An Introduction to the Mechanics of Fluids
An Introduction to the Mechanics of Fluids
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Hence, for the whole triangle of area A= bp/2, we have (Axis through Vertex) J P p 4 A^ (14) By means of the parallel axes theorem, we can now pass from this result to the moment of inertia J about the axis through the centroid and still parallel to the base CA.
Then Jo=Ji-A(§^) 2 = A^(i_|) or, (Axis through Centroid) Jo=^A^> 2 (15) Taking now the base CA as axis and writing J 2 for the correspond- ing moment of inertia, we have j2=Jo + A(f) 2 =A^.^+^)or, (Axis along base) h=iw >6) 62 MECHANICS... OF FLUIDS 55. Moments of Inertia of a Triangle about Axes Perpendicular to a Side. — Still referring to Fig. 35, let us now determine the moment of inertia Ii about the axis BX perpendicular to the side CA. It will obviously be the difference between those of the triangles ABM and CMB, each about BX. But these values can be written down from the dimensions by application of equation (16). Thus, writing d for the length MC, we have . I, -|. Tf±fl. CH-^--i^, Ji (Axis through Vertex) Ii= -r(b 2 + sbd+sd 2 ) .
What to read after An Introduction to the Mechanics of Fluids?
You can find similar books in the "Read Also" column, or choose other free books by Edwin H Edwin Henry Barton to read onlineMoreLess
To quickly assess the difficulty of the text, read a short excerpt:
Hence, for the whole triangle of area A= bp/2, we have (Axis through Vertex) J P p 4 A^ (14) By means of the parallel axes theorem, we can now pass from this result to the moment of inertia J about the axis through the centroid and still parallel to the base CA.
Then Jo=Ji-A(§^) 2 = A^(i_|) or, (Axis through Centroid) Jo=^A^> 2 (15) Taking now the base CA as axis and writing J 2 for the correspond- ing moment of inertia, we have j2=Jo + A(f) 2 =A^.^+^)or, (Axis along base) h=iw >6) 62 MECHANICS... OF FLUIDS 55. Moments of Inertia of a Triangle about Axes Perpendicular to a Side. — Still referring to Fig. 35, let us now determine the moment of inertia Ii about the axis BX perpendicular to the side CA. It will obviously be the difference between those of the triangles ABM and CMB, each about BX. But these values can be written down from the dimensions by application of equation (16). Thus, writing d for the length MC, we have . I, -|. Tf±fl. CH-^--i^, Ji (Axis through Vertex) Ii= -r(b 2 + sbd+sd 2 ) .
What to read after An Introduction to the Mechanics of Fluids?
You can find similar books in the "Read Also" column, or choose other free books by Edwin H Edwin Henry Barton to read onlineMoreLess
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