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Sagot :
The displacement at the end of A is 1.09 mm
According to the question,
Rod's diameter, d= 20 mm
Stiffness constant, k = 55 M-N/m or,
= 55 × 10⁶ N/m
Force, f = 60 kN or,
= 60 × 10³ N
Young's modulus, E = 200 Gpa or,
= 200 × 10⁹ pa
We know the formula,
A=[tex]\pi d^2 /4[/tex]
By substituting the values,
= [tex]\pi (0.02)^{2} /4[/tex]
=0.0003 m³
length, L =A×E/K
=1.14 m
hence,
The displacement be:
60×10³×1.14/0.0003×200×10³×10³×10³=1.09 mm
Displacement is 1.09 mm
Learn more about force here:
brainly.com/question/25573309
#4264
The displacement of end a is 1.09mm
given:
Rod diameter,d=20mm
stiffness constant,k=55mn/m
force,f=60kn
young modulus,E= 200gpa
we know the formula
[tex]\pi \: d {}^{2} /4[/tex]
By substituting values in the above formula
[tex] = \pi(0.02) {}^{2} /4[/tex]
[tex] = 0.0003m {}^{3} [/tex]
length L=A×E/K
=1.14m
Hence, displacement is
[tex]60 \times 10 {}^{3} \times 1.14 \times/0.0003 \times 200 \times 10 {}^{3} \times 10 {}^{3} \times 10 {}^{3} [/tex]
=1.09mm
learn more about displacement from here: https:/brainly.com/question/28167908
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