Tarea 1_MI
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4.11 (a) Write a simple computer program that gives the shear stress of a screw dislocation as a function of the perpendicular distance from the dislocation (see Eq. 4.8). Assuming the shear modulus of iron is 86 GPa and the Burgers vector is 0.248 nm, use the program to Unlock to An obtain the shear stress at the following valuesAccess of r : 50, 100, 150, and 200 nm, respectively. Plot the resulting versus r data, and with the aid of this curve, determine the distance from the dislocation where is 4000 psi, the shear stress at which an iron crystal will begin to undergo slip.
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%Problema 4.11 %a) Tarea 1_MI clear all clc shear_modulus=86; %[GPa] miu=86*145037.73773; %Shear modulus[psi]Access Now b=0.248; %Burgers vector [nm] r=50:10:150; %Radius of the Burgers circuitI [nm] No thanks, don't want my exclusive trial tau=(miu*b)./(2*pi.*r); %Shear stress [psi] r_4000=(miu*b)/(2*pi*4000)%distance r_4000=(miu*b)/(2*pi*4000) %distance [nm] from the dislocation where tau is 4000 psi n_b=r_4000/b %b)To how many Burgers vectors does this distance correspond? plot(r,tau,'linewidth' plot(r,tau,'linewidth',2) ,2) ylabel('Shear ylabel('Shear stress [psi]') [psi]') xlabel('Radius xlabel('Radius of the Burgers circuit [nm]') [nm]') title('\tau title('\tau vs r') r') grid on hold on plot(r_4000,4000,'*' plot(r_4000,4000,'*'))
The distance from the dislocation where
is 4000 psi is [nm]:
= vs
10000
r
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9000
8000
]i s [p
Tarea 1_MI
7000
s s er t s
Access Now
r a e h
6000
S
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Radius of the Burgers circuit [nm]
(b) To how many Burgers vectors does this distance correspond?
130
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4.17 (a) Consider Eq. 4.19, which gives the strain energy per unit length of a screw dislocation. Assume that one has a very large square array of long, straight, parallel screw dislocations of alternating signs so that the effective outer radius r of the strain field of a Unlock An dislocation may be taken as . With theAccess aid of to a computer determine the strain energy ’
1/2√
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per unit length of a screw dislocation as a function of the dislocation density , between =1011 and =1018 m/m3.
%Problema 4.17 clear all clc Tarea 1_MI r_rho=316227.766:100000:10000000000;%sqrt(rho) [m] r_p=1./(2.*r_rho) %The effective outer radius [m] miu=86*(10^9); %Shear modulus[Pa] Access Now b=0.248/(10^9); %Burgers vector [m] r_o=b/4; %Inner radius that excludes the dislocation core [m] alpha=4 alpha=4; %Constant No thanks, I don't want my exclusive trial W_s=((miu.*(b.^2))./(4.*pi)).*(log(r_p./r_o))%The strain energy per unit length of this screw dislocation [J/m]
Shown below assessments of some values:
[]
[ ]
Ws
1011
4.2708e-09
1012
3.7862e-09
1013
3.3016e-09
1014
2.8170e-09
1015
2.3324e-09
1016
1.8478e-09
1017
1.3632e-09
1018
8.7864e-10
(b) Plot the line energy against the dislocation density. Assume that nm.
=86 GPa and b=0.248
%Problema 4.17 Unlock Access to An clear all clc r_rho=316227.766:100000:1000000000;%sqrt(rho) [m] r_p=1./(2.*r_rho) %The effective outer radius [m] miu=86*(10^9); %Shear modulus[Pa] b=0.248/(10^9); %Burgers vector [m] r_o=b/4; %Inner radius that excludes the dislocation core [m] alpha=4 Tarea 1_MI alpha=4; %Constant W_s=((miu.*(b.^2))./(4.*pi)).*(log(r_p./r_o))%The strain energy per unit length of this screw dislocation [J/m] Access Now rho=r_rho.^2; plot(rho,W_s,'linewidth',2) xlabel('Dislocation density \rho [m/m^3]') No thanks, I don't want my exclusive trial ylabel('Strain energy per unit length of this screw dislocation Ws [J/m]') title('Ws vs \rho ') grid on
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4.5
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Ws vs ;
-9
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Dislocation density ; [m/m ]
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Ws vs r
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W n oi
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t a c ol
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is d w er c
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s is th f o
Tarea 1_MI
2.5
th g n el it n
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u
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r e p y gr
1.5
e
No thanks, I don't want my exclusive trial n e ni ar t S
1
0.5 0
0.2
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Effective outer radius r´ [m]
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(c) Now plot the energy per unit volume as a function of , assuming that the former can be equated to w, where w is the energy per unit length of the screw dislocations. %Problema 4.17 clc clear all r_rho=316227.766:1000000:1000000000; %sqrt(rho) [m] r_p=1./(2.*r_rho) %The effective outer radius [m] miu=86*(10^9); %Shear modulus[Pa] b=0.248/(10^9); %Burgers vector [m] r_o=b/4; %Inner radius that excludes the dislocation core [m] alpha=4 alpha=4; %Constant W_s=((miu.*(b.^2))./(4.*pi)).*(log(r_p./r_o));%The strain energy per unit length of this screw dislocation [N] rho=r_rho.^2; rhoWs=rho.*W_s plot(rho,rhoWs,'linewidth',2) xlabel('\rho [m/m^3]') ylabel('\rho Ws [J/m^3]') grid on
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# 10
8
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7
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Tarea 1_MI ]
3
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m/ J[ s
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W 4 ;
No thanks, I don't want my exclusive trial
3
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