Eample Problem F.: (Beer & Johnston Eample 9-) Determine the mass moment of inertia with respect to: (a) its longitudinal ais (-ais) (b) the y-ais SOLUTION: a) Mass moment of inertia about the -ais: Step : Choose appropriate differential of mass r = a h dm = ρd πr = ρπr d a dm = ρπ d h Step : Compute mass moment of inertia of differential element with respect to desired ais For a circular disk, di = r dm a di = a ρπ d h h di = a ρπ h d
Step : Integrate and calculate the moment of inertia Integrating from = 0 to = h, I h 0 0 5 d ρπ a a h di = ρπ = h h 5 I = ρπa h 0 Total mass of the cone: m = ρ( πa ) h ( ρπ ) I = ρπa h = a a h 0 0 I = ma 0 b) Mass moment of inertia about the y-ais Step : The same differential element is used a dm = ρπ d h Step : Apply the parallel-ais theorem to determine the moment of inertia of the differential element di = di + dm y y' For a circular plate, the mass moment of inertia with respect to an ais plane of the surface is: IAA' = mr Therefore, diy = r dm + dm = ( r + ) dm Substituting the epressions for r and dm into the above equation, a a diy = + ρπ d h h AA ' parallel to the di y a a = ρπ d + h h
Step : Integrating from = 0 to = h, I y h a a a a h = ρπ + = + h h h h 5 0 5 d ρπ Introducing the total mass of the cone: m = ρ( πa ) h we re-write I y as Iy = 5( a + h ) ρπa h Iy = m a + h 5
Eample Problem F.: Final Eam Spring 00 L 759.5 y Top 0mm di 5mm 00mm = 50mm A beam is made from a cylindrical tube and two L-Sections, which are welded to a plate as shown below. For the L 759.5 properties, refer to the attached beam table. Bottom d0 = 70mm Goals: a. Find the location of the neutral ais (y ) as measured from the bottom of the beam b. Calculate the second moment of area for this section ( I ) c. Calculate the maimum bending stress if the section is subjected to a moment of + 900 0 N-m (recall the sign convention for positive bending moment). Also identify if this maimum stress is tensile or compressive. Solution Part a: For an elastic beam, the location of the neutral ais passes through the centroid of the section. For the composite area, the y coordinate of the centroid is given by: Ay i i y = Atotal Portion : Cylindrical tube A = π 5 5 y = 5mm Portion : Plate A = ( 0)( 00) = 8000mm y = ( 00 + 70) = 70mm
5 Portions : L-Sections A = 0mm y = ( 70.7 ) = 5.mm Therefore, Ay i i Ay + Ay + Ay 998057 y = = = A A A A.9 total y =.9mm = 5mm + + Part b: Using the parallel-ais theorem, the second moment of area of the section about I = ( I + d { da} ) ' ' Portion : Cylindrical tube π Moment of inertia of a circle respect to ais -: I = r d0 5 5 π I = + y A π I = + Portion : Plate 70 5 5.9 I = ( 0 )( 00 ) + ( 70 y ) A I = ( 0 )( 00 ) + ( 5. ) A Portions : L-Sections I = 0.8 0 + ( 70.9.7 ) A I = 0.8 0 + ( 9.) A Therefore, I = I + r da = I + I + I I = 7.855 0 A ' ' 5 m is:
Part c: The maimum bending stress can be calculated using the elastic fleural formula: Mbz c σ m = I The maimum distance from the neutral ais is c = y c =.9mm Therefore, the maimum bending stress is: σ Mbz c 900 0 ( 0.9) m = = I 5 7.855 0 σ =.89 0 Pa m Since the bending moment is positive, the beam bends concave upwards. The maimum stress is occurring at the bottom of the section therefore it is tensile