Like this document? Why not share!

- Centroids moments of inertia by coolzero2012 63246 views
- Unsymmetrical bending and shear centre by Yatin Singh 9184 views
- Hibbeler chapter10 by ahmedalnamer 96235 views
- Unsymmetrical bending.ppt by Venkatesh Ca 4964 views
- Unsymmetrical bending by flyersfan376 10603 views
- MOMENT OF INERTIA by Azizul Izham 26322 views

12,338

-1

-1

Published on

No Downloads

Total Views

12,338

On Slideshare

0

From Embeds

0

Number of Embeds

0

Shares

0

Downloads

196

Comments

0

Likes

1

No embeds

No notes for slide

- 1. Lecture 5 – Moment of Inertia of Non-symmetric Shapes In general, most cross-sectional shapes of structural members are symmetric (i.e., mirror image on both sides of both neutral axes). The determination of section properties for these symmetric shapes involves plugging in numbers into the formulas as discussed in Lecture 4. Non-symmetric shapes are those that are NOT mirror image on both sides of one or both of the neutral axes. Examples of non-symmetric shapes are as follows: Angles T-Sections Channels Sections w/ unsymmetric holes The procedure for determining the moment of inertia of these non-symmetric shapes involves two steps: 1) Determine location of centroid of entire shape, y and x y= Σ( A piece y piece ) ΣA piece where: Apiece = area of the individual piece ypiece = dist. to centroid of the individual piece x= Σ( A piece x piece ) ΣA piece 2) Determine the transformed moment of inertia, It by using the “Parallel Axis Theorem” I t = ∑ ( I + Ad 2 ) piece where: I = moment of inertia of the piece A = area of the piece d = y – ypiece Lecture 5 - Page 1 of 8
- 2. Example 1 GIVEN: A “T” shaped beam using a nominal wood 2x8 web with a 2x4 top flange REQUIRED: Determine a) The location of the neutral axis y b) The transformed moment of inertia about the strong axis c) The moment of inertia about the weak axis Piece “B” 3½” 1½” 7¼” yB y Piece “A” (centered under piece “B”) yA Datum 1½” a) Location of neutral axis y: Piece Area (1.5”)(7.25”) = 10.88 in2 A B (1.5”)(3.5”) = 5.25 in2 16.13 in2 Totals: y= y ½(7.25”) = 3.625” 7.25” + ½(1.5”) = 8” Σ( A piece y piece ) ΣA piece = 81.44 in3 16.13 in2 y = 5.05 in. Lecture 5 - Page 2 of 8 (Area)y (10.88)(3.625”) = 39.44 in3 (5.25 in2)(8”) = 42 in3 81.44 in3
- 3. b) Determine transformed moment of inertia about the strong “x” axis: Area Piece A 10.88 in 3.625” 39.44 in B 5.25 in2 8” 42 in3 (3.5" )(1.5" ) 3 = 0.98in 4 12 Totals: 16.13 in2 81.44 in3 Ad2 48.61 in4 2 y (Area)y 3 I (1.5" )(7.25" ) 3 = 47.63in 4 12 d = y-ypiece 3.625 - 5.05 = -1.425” 8 - 5.05 = 2.95” (10.88)(-1.425)2 = 22.09 in4 (5.25)(2.95)2 = 45.69 in4 67.78 in4 I x = ∑ ( I + Ad 2 ) piece = 48.61 in4 + 67.78 in4 Ix= 116.39 in4 c) Determine the moment of inertia about the weak “y” axis: Piece “B” 3½” 1½” Neutral axis of piece “A” and “B” 7¼” Piece “A” (centered under piece “B”) 1½” Since the neutral axis of both pieces line–up over each other, the total moment of inertia is the sum of the moment of inertias of the pieces. Iy = IA + IB Iy = (7.25)(1.5) 3 (1.5)(3.5) 3 + = Iy = 7.4 in4 12 12 Lecture 5 - Page 3 of 8
- 4. Example 2 GIVEN: A steel W18x35 “I” beam reinforced with a ½” x 8” steel plate welded to the bottom flange of the beam as shown below REQUIRED: Determine: a) The location of the neutral axis “y” b) The transformed moment of inertia about the strong axis c) The moment of inertia about the weak axis d) The section modulus about the strong axis e) The radius of gyration about the strong axis Piece 2 W18x35 steel beam (Area = 10.3 in2) (Ix = 510 in4) (Iy =15.3 in4) 8.85” From textbook 17.70” 8.85” y Y2 Datum 0.5” Piece 1 Y1 = 0.25” ½” x 8” steel plate welded to center of bottom flange of beam Make a Table as shown below: Piece 1 Area 4 in2 0.25” (Area)y 1 in3 2 10.3 in2 9.35” 96.3 in3 510 in4 Totals: 14.3 in2 97.3 in3 510.08 in4 y I (8" )(0.5" ) 3 = 0.08in 4 12 d = y-ypiece 6.80”-0.25” = 6.55” (4)(6.55)2 = 171.6 in4 6.80”–9.35” = -2.55” (10.3)(-2.55)2 = 66.98 in4 a) Determine location of neutral axis “y”: y= Σ( A piece y piece ) ΣA piece = 97.3in 3 14.3in 2 y = 6.80” Lecture 5 - Page 4 of 8 Ad2 238.58 in4
- 5. b) Determine Transformed Moment of Inertia about Strong Axis: I x = ∑ ( I + Ad 2 ) piece = 510.08 in4 + 238.58 in4 Ix= 748.66 in4 c) Determine Moment of Inertia about Weak Axis: W18x35 steel beam (Area = 10.3 in2) (Ix = 510 in4) (Iy =15.3 in4) 4” 4” 8” Since the neutral axis of both pieces line–up over each other, the total moment of inertia is the sum of the moment of inertias of the pieces. Iy = I1 + I2 (0.5" )(8" ) 3 Iy = + 15.3in 4 12 Iy = 36.63 in4 Lecture 5 - Page 5 of 8
- 6. d) Determine the Section Modulus about Strong Axis: S strong = = I strong y strong 748.66in 4 6.80" Sstrong = 110.1 in3 e) Determine Radius of Gyration about Strong Axis: rstrong = rstrong = I s trong Atot 748.66in 4 14.3in 2 rstrong = 7.24” Lecture 5 - Page 6 of 8
- 7. Example 3 GIVEN: Repeat Example 1, using a 2x4 top flange and a 2x8 web. REQUIRED: Using AutoCAD, determine the following: a) The location of the neutral axis y b) The transformed moment of inertia about the strong axis d) The moment of inertia about the weak axis Piece “B” 3½” 1½” 7¼” yB y Piece “A” (centered under piece “B”) yA Datum 1½” Lecture 5 - Page 7 of 8
- 8. Using AutoCAD, draw the shape shown and make a “REGION” out of it: Results are exactly the same as in Example 1 Lecture 5 - Page 8 of 8

No public clipboards found for this slide

×
### Save the most important slides with Clipping

Clipping is a handy way to collect and organize the most important slides from a presentation. You can keep your great finds in clipboards organized around topics.

Be the first to comment