Т2 KN KN MPa MPA MPa MPa
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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PLEASE EXPLAIN

Transcribed Image Text:A steel plate of dimensions 2.5 x 1.5 x 0.08 m and weighing 23.1 KN is hoisted by steel cables with lengths L₁ = 3.1 m and L₂ = 4.0 m that are each attached to the plate by a clevis and pin (see figure).
T₁ =
T₂ =
Taver, 1
=
Clevis L₁
and pin 1
=
Taver, 2
%b, 1 =
%b, 2 =
a = 0.6 m
Center of gravity
of plate
Ⓡ
The pins through the clevises are 26 mm in diameter and are located 2 m apart. The orientation angles are measured to be 0 = 101.11° and a = 49.51⁰.
For these conditions, first determine the cable forces T₁ and T₂ (both in kN), then find the average shear stress Taver (in MPa) in both pin 1 and pin 2, and then the average bearing stress (in MPa) between the steel plate and each pin. Ignore the weight of the cables.
(Enter the magnitudes. Assume the lengths L₁ and L₂ are measured from the loop at the top of the figure to the center of their respective pin. Due to the nature of this problem, do not use rounded intermediate values in your calculations including answers submitted in
WebAssign.)
KN
KN
MPa
MPa
MPa
βι ¦ βι
Ꮎ
MPa
b=1.0 m-
2.0 m
Clevis
and pin 2
Steel plate
(2.5 x 1.5 x 0,08 m)
Expert Solution

Step 1: Introduction
Given data as per question
dimensions of steel plate = 2.5 x 1.5 x 0.08 m
weight of the steel plate= 23.1 kN
L1 = 3.2 m
L2 = 4 m
The orientation angles are
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