RCH 1 Note Set 9. S015abn Moments of nertia Notation: b d d d h c Jo O = name for area = name for a (base) width = calculus smbol for differentiation = name for a difference = name for a depth = difference in the direction between an area centroid ( ) and the centroid of the composite shape ( ˆ ) = difference in the direction between an area centroid ( ) and the centroid of the composite shape ( ŷ ) = name for a height = moment of inertia about the centroid = moment of inertia of a component about the centroid = moment of inertia with respect to an -ais = moment of inertia with respect to a -ais = polar moment of inertia, as is J = name for reference origin ro r r tf tw ˆ ŷ P L = polar radius of gration = radius of gration with respect to an -ais = radius of gration with respect to a -ais = thickness of a flange = thickness of web of wide flange = horizontal distance = the distance in the direction from a reference ais to the centroid of a shape = the distance in the direction from a reference ais to the centroid of a composite shape = vertical distance = the distance in the direction from a reference ais to the centroid of a shape = the distance in the direction from a reference ais to the centroid of a composite shape = plate smbol = smbol for integration = summation smbol The cross section shape and how it resists bending and twisting is important to understanding beam and column behavior. Definition: Moment of nertia; the second area moment d d We can define a single integral using a narrow strip: d = d for,, strip is parallel to for, strip is parallel to * can be negative if the area is negative (a hole or subtraction). el d shape that has area at a greater distance awa from an ais through its centroid will have a larger value of. 1
RCH 1 Note Set 9. S015abn Just like for center of gravit of an area, the moment of inertia can be determined with respect to an reference ais. Definition: Polar Moment of nertia; the second area moment using polar coordinate aes J o J o r d d d Definition: Radius of Gration; the distance from the moment of inertia ais for an area at which the entire area could be considered as being concentrated at. r r r radius of gration in radius of gration in J o ro polar radius of gration, and ro = r + r pole o r The Parallel-is Theorem The moment of inertia of an area with respect to an ais not through its centroid is equal to the moment of inertia of that area with respect to its own parallel centroidal ais plus the product of the area and the square of the distance between the two aes. d d d -d d d d d B d d B ais through centroid at a distance d awa from the other ais ais to find moment of inertia about but d 0, because the centroid is on this ais, resulting in: c d (tet notation) or d where c (or )is the moment of inertia about the centroid of the area about an ais and d is the distance between the parallel aes
RCH 1 Note Set 9. S015abn Similarl d Moment of inertia about a ais J o J c d Polar moment of nertia ro rc d Polar radius of gration r Radius of gration r d * can be negative again if the area is negative (a hole or subtraction). ** f is not given in a chart, but are: YOU MUST CLCULTE & WTH d Composite reas: where is the moment of inertia about the centroid of the component area d Basic Steps d is the distance from the centroid of the component area to the centroid of the composite area (ie. d = - ) ŷ 1. Draw a reference origin.. Divide the area into basic shapes. Label the basic shapes (components) 4. Draw a table with headers of Component, rea,,,,,, d, d,, d, d 5. Fill in the table values needed to calculate ˆ and 6. Fill in the rest of the table values. ŷ for the composite 7. Sum the moment of inertia ( s) and d columns and add together.
RCH 1 Note Set 9. S015abn Geometric Properties of reas about bottom left rea = bh = b/ = h/ Triangle b 1 ' 6b h rea = bh b h rea = r = 0 = 0 d 4 = = 0.1098r 4 4 r 8 = 0.0549r 4 = 0.0549r 4 rea = r d = 0 rea = r d 4 = 4r = 4r 8 = 4r 16 rea = = 0 = 0 ab = 16ah 175 = 4a h 15 rea = = 0 4ah = h 5 = 7ah 100 rea = ah = a h 80 = a = h 4 10 4
RCH 1 Note Set 9. S015abn Eample 1 (pg 57) 1 ˆ.05" Find the moments of inertia ( ˆ =.05, ŷ = 1.05 ). 1 ˆ 1.05" Eample (pg 5) 5
RCH 1 Note Set 9. S015abn Eample Determine the moments of inertia about the centroid of the shape. Solution: There is no reference origin suggested in figure (a), so the bottom left corner is good. n figure (b) area will be a complete rectangle, while areas C and are "holes" with negative area and negative moment of inertias. o rea = 00 mm 100 mm = 0000 mm = (00 mm)(100 mm) /1 = 16.667 10 6 mm 4 = (00 mm) (100 mm)/1 = 66.667 10 6 mm 4 rea B = -(0 mm) = -87.4 mm = = - (0 mm) 4 /4 = -0.66 10 6 mm 4 rea C = -1/(50 mm) = 97.0 mm = - (50 mm) 4 /8 = -.454 10 6 mm 4 = -0.1098(50 mm) 4 = -0.686 10 6 mm 4 rea D = 100 mm 00 mm 1/ = 10000 mm = (00 mm)(100 mm) /6 = 5.556 10 6 mm 4 = (00 mm) (100 mm)/6 =. 10 6 mm 4 shape (mm ) (mm) (mm ) (mm) (mm ) 0000 100 000000 50 1000000 B -87.4 150-44115 50-1417 C -96.99 1.066-8. 50-19650 D 10000 66.66667 666666.7 1. 1 45.58 15918 199561 15918mm ˆ 45.58mm 199561 mm ˆ 45.58 mm 9. 9 mm 85. 8 mm shape (mm 4 ) d (mm) d (mm 4 ) (mm 4 ) d (mm) d (mm 4 ) 16666667 5.8 56800 66666667-7.1 100800 B -6617 5.8-6751.7-6617 -57.1-91859.09 C -45469 5.8-50988.51-68650 71.6794-0176595. D 5555556-47.5 594177.8 6. 6881876.09 1911680 95707.5 87566466-1505111.9 So, = 1911680 + 95707.5 = 58701918 = 58.7 10 6 mm 4 = 87566466 +-1505111. = 45705 = 66.1 10 6 mm 4 6
RCH 1 Note Set 9. S015abn Eample 4 (pg 58) 7
RCH 1 Note Set 9. S015abn Eample 5 (pg 49)*.8 lso determine the moment of inertia about both major centroidal aes. 0. shape (in ) (in) (in ) (in) (in ) channel 6.09 0 0.00 9.694 59.04 left plate 5 -.5-16.5 5.11 5.55 right plate 5.5 16.5 5.11 5.55 wide flange 4.71 0 0.00 0 0.00 0.80 0.00 110.14 0 in ˆ 0.8 in 110.14 in ˆ 0.8 in 0 in 5.95 in shape (in 4 ) d (in) d (in 4 ) (in 4 ) d (in) d (in 4 ) channel.880-4.99 117.849 19.000 0.000 0.000 left plate 41.667 0.185 0.171 0.104.50 5.81 right plate 41.667 0.185 0.171 0.104 -.50 5.81 wide flange.800 5.95 1.054 10.000 0.000 0.000 90.01 50.45.08 105.65 = 90.01 + 50.45 = 40.59 = 40. in 4 =.08 + 105.65 = 7.8 = 7.8 in 4 8