Determine the volume and the surface area of the chain link shown, which is made from a 6-mm-diameter bar, if R = 10 mm and L = 30 mm.
Answers
Explanation:
The area A and circumference C of the cross section of the bar are
A=\frac { \pi }{ 4 } { d }^{ 2 }
4
π
d
2
and C=\piπ d
Also, the semicircular ends of the link can be obtained by rotating the cross section through a horizontal
semicircular arc of radius R. Now, applying the theorems of Pappus-Guldinus, we have for the volume V,
V=2\left( { V }_{ Side } \right) +2\left( { V }_{ end } \right)(V
Side
)+2(V
end
)=2\left( AL \right) +2\left( \pi RA \right) \\(AL)+2(πRA)
=2\left( L+\pi R \right)(L+πR) A
or V=2\left[ 30mm+\pi \left( 10mm \right) \right] \left[ \frac { \pi }{ 4 } { \left( 6mm \right) }^{ 2 } \right][30mm+π(10mm)][
4
π
(6mm)
2
]
=3470m{ m }^{ 3 }m
3
or V=3470m{ m }^{ 3 }m
3
For the area A,
A=2\left( { A }_{ Side } \right) +2\left( { A }_{ end } \right)(A
Side
)+2(A
end
) =2\left( CL \right) +2\left( \pi RC \right) \\(CL)+2(πRC)
=2\left( L+\pi R \right)(L+πR) C
or A=2\left[ 30mm+\pi \left( 10mm \right) \right] \left[ \pi { \left( 6mm \right) } \right][30mm+π(10mm)][π(6mm)]
=2320m{ m }^{ 2 }m
2
or A=2320m{ m }^{ 2 }m
2