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CVENG324: PRINCIPLES OF STEEL DESIGN INSTRUCTOR: ENGR. JOSHUA FERDINAND B. VILLAFUERTE MODULE 1: PROPERTIES OF SECTION PART II SHAPE STRUCTURAL SHAPES Structural steels are available of many shapes. The dimension and weight must be added to the designation to uniquely identify the shape. Example: W 40 x 436 DESIGNATION Wide Flange Beam W American Standard Beam S Bearing Piles HP Miscellaneous (those that cannot be classified as W, S, or HP) M Channel C Angle L Structural Tee (cut from W or S or M) Structural Tubing WT or ST TS W-Shape Pipe Pipe Depth = 40 inches (100 mm) Plate PL Weight = 436 lb/ft (640 kg/m) Bar Bar STRUCTURAL STEEL SHAPES WIDE FLANGE SECTIONS Designation W (Nominal Depth x Mass per unit length) W 44 X 290 44 – Nominal depth = 44 inches 290 – Mass per unit length = 290 lb/ft STRUCTURAL STEEL SHAPES AMERICAN STANDARD SECTIONS Designation S (Nominal Depth x Mass per unit length) S 44 X 290 44 – Nominal depth = 44 inches 290 – Mass per unit length = 290 lb/ft STRUCTURAL STEEL SHAPES BEARING PILES Designation HP (Nominal Depth x Mass per unit length) HP 44 X 290 44 – Nominal depth = 44 inches 290 – Mass per unit length = 290 lb/ft STRUCTURAL STEEL SHAPES MISCELLANEOUS (those that cannot be classified as W, S, or HP) Designation M (Nominal Depth x Mass per unit length) M 44 X 290 44 – Nominal depth = 44 inches 290 – Mass per unit length = 290 lb/ft STRUCTURAL STEEL SHAPES CHANNEL SECTIONS Designation C (Nominal Depth x Mass per unit length) C 44 X 290 44 – Nominal depth = 44 inches 290 – Mass per unit length = 290 lb/ft STRUCTURAL STEEL SHAPES ANGLE SECTIONS Designation L (Depth x Base x Thickness) π π L8x8x1 8 – Depth 8 – Base 1 1 - Thickness 8 STRUCTURAL STEEL SHAPES DOUBLE ANGLE SECTIONS Designation L (Deepth x Base x Thickness) L8x8x1 π π 8 – Depth 8 – Base 1 8 1 - Thickness STRUCTURAL STEEL SHAPES DOUBLE ANGLE SECTIONS Designation L (Depth x Base x Thickness) L8x8x1 π π 8 – Depth 8 – Base 1 8 1 - Thickness STRUCTURAL STEEL SHAPES DOUBLE ANGLE SECTIONS Designation L (Depth x Base x Thickness) L8x8x1 π π 8 – Depth 8 – Base 1 8 1 - Thickness STRUCTURAL STEEL SHAPES STRUCTURAL TEE SECTIONS (cut from W shapes) Designation WT (Nominal Depth x Mass per unit length) WT 44 X 290 44 – Nominal depth = 44 inches 290 – Mass per unit length = 290 lb/ft STRUCTURAL STEEL SHAPES STRUCTURAL TEE SECTIONS (cut from M shapes) Designation MT (Nominal Depth x Mass per unit length) WT 44 X 290 44 – Nominal depth = 44 inches 290 – Mass per unit length = 290 lb/ft STRUCTURAL STEEL SHAPES STRUCTURAL TEE SECTIONS (cut from S shapes) Designation ST (Nominal Depth x Mass per unit length) ST 44 X 290 44 – Nominal depth = 44 inches 290 – Mass per unit length = 290 lb/ft STRUCTURAL STEEL SHAPES COMBINATION SECTIONS W SHAPES AND CHANNELS STRUCTURAL STEEL SHAPES COMBINATION SECTIONS S SHAPES AND CHANNELS STRUCTURAL STEEL SHAPES COMBINATION SECTIONS TWO CHANNELS STRUCTURAL STEEL SHAPES COMBINATION SECTIONS CHANNELS AND ANGLES (Long leg of angle turned out) STRUCTURAL STEEL SHAPES COMBINATION SECTIONS CHANNELS AND ANGLES (Short leg of angle turned out) STRUCTURAL STEEL SHAPES STRUCTURAL STEEL SHAPES SAMPLE PROBLEM # 4: MAY 2014 CE BOARD EXAM PROBLEM A built-up steel beam composed of an W shape (W 21 x 83) and a channel (C 15 x 50) bolted together as shown is subjected to a moving concentrated load P. The beam carries a 100-mm-thick concrete slab having an effective width of 1.50 m. Determine the properties of the composite section. Y π¨π π¨π π» = π(π€π βπππ) + π‘π€(πβπππππ) π» = 544.30 ππ + 18.20 ππ π― = πππ. ππ ππ π΄πβπππππ = 9,484 ππ2 π¦πβπππππ = 20.27 ππ π΄π€π = 15,677.00 ππ2 ππ€π βπππ = 544.30 ππ X π΄π‘ππ‘ππ = π΄πβπππππ + π΄π€πππ ππππππ π΄π‘ππ‘ππ = 9,484 ππ2 + 15,677.00 ππ2 π¨πππππ = ππ, πππ πππ Centroid from the Y-axis: πΏπ = ? π΄π‘ππ‘ππ ππ = σ(π΄ π₯ π) π΄π‘ππ‘ππ ππ = π΄πβπππππ ππβπππππ + π΄π€πππ ππππππ ππ€πππ ππππππ πΏπ = π Centroid from the X-axis: ππ = ? π΄π‘ππ‘ππ ππ = σ(π΄ π₯ π) π΄π‘ππ‘ππ ππ = π΄πβπππππ ππβπππππ + π΄π€πππ ππππππ ππ€πππ ππππππ (25,161 ππ2 )(ππ ) = 9,484 ππ2 (20.27 ππ) 544.30 + 15,677.00 ππ2 ( 2 ππ + 18.20 ππ) ππ = πππ. ππ ππ Y π¨π Distance of the Centroid of Area 1 and Area 2 from the Geometric Center: ππβπππππ ππ ππ€π βπππ 2 NA of Channel π π X π π π¨π NA of W shape ππ€π βπππ π ππ = ππ − ππβπππππ π ππ = ππ¦1 = 188.55 ππ − 20.27 ππ ππ¦2 = π ππ = πππ. ππ ππ π ππ = πππ. ππ ππ 2 544.30 2 + π‘π€ − ππ ππ + 18.20 − 188.55ππ Moment of Inertia with respect to X-axis: πΌπ₯ = ΰ·(πΌππ₯ + π΄π 2 ) πΌπ₯ = πΌπΆππ¦ + π΄πΆ (π1 )2 + (πΌπππ₯ + π΄π (π2 )2 ) ππ = πππ. ππ ππ π¨πππππ = ππ, πππ πππ πΏπ = π ππ = πππ. ππ ππ πΌπ₯ = [4.579π₯106 ππ4 + 9,484 ππ2 168.28 ππ 2 ] +[761.704π₯106 ππ4 + 15,677 ππ2 101.80 ππ 2 ] π°π = π, πππ. ππππππ πππ Moment of Inertia with respect to Y-axis: πΌπ¦ = ΰ·(πΌππ¦ + π΄π 2 ) πΌπ¦ = πΌπΆππ₯ + π΄πΆ (ππ₯1 )2 + (πΌπππ¦ + π΄π (ππ₯2 )2 ) πΌπ¦ = 168.157π₯106 ππ4 + 0 + 33.881π₯106 ππ4 + 0 π°π = πππ. πππππππ πππ Y π¨π X Polar Moment of Inertia with respect to X and Y axes, JO : π±πΆ = π° πΏ + π° π π½π = 1,197.32 π₯ 106 +πππ.πππ π₯ 106 π¨π π±πΆ = ππππ. πππ π πππ π¦π¦π Radius of Gyration, ππ ππ§π ππ π¨πππππ = ππ, πππ πππ πΏπ = π ππ = πππ. ππ ππ π°π = π, πππ. ππππππ πππ π°π = πππ. πππππππ πππ ππ = π°π π¨ ππ₯ = 1,197.32π₯106 ππ4 25,161 ππ2 ππ = πππ. ππ π¦π¦ ππ = π°π π¨ ππ¦ = 202.038π₯106 ππ4 25,161 ππ2 ππ = ππ. ππ π¦π¦ Y π¨π π― = πππ. ππ ππ π¨π X Section Modulus, πΊπ ππ§π πΊπ πΌπ₯ ππ₯ = π 1,197.32π₯106 ππ4 ππ₯ = 562.50 ππ − 188.55 ππ πΊπ = π. ππππ π± πππ π¦π¦π π¨πππππ = ππ, πππ πππ πΏπ = π ππ = πππ. ππ ππ π°π = π, πππ. ππππππ πππ π°π = πππ. πππππππ πππ ππ¦ = πΌπ¦ π 202.038π₯106 ππ4 ππ¦ = 381 ππ 2 πΊπ = π. πππππππ π¦π¦π SAMPLE PROBLEM # 5: Determine the properties of the double angle shown in the figure. The section is made of 2L150 x 90 x 12 with long legs back-to-back and spacing s = 6mm. B = 90 mm B = 90 mm X = 21.16 mm X = 21.16 mm π΄π‘ππ‘ππ = π΄1 + π΄2 π΄π‘ππ‘ππ = 2 π₯ (2,751.45 ππ2 ) π¨πππππ = π, πππ. ππ πππ Y = 50.82 mm H = 150 mm ππ± = ? πΌπ₯ = 2 π₯ (6.273π₯106 ππ4 ) π°π = ππ. πππππππ πππ πΌπ¦ = ΰ·(πΌππ¦ + π΄π 2 ) πΌπ¦ = 2 π₯ (πΌππ¦ + π΄(π)2 ) πΌπ¦ = 2 π₯ (1,708,656.95 ππ4 + (2,751.45 ππ2 )(21.16ππ + 3ππ)2 ) s = 6mm π°π = π, πππ, πππ πππ π΄π‘ππ‘ππ = 5,502.90 ππ2 πΌπ₯ = 12.546π₯106 ππ4 πΌπ¦ = 6.629π₯106 ππ4 Polar Moment of Inertia with respect to X and Y axes, JO : B = 90 mm B = 90 mm X = 21.16 mm π±πΆ = π°πΏ + π°π X = 21.16 mm π½π = 12.546π₯106 ππ4 + 6.629π₯106 ππ4 π±πΆ = ππ. ππππ±πππ π¦π¦π Section Modulus, πΊπ ππ§π πΊπ 12.546π₯106 ππ4 πΌπ₯ = ππ₯ = 150 ππ − 50.82 ππ π Y = 50.82 mm H = 150 mm πΌπ¦ ππ¦ = π 6.629π₯106 ππ4 = 90 ππ + 3ππ = πππ, πππ π¦π¦π = ππ, πππ π¦π¦π Radius of Gyration, ππ ππ§π ππ ππ = s = 6mm ππ = π°π π¨ π°π π¨ = 12.546π₯106 ππ4 5,502.90 ππ2 = ππ. πππ π¦π¦ = 6.629π₯106 ππ4 5,502.90 ππ2 = ππ. πππ π¦π¦ SEATWORK # 1: Y X NOVEMBER 2013 CE BOARD EXAM QUESTION: The deck of a bridge consist of ribbed metal deck with 100 mm concrete slab on top. The superstructure supporting the deck is made of wide flange steel beams strengthened by cover plate 16 mm x 260 mm one at a top and one at the bottom. Properties of W 850 x 185: A = 23,750 ππ2 d = 850 mm πππππππ = 290 ππ π‘ππππππ = 20 ππ π‘π€ππ = 15 mm πΌπ₯ = 2,662 π₯ 106 ππ4 πΌπ¦ = 81.52 π₯ 106 ππ4 Determine the following: a. Moment of Inertia along X and Y axes. b. Polar Moment of Inertia with respect to X and Y axes. c. Section Modulus, ππ₯ and ππ¦ d. Radius of Gyration, ππ₯ and ππ¦ SEATWORK # 1: Y MAY 2015 CE BOARD EXAM QUESTION: The column shown in the figure is made from four 300 mm x 16 mm A36 Steel. Determine the following: a. Moment of Inertia along X and Y axes. b. Polar Moment of Inertia with respect to X and Y axes. c. Section Modulus, ππ₯ and ππ¦ d. Radius of Gyration, ππ₯ and ππ¦ X
