Chapter 2. Design of Beams – Flexure and Shear2.2 Flexural Deflection of Beams – ServiceabilityExample 2.2 Design a simply supported beam subjected to uniformly distributed dead load of 450 lbs/ft. and a uniformly distributed live load of 550 lbs/ft. The dead load does not include the self-weight of the beam.Note that the serviceability design criteria controlled the design and the sectionTherefore, service-load deflection = ? = ?d + ?c2.3 Local buckling of beam section – Compact and SectionIn CE405 we will design all beam sections to be compact from a local buckling standpoint2.4 Lateral-Torsional Buckling2.4.2 Moment Capacity of beams subjected to non-uniform bending momentsExample 2.4Example 2.5SpanCE 405: Design of Steel Structures – Prof. Dr. A. Varma Chapter 2. Design of Beams – Flexure and Shear 2.1 Section force-deformation response & Plastic Moment (Mp) • A beam is a structural member that is subjected primarily to transverse loads and negligible axial loads. • The transverse loads cause internal shear forces and bending moments in the beams as shown in Figure 1 below. wPV(x)M(x)xwPV(x)M(x)x Figure 1. Internal shear force and bending moment diagrams for transversely loaded beams. • These internal shear forces and bending moments cause longitudinal axial stresses and shear stresses in the cross-section as shown in the Figure 2 below. V(x)M(x)ydbεεσσdF = σbdyV(x)M(x)ydbεεσσdF = σbdy Curvature = φ = 2ε/d (Planes remain plane)∫σ=+−2/d2/ddybF ydybM2/d2/d∫σ=+− Figure 2. Longitudinal axial stresses caused by internal bending moment. 1CE 405: Design of Steel Structures – Prof. Dr. A. Varma • Steel material follows a typical stress-strain behavior as shown in Figure 3 below. σyεyεuσuσεσyεyεuσuσε Figure 3. Typical steel stress-strain behavior. • If the steel stress-strain curve is approximated as a bilinear elasto-plastic curve with yield stress equal to σy, then the section Moment - Curvature (M-φ) response for monotonically increasing moment is given by Figure 4. MyMpA: Extreme fiber reaches εy B: Extreme fiber reaches 2εy C: Extreme fiber reaches 5εyD: Extreme fiber reaches 10εy E: Extreme fiber reaches infinite strainABCEDCurvature, φSectio n Mo me nt, Mσyσyσyσyεyεyσyσyσyσyσyσy2εy2εy5εy5εy10εy10εyA B C D EMyMpA: Extreme fiber reaches εy B: Extreme fiber reaches 2εy C: Extreme fiber reaches 5εyD: Extreme fiber reaches 10εy E: Extreme fiber reaches infinite strainABCEDCurvature, φSectio n Mo me nt, Mσyσyσyσyεyεyσyσyσyσyσyσy2εy2εy5εy5εy10εy10εyA B C D Eσyσyσyσyσyσyσyσyεyεyεyεyσyσyσyσyσyσyσyσyσyσyσyσy2εy2εy2εy2εy5εy5εy5εy5εy10εy10εy10εy10εyA B C D E Figure 4. Section Moment - Curvature (M-φ) behavior. 2CE 405: Design of Steel Structures – Prof. Dr. A. Varma • In Figure 4, My is the moment corresponding to first yield and Mp is the plastic moment capacity of the cross-section. - The ratio of Mp to My is called as the shape factor f for the section. - For a rectangular section, f is equal to 1.5. For a wide-flange section, f is equal to 1.1. • Calculation of Mp: Cross-section subjected to either +σy or -σy at the plastic limit. See Figure 5 below. Plastic centroid.A1A2σyσyσyA1σyA2y1y2Plastic centroid.A1A2σyσyσyA1σyA2y1y2 (a) General cross-section (b) Stress distribution (c) Force distribution 221121y212y1yAofcentroidyAofcentroidy,Where)yy(2AM2/AAA0AAF==+×σ=∴==∴=σ−σ=221121y212y1yAofcentroidyAofcentroidy,Where)yy(2AM2/AAA0AAF==+×σ=∴==∴=σ−σ= (d) Equations Figure 5. Plastic centroid and Mp for general cross-section. • The plastic centroid for a general cross-section corresponds to the axis about which the total area is equally divided, i.e., A1 = A2 = A/2 - The plastic centroid is not the same as the elastic centroid or center of gravity (c.g.) of the cross-section. 3CE 405: Design of Steel Structures – Prof. Dr. A. Varma - As shown below, the c.g. is defined as the axis about which A1y1 = A2y2. c.g. = elastic N.A.A1, y1A2, y2About the c.g. A1y1= A2y2y1y2c.g. = elastic N.A.A1, y1A2, y2About the c.g. A1y1= A2y2y1y2 • For a cross-section with at-least one axis of symmetry, the neutral axis corresponds to the centroidal axis in the elastic range. However, at Mp, the neutral axis will correspond to the plastic centroidal axis. • For a doubly symmetric cross-section, the elastic and the plastic centroid lie at the same point. • Mp = σy x A/2 x (y1+y2) • As shown in Figure 5, y1 and y2 are the distance from the plastic centroid to the centroid of area A1 and A2, respectively. • A/2 x (y1+y2) is called Z, the plastic section modulus of the cross-section. Values for Z are tabulated for various cross-sections in the properties section of the LRFD manual. • φ Mp = 0.90 Z Fy - See Spec. F1.1 where, Mp = plastic moment, which must be ≤ 1.5 My for homogenous cross-sections My = moment corresponding to onset of yielding at the extreme fiber from an elastic stress distribution = Fy S for homogenous cross-sections and = Fyf S for hybrid sections. Z = plastic section modulus from the Properties section of the AISC manual. S = elastic section modulus, also from the Properties section of the AISC manual. 4CE 405: Design of Steel Structures – Prof. Dr. A. Varma Example 2.1 Determine the elastic section modulus, S, plastic section modulus, Z, yield moment, My, and the plastic moment Mp, of the cross-section shown below. What is the design moment for the beam cross-section. Assume 50 ksi steel. 12 in.16 in.15 in.0.75 in.1.0 in.F1WF2tw= 0.5 in.12 in.16 in.15 in.0.75 in.1.0 in.F1WF212 in.16 in.15 in.0.75 in.1.0 in.F1WF2tw= 0.5 in. • Ag = 12 x 0.75 + (16 - 0.75 - 1.0) x 0.5 + 15 x 1.0 = 31.125 in2 Af1 = 12 x 0.75 = 9 in2 Af2 = 15 x 1.0 = 15.0 in2 Aw = 0.5 x (16 - 0.75 - 1.0) = 7.125 in2 • distance of elastic centroid from bottom = y .in619.6125.315.015125.8125.7)2/75.016(9y =×+×+−×= Ix = 12×0.753/12 + 9.0×9.0062 + 0.5×14.253/12 + 7.125×1.5062 + 15.0×13/12 + 156.119×2 = 1430 in4 Sx = Ix / (16-6.619) = 152.43 in3 My-x = Fy Sx = 7621.8 kip-in. = 635.15 kip-ft. • distance of plastic centroid from bottom = py .in125.2y5625.152125.31)0.1y(5.00.10.15pp=∴==−×+×∴ 5CE 405: Design of Steel Structures – Prof. Dr. A. Varma y1=centroid of top half-area about plastic centroid =