Respuesta :
Answer:
principal stress σ = 46.46 MPa and -5.72 MPa
maximum shear stress = 26.08 MPa
Explanation:
given data
diameter = 5 cm = 50 mm
axial load = 80 kN = 80000 N
torque = 400 Nm - 400000 N-mm
to find out
principal stresses and shear stress
solution
we find first axial stress that is force / area
axial stress = 80000 / ( π/4×50² )
axial stress σ = 40.74 MPa
and
shear stress = torque × radius / area
shear stress = 400000 × 25 / ( π/4×50² )
shear stress τ = 16.3 MPa
so
principal stress will be
principal stress = σ/2 ± [tex]\sqrt{(\sigma/2^{2}+\tau^{2} )}[/tex] ..........1
principal stress σ = 40.74 /2 ± [tex]\sqrt{(40.74/2^{2}+ 16.3^{2} )}[/tex]
principal stress σ = 46.46 MPa and -5.72 MPa
and
maximum shear stress is
maximum shear stress = [tex]\sqrt{(\sigma/2^{2}+\tau^{2} )}[/tex] ..........2
maximum shear stress = [tex]\sqrt{(40.74/2^{2}+ 16.3^{2} )}[/tex]
maximum shear stress = 26.08 MPa
Answer:
principle stress =46.46 MPa, -5.72 MPa
maximu shear stress 26.08 MPa
Explanation:
d = 5 cm = 50 mm
axial load = 80 kN
torque = 400Nm
we know that
axial stress = force/area
[tex]= \frac{80*10^3}{\frac{\pi}{4} 50^2}[/tex]
= 40.74 MPa
shear stress[tex] = \frac{ Tr}{J} [/tex]
[tex] = \frac{400*10^3 *25mm}{\frac{\pi}{32}50^4}[/tex]
= 16.3 MPa
principle stress is given as
[tex]\sigma_p = \frac{\sigma_a}{2} \pm \sqrt{ (\frac{\sigma_a}{2})^2 +z^2}[/tex]
[tex]=\frac{40.74}{2} \pm \sqrt{ (\frac{40.74}{2})^2 +16.3^2}[/tex]
=46.46 MPa, -5.72 MPa
b) maximu shear stress
[tex]\sigma_p = \sqrt{ (\frac{\sigma_a}{2})^2 +z^2}[/tex]
[tex]= \sqrt{ (\frac{40.74}{2})^2 +16.3^2}[/tex]
= 26.08 MPa
