Electrical strain gauges were applied to a notched specimen to determine the stresses in the notch. The results were εx=0.0019 a
1 answer:
You might be interested in
Answer & Explanation:
function Temprature
NYC=[33 33 18 29 40 55 19 22 32 37 58 54 51 52 45 41 45 39 36 45 33 18 19 19 28 34 44 21 23 30 39];
DEN=[39 48 61 39 14 37 43 38 46 39 55 46 46 39 54 45 52 52 62 45 62 40 25 57 60 57 20 32 50 48 28];
%AVERAGE CALCULATION AND ROUND TO NEAREST INT
avgNYC=round(mean(NYC));
avgDEN=round(mean(DEN));
fprintf('\nThe average temperature for the month of January in New York city is %g (F)',avgNYC);
fprintf('\nThe average temperature for the month of January in Denvar is %g (F)',avgDEN);
%part B
count=1;
NNYC=0;
NDEN=0;
while count<=length(NYC)
if NYC(count)>avgNYC
NNYC=NNYC+1;
end
if DEN(count)>avgDEN
NDEN=NDEN+1;
end
count=count+1;
end
fprintf('\nDuring %g days, the temprature in New York city was above the average',NNYC);
fprintf('\nDuring %g days, the temprature in Denvar was above the average',NDEN);
%part C
count=1;
highDen=0;
while count<=length(NYC)
if NYC(count)>DEN(count)
highDen=highDen+1;
end
count=count+1;
end
fprintf('\nDuring %g days, the temprature in Denver was higher than the temprature in New York city.\n',highDen);
end
%output
check the attachment for additional Information
Answer:
a. ε₁=-0.000317
ε₂=0.000017
θ₁= -13.28° and θ₂=76.72°
b. maximum in-plane shear strain =3.335 *10^-4
Associated average normal strain ε(avg) =150 *10^-6
θ = 31.71 or -58.29
Explanation:
ε₁=-0.000317
ε₂=0.000017
To determine the orientation of ε₁ and ε₂
θ= -13.28° and 76.72°
To determine the direction of ε₁ and ε₂
=-0.000284 -0.0000335 = -0.000317 =ε₁
Therefore θ₁= -13.28° and θ₂=76.72°
b. maximum in-plane shear strain
=3.335 *10^-4
ε(avg) =150 *10^-6
orientation of γmax
θ = 31.71 or -58.29
To determine the direction of γmax
= 1.67 *10^-4