
(a)
Interpretation:
The cross-sectional area of the hoist of a cable is 80mm2. Crate of 500 kg is lifted by this hoist and the length of cable is 30 m. It is assumed that the deformation is elastic. The stress of cable is to be determined.
Concept Introduction:
Following formulas will be used for the calculation of stress of cable.
And
(b)
Interpretation:
The cross-sectional area of the hoist of a cable is 80mm2. Crate of 500 kg is lifted by this hoist and the length of cable is 30 m. It is assumed that the deformation is elastic. Elongation of cable is to be determined considering that it is made from steel. And modulus of elasticity (E) is 200GPa.
Concept Introduction:
Concept of Hooke's law will be used. Hooke's law gives the following relation-
(c)
Interpretation:
The cross-sectional area of the hoist of a cable is 80mm2. Crate of 500 kg is lifted by this hoist and the length of cable is 30 m. It is assumed that the deformation is elastic. Elongation of cable is to be determined considering that it is made from polypropylene. And modulus of elasticity (E) is 1.2GPa.
Concept Introduction:
Concept of Hooke's law will be used. Hooke's law gives the following relation-

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Chapter 6 Solutions
Essentials of Materials Science and Engineering, SI Edition
- Given an existing two-story steel structure with interior columns spaced as shown in Fig.2. The columns are spaced at 18 ft in the North-South direction and at 30 ft in the East-West direction. An interior lower-story column is to be removed by adding newsteel girder as shown in Fig. 4. The floor dead loads and the roof dead loads are 70 psfand 18 psf respectively. The floor live loads and the roof live loads are 50 psf and20 psf respectively. All existing steel materials are ASTM A36 steel (Fy=36 ksi). Newgirder is ASTM A992 steel (Fy= 50 ksi). All columns are W8x31. Use the LRFD Method.Assumptions:1- The loads given include column and beam self weights.2- Existing beam and new girder are simply supported at both ends.3- New girder top flange is laterally braced at mid span and at girder ends only.4- Columns are continuous from foundation to roof and are prevented from sway atfloor level and at roof level in both directions.5- Columns are pin supported at foundation, at floor level,…arrow_forwardI have this fsk function code: function [x]=fsk_encode(b,s,f0,f1,N,Fs,K) % b= bit sequence vector % s(1)= output level for 0 % s(2)= output level for 1 % N= length of bit sequence % Fs= Sampling frequency y=zeros(1,N*K); %Setup output vector %for each bit calculatee the rando samples for n=1:N for k=1:K t = (k - 1) / Fs; if(b(n)==0) y((n-1)*K+k)=cos(2*pi*f0*t); % pulse=0 else y((n-1)*K+k)=cos(2*pi*f1*t); % pulse=1 end end x=y; %set output end And this is another code that calls the function in order to get the power density spectrum: clc;clear; % EE 382 Communication Systems- Lab 8 % Plots the power spectrum of the ASK modulation % First specify some parameters N=256; % number of bits per realization M=100; % number of realizations in the ensemble T=0.001; % bit duration in seconds delf =2e+3; fc=10e+3; f0=fc-delf; f1=fc+delf; Fs=8*f1; % sampling frequency (this is needed to calibrate the frequency axis) K=(T/(1/Fs)); % Define arrays for bit sequences and sampled waveforms…arrow_forwardThe uniform rods have a mass per unit length of 10kg/m . (Figure 1)If the dashpot has a damping coefficient of c=50N⋅s/m , and the spring has a stiffness of k=600N/m , show that the system is underdamped, and then find the pendulum's period of oscillation.arrow_forward
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