Why is the net area for AB (0.045-0.0102) * (0.012) ? Why is AB in tension and BC in compression Problem 1.601.60 Two horizontal 20 kN forces are applied to pin B of the assemblyshown. Knowing that a pin of 20 mm diameter is used at each connection, de-termine the maximum value of the average normal stress (a) in link AB, (b) inlink BC.12 mnBUse joint B as free body.45 mm20 kN20 kN12 mmA60°40 kN45 nim45°FecFee60°45040 EN.Law of SinesForce triangleFAB40Feesin 60sin 45°sin 95°= 28.4 kNFec = 34.8 KNLink AB isa tension m emberMin imum section at pinAnet = (o-045-0:o2)(0.012) = 300xiom

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Why is the net area for AB (0.045-0.0102) * (0.012) ? Why is AB in tension and BC in compression

Elements Of Electromagnetics

7th Edition

ISBN:9780190698614

Author:Sadiku, Matthew N. O.

Publisher:Sadiku, Matthew N. O.

ChapterMA: Math Assessment

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Work and Heat Transfer

It is generally related to thermal engineering which tells about the use, creation, conversion and transfer of heat energy between the systems. It is also considered that to transfer heat from one source to another, transfer of masses is also required. There are various mechanisms by which, transfer of heat takes place i.e. thermal conduction, convection, radiation or by change of phase. All these mechanisms have their own specific features which sometimes occur simultaneously within the same system.

Question

Why is the net area for AB (0.045-0.0102) * (0.012) ? Why is AB in tension and BC in compression

Problem 1.60
1.60 Two horizontal 20 kN forces are applied to pin B of the assembly
shown. Knowing that a pin of 20 mm diameter is used at each connection, de-
termine the maximum value of the average normal stress (a) in link AB, (b) in
link BC.
12 mn
B
Use joint B as free body.
45 mm
20 kN
20 kN
12 mm
A
60°
40 kN
45 nim
45°
Fec
Fee
60°
450
40 EN.
Law of Sines
Force triangle
FAB
40
Fee
sin 60
sin 45°
sin 95°
= 28.4 kN
Fec = 34.8 KN
Link AB is
a tension m ember
Min imum section at pin
Anet = (o-045-0:o2)(0.012) = 300xiom

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