = 0.31 18mm Dia. Young's Modulus = 210 GPa Steel Aluminium Bar Bar Common Connection Aluminium Bar: P = 190 kN Poisson's Ratio = 0.34 Young's Modulus = 68 GPa Fig QB1
DR LARISA MALYSHEVA QB1 DIRECT STRESS Two round solid bars of different materials are connected (in parallel) and have a common tensile force, P, of 190 kN is applied,ras shown in Figure QB1. For each bar, determine the: a) b) Stress. Strain and Strain in the diameter. Use the following information. Clearly show all your derivations. The original length of each bar = 1.7m. Steel Bar: 32mm Dia. Poisson's Ratio = 0.31 18mm Dia. Young's Modulus = 210 GPa Steel Aluminium Bar Bar Common Connection Aluminium Bar: P = 190 kN Poisson's Ratio = 0.34 Young's Modulus = 68 GPa Fig QB1
Elements Of Electromagnetics
7th Edition
ISBN:9780190698614
Author:Sadiku, Matthew N. O.
Publisher:Sadiku, Matthew N. O.
ChapterMA: Math Assessment
Section: Chapter Questions
Problem 1.1MA
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Transcribed Image Text:SECTION B-
STRUCTURES-DR LARISA MALYSHEVA
QB1 DIRECT STRESS
Two round solid bars of different materials are connected (in parallel) and have
a common
tensile force, P, of 190 kN is applied,ras shown in Figure QB1. For each bar, determine the:
a)
b)
Stress.
Strain and Strain in the diameter.
Use the following information.
Clearly show all your derivations.
The original length of each bar = 1.7m.
%3D
Steel Bar:
32mm Dia.
Poisson's Ratio = 0.31
18mm Dia.
Young's Modulus = 210 GPa
%3D
Steel
Aluminium
Bar
Bar
Common Connection Aluminium Bar:
P = 190 kN
Poisson's Ratio = 0.34
Young's Modulus = 68 GPa
Fig QB1

Transcribed Image Text:QA5- GENERAL PLANAR MOTION (USING NEWTON'S 2ND LAW & EULER'S EQUATION)
For the purposes of calculating its inertia, the body of a micro-satellite can be considered to be a
uniform rod of length 1 m and mass 15 kg. The satellite is required to move such that its centre of
mass, G, has a linear acceleration of 1.5 m/s in a direction perpendicular to the satellite (rod), and it
has an angular acceleration of 2 rad/s about an axis perpendicular to both the satellite and to the
direction of the linear acceleration of G. Find a single thrust vector (force) that will cause this, and
where on the satellite this must be applied.
Hint: Draw a free-body diagram of the satellite (rod), clearly showing the thrust vector, linear
acceleration, and angular acceleration. Due to Newton's 2nd law, the thrust vector must act in the
same direction as the linear acceleration.
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