A beam is a structural member that is subjected primarily to transverse loads and negligible


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1 Chapter. Design of Beams Flexure and Shear.1 Section forcedeformation response & Plastic Moment (M p ) 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. w P x V(x) M(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 crosssection as shown in the Figure below. ε σ df = σ bdy d y M(x) V(x) b Curvature = φ = ε/d F ε d / = + d / σ b dy σ M d / = + d / σ b dy y (Planes remain plane) Figure. Longitudinal axial stresses caused by internal bending moment. 1
2 Steel material follows a typical stressstrain behavior as shown in Figure 3 below. σ u σ ε y ε Figure 3. Typical steel stressstrain behavior. ε u If the steel stressstrain curve is approximated as a bilinear elastoplastic curve with yield stress equal to, then the section Moment  Curvature (Mφ) response for monotonically increasing moment is given by Figure 4. M p B C D E Section Moment, M M y A ε y ε y 5ε y 10ε y ε y ε y 5ε y Curvature, φ A: Extreme fiber reaches ε y B: Extreme fiber reaches ε y C: Extreme fiber reaches 5ε y D: Extreme fiber reaches 10ε y E: Extreme fiber reaches infinite strain Figure 4. Section Moment  Curvature (Mφ) behavior. 10ε y A B C D E
3 In Figure 4, M y is the moment corresponding to first yield and M p is the plastic moment capacity of the crosssection.  The ratio of M p to M y is called as the shape factor f for the section.  For a rectangular section, f is equal to 1.5. For a wideflange section, f is equal to 1.1. Calculation of M p : Crosssection subjected to either + or  at the plastic limit. See Figure 5 below. A 1 Plastic centroid. A 1 A y 1 y A (a) General crosssection (b) Stress distribution (c) Force distribution F = σ A 1 y A 1 = A y σ A 1 y = A / = 0 A M = σ y (y 1 + y ) Where, y = centroid of = centroid of A 1 A (d) Equations Figure 5. Plastic centroid and M p for general crosssection. The plastic centroid for a general crosssection corresponds to the axis about which the total area is equally divided, i.e., A 1 = A = A/  The plastic centroid is not the same as the elastic centroid or center of gravity (c.g.) of the crosssection. 3
4  As shown below, the c.g. is defined as the axis about which A 1 y 1 = A y. y y 1 c.g. = elastic N.A. A 1, y 1 About the c.g. A 1 y 1 = A y A, y For a crosssection with atleast one axis of symmetry, the neutral axis corresponds to the centroidal axis in the elastic range. However, at M p, the neutral axis will correspond to the plastic centroidal axis. For a doubly symmetric crosssection, the elastic and the plastic centroid lie at the same point. M p = x A/ x (y 1 +y ) As shown in Figure 5, y 1 and y are the distance from the plastic centroid to the centroid of area A 1 and A, respectively. A/ x (y 1 +y ) is called Z, the plastic section modulus of the crosssection. Values for Z are tabulated for various crosssections in the properties section of the LRFD manual. φ M p = 0.90 Z F y  See Spec. F1.1 where, M p = plastic moment, which must be 1.5 M y for homogenous crosssections M y = moment corresponding to onset of yielding at the extreme fiber from an elastic stress distribution = F y S for homogenous crosssections and = F yf 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. 4
5 Example.1 Determine the elastic section modulus, S, plastic section modulus, Z, yield moment, M y, and the plastic moment M p, of the crosssection shown below. What is the design moment for the beam crosssection. Assume 50 ksi steel. 1 in. F in. W t w = 0.5 in. 16 in. F 1.0 in. 15 in. A g = 1 x ( ) x x 1.0 = in A f1 = 1 x 0.75 = 9 in A f = 15 x 1.0 = 15.0 in A w = 0.5 x ( ) = 7.15 in distance of elastic centroid from bottom = y 9 ( / ) y = = in I x = / / / = 1430 in 4 S x = I x / ( ) = in 3 M yx = F y S x = kipin. = kipft. distance of plastic centroid from bottom = y p (y y p =.15 in. p ) = =
6 y 1 =centroid of top halfarea about plastic centroid = = in y =centroid of bottom halfarea about plas. cent. = = in. Z x = A/ x (y 1 + y ) = x ( ) = in 3 M px = Z x F y = x 50 = kipin. = kipft. Design strength according to AISC Spec. F1.1= φ b M p = 0.9 x = kipft. Check = M p 1.5 M y Therefore, kipft. < 1.5 x = kipft.  OK! Reading Assignment 6
7 . Flexural Deflection of Beams Serviceability Steel beams are designed for the factored design loads. The moment capacity, i.e., the factored moment strength (φ b M n ) should be greater than the moment (M u ) caused by the factored loads. A serviceable structure is one that performs satisfactorily, not causing discomfort or perceptions of unsafety for the occupants or users of the structure.  For a beam, being serviceable usually means that the deformations, primarily the vertical slag, or deflection, must be limited.  The maximum deflection of the designed beam is checked at the servicelevel loads. The deflection due to servicelevel loads must be less than the specified values. The AISC Specification gives little guidance other than a statement in Chapter L, Serviceability Design Considerations, that deflections should be checked. Appropriate limits for deflection can be found from the governing building code for the region. The following values of deflection are typical maximum allowable total (service dead load plus service live load) deflections. Plastered floor construction L/360 Unplastered floor construction L/40 Unplastered roof construction L/180 In the following examples, we will assume that local buckling and lateraltorsional buckling are not controlling limit states, i.e, the beam section is compact and laterally supported along the length. 7
8 Example. 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 selfweight of the beam. Step I. Calculate the factored design loads (without selfweight). w U = 1. w D w L = 1.4 kips / ft. M U = w u L / 8 = 1.4 x 30 / 8 = kipft. Step II. Select the lightest section from the AISC Manual design tables. From page of the AISC manual, select W16 x 6 made from 50 ksi steel with φ b M p = kipft. Step III. Add selfweight of designed section and check design w sw = 6 lbs/ft Therefore, w D = 476 lbs/ft = lbs/ft. w u = 1. x x 0.55 = kips/ft. Therefore, M u = x 30 / 8 = kipft. < φ b M p of W16 x 6. OK! Step IV. Check deflection at service loads. w = kips/ft. = 1.06 kips/ft. = 5 w L 4 / (384 E I x ) = 5 x (1.06/1) x (30 x 1) 4 / (384 x 9000 x 301) =.14 in. > L/360  for plastered floor construction Step V. Redesign with serviceload deflection as design criteria L /360 = 1.0 in. > 5 w L 4 /(384 E I x ) Therefore, I x > in 4 8
9 Select the section from the moment of inertia selection tables in the AISC manual. See page select W1 x 44. W1 x 44 with I x = 843 in 4 and φ b M p = 358 kipft. (50 ksi steel). Deflection at service load = = in. < L/360  OK! Note that the serviceability design criteria controlled the design and the section Example.3 Design the beam shown below. The unfactored dead and live loads are shown in the Figure. 10 kips (live load) 0.67 k/ft. (dead load) 0.75 k/ft. (live load) 15 ft. 30 ft. Step I. Calculate the factored design loads (without selfweight). w u = 1. w D w L = 1. x x 0.75 =.004 kips / ft. P u = 1. P D P L = 1. x x 10 = 16.0 kips M u = w U L / 8 + P U L / 4 = = kipft. Step II. Select the lightest section from the AISC Manual design tables. From page of the AISC manual, select W1 x 44 made from 50 ksi steel with φ b M p = kipft. Selfweight = w sw = 44 lb/ft. 9
10 Step III. Add selfweight of designed section and check design w D = = kips/ft w u = 1. x x 0.75 =.0568 kips/ft. Therefore, M u =.0568 x 30 / = kipft. < φ b M p of W1 x 44. OK! Step IV. Check deflection at service loads. Service loads Distributed load = w = = kips/ft. Concentrated load = P = D + L = kips = 10 kips Deflection due to uniform distributed load = d = 5 w L 4 / (384 EI) Deflection due to concentrated load = c = P L 3 / (48 EI) Therefore, serviceload deflection = = d + c = 5 x x / (384 x 9000 x 1 x 843) + 10 x / (48 x 9000 x 843) = = 1.49 in. Assuming unplastered floor construction, max = L/40 = 360/40 = 1.5 in. Therefore, < max  OK! 10
11 .3 Local buckling of beam section Compact and Noncompact M p, the plastic moment capacity for the steel shape, is calculated by assuming a plastic stress distribution (+ or  ) over the crosssection. The development of a plastic stress distribution over the crosssection can be hindered by two different length effects: (1) Local buckling of the individual plates (flanges and webs) of the crosssection before they develop the compressive yield stress. () Lateraltorsional buckling of the unsupported length of the beam / member before the crosssection develops the plastic moment M p. M M Figure 7. Local buckling of flange due to compressive stress (σ) The analytical equations for local buckling of steel plates with various edge conditions and the results from experimental investigations have been used to develop limiting slenderness ratios for the individual plate elements of the crosssections. See Spec. B5 (page ), Table B5.1 ( ) and Page of the AISCmanual Steel sections are classified as compact, noncompact, or slender depending upon the slenderness (λ) ratio of the individual plates of the crosssection. 11
12  Compact section if all elements of crosssection have λ λ p  Noncompact sections if any one element of the crosssection has λ p λ λ r  Slender section if any element of the crosssection has λ r λ It is important to note that:  If λ λ p, then the individual plate element can develop and sustain for large values of ε before local buckling occurs.  If λ p λ λ r, then the individual plate element can develop but cannot sustain it before local buckling occurs.  If λ r λ, then elastic local buckling of the individual plate element occurs. Compressive axial stress, σ Compact NonCompact Slender Effective axial strain, ε Figure 8. Stressstrain response of plates subjected to axial compression and local buckling. Thus, slender sections cannot develop M p due to elastic local buckling. Noncompact sections can develop M y but not M p before local buckling occurs. Only compact sections can develop the plastic moment M p. All rolled wideflange shapes are compact with the following exceptions, which are noncompact. 1
13  W40x174, W14x99, W14x90, W1x65, W10x1, W8x10, W6x15 (made from A99) The definition of λ and the values for λ p and λ r for the individual elements of various crosssections are given in Table B5.1 and shown graphically on page For example, Section Plate element λ λ p λ r Wideflange Flange b f /t f 0.38 E / Fy 0.38 E / FL Web h/t w 3.76 E / Fy 5.70 E / Fy Channel Flange b f /t f 0.38 E / Fy 0.38 E / FL Web h/t w 3.76 E / Fy 5.70 E / Fy Square or Rect. Box Flange (b3t)/t 1.1 E / Fy 1.40 E / Fy Web (b3t)/t 3.76 E / Fy 5.70 E / Fy In CE405 we will design all beam sections to be compact from a local buckling standpoint 13
14 .4 LateralTorsional Buckling The laterally unsupported length of a beammember can undergo lateraltorsional buckling due to the applied flexural loading (bending moment). (a) M (b) M M M Figure 9. Lateraltorsional buckling of a wideflange beam subjected to constant moment. 14
15 Lateraltorsional buckling is fundamentally similar to the flexural buckling or flexuraltorsional buckling of a column subjected to axial loading.  The similarity is that it is also a bifurcationbuckling type phenomenon.  The differences are that lateraltorsional buckling is caused by flexural loading (M), and the buckling deformations are coupled in the lateral and torsional directions. There is one very important difference. For a column, the axial load causing buckling remains constant along the length. But, for a beam, usually the lateraltorsional buckling causing bending moment M(x) varies along the unbraced length.  The worst situation is for beams subjected to uniform bending moment along the unbraced length. Why?.4.1 Lateraltorsional buckling Uniform bending moment Consider a beam that is simplysupported at the ends and subjected to fourpoint loading as shown below. The beam centerspan is subjected to uniform bending moment M. Assume that lateral supports are provided at the load points. P P L b Laterally unsupported length = L b. If the laterally unbraced length L b is less than or equal to a plastic length L p then lateral torsional buckling is not a problem and the beam will develop its plastic strength M p. L p = 1.76 r y x E / Fy  for I members & channels (See Pg ) 15
16 If L b is greater than L p then lateral torsional buckling will occur and the moment capacity of the beam will be reduced below the plastic strength M p as shown in Figure 10 below. Z x F y = M p M n = M p L b L M n = M p ( M p M r ) L r L p p S x (F y 10) =M r Moment Capacity, M n M n = π EI L b y π EC GJ + L b w L p L Lrr Unbraced length, L b Figure 10. Moment capacity (M n ) versus unsupported length (L b ). As shown in Figure 10 above, the lateraltorsional buckling moment (M n = M cr ) is a function of the laterally unbraced length L b and can be calculated using the equation: M n = M cr = π L b E I y π E G J + L b I y C w where, M n = moment capacity L b = laterally unsupported length. M cr = critical lateraltorsional buckling moment. E = 9000 ksi; G = 11,00 ksi I y = moment of inertia about minor or yaxis (in 4 ) J = torsional constant (in 4 ) from the AISC manual pages. C w = warping constant (in 6 ) from the AISC manual pages. 16
17 This equation is valid for ELASTIC lateral torsional buckling only (like the Euler equation). That is it will work only as long as the crosssection is elastic and no portion of the crosssection has yielded. As soon as any portion of the crosssection reaches the yield stress F y, the elastic lateral torsional buckling equation cannot be used.  L r is the unbraced length that corresponds to a lateraltorsional buckling moment M r = S x (F y 10).  M r will cause yielding of the crosssection due to residual stresses. When the unbraced length is less than L r, then the elastic lateral torsional buckling equation cannot be used. When the unbraced length (L b ) is less than L r but more than the plastic length L p, then the lateraltorsional buckling M n is given by the equation below:  If L p L b L r, then L M n = M p ( M p M r ) L b r L L p p  This is linear interpolation between (L p, M p ) and (L r, M r )  See Figure 10 again. 17
18 .4. Moment Capacity of beams subjected to nonuniform bending moments As mentioned previously, the case with uniform bending moment is worst for lateral torsional buckling. For cases with nonuniform bending moment, the lateral torsional buckling moment is greater than that for the case with uniform moment. The AISC specification says that:  The lateral torsional buckling moment for nonuniform bending moment case = C b x lateral torsional buckling moment for uniform moment case. C b is always greater than 1.0 for nonuniform bending moment.  C b is equal to 1.0 for uniform bending moment.  Sometimes, if you cannot calculate or figure out C b, then it can be conservatively assumed as 1.0. C b =.5 M max 1.5 M + 3 M A max + 4 M B + 3 M c where, M max = magnitude of maximum bending moment in L b M A = magnitude of bending moment at quarter point of L b M B = magnitude of bending moment at half point of L b M C = magnitude of bending moment at threequarter point of L b The moment capacity M n for the case of nonuniform bending moment  M n = C b x {M n for the case of uniform bending moment} M p  Important to note that the increased moment capacity for the nonuniform moment case cannot possibly be more than M p.  Therefore, if the calculated values is greater than M p, then you have to reduce it to M p 18
19 M p Moment Capacity, M n M r C b = 1.5 C b = 1. C b = 1.0 L p L r.5 Beam Design Unbraced length, L b Figure 11. Moment capacity versus L b for nonuniform moment case. Example.4 Design the beam shown below. The unfactored uniformly distributed live load is equal to 3 kips/ft. There is no dead load. Lateral support is provided at the end reactions. w L = 3 kips/ft. 4 ft. Lateral support / bracing Step I. Calculate the factored loads assuming a reasonable selfweight. Assume selfweight = w sw = 100 lbs/ft. Dead load = w D = = 0.1 kips/ft. Live load = w L = 3.0 kips/ft. Ultimate load = w u = 1. w D w L = 4.9 kips/ft. Factored ultimate moment = M u = w u L /8 = kipft. 19
20 Step II. Determine unsupported length L b and C b There is only one unsupported span with L b = 4 ft. C b = 1.14 for the parabolic bending moment diagram, See values of C b shown in Figure. Step III. Select a wideflange shape The moment capacity of the selected section φ b M n > M u (Note φ b = 0.9) φ b M n = moment capacity = C b x (φ b M n for the case with uniform moment) φ b M p  Pages in the AISCLRFD manual, show the plots of φ b M n L b for the case of uniform bending moment (C b =1.0)  Therefore, in order to select a section, calculate M u /C b and use it with L b to find the first section with a solid line as shown in class.  M u /C b = 354.4/1.14 = kipft.  Select W16 x 67 (50 ksi steel) with φ b M n =357 kipft. for L b = 4 ft. and C b =1.0  For the case with C b = 1.14, φ b M n = 1.14 x 357 = kipft., which must be φ b M p = 491 kipft. OK! Thus, W16 x 67 made from 50 ksi steel with moment capacity equal to kipft. for an unsupported length of 4 ft. is the designed section. Step IV. Check for local buckling. λ = b f / t f = 7.7; Corresponding λ p = 0.38 (E/Fy) 0.5 = 9.19 Therefore, λ < λ p  compact flange λ = h/t w = 34.4; Corresponding λ p = 3.76 (E/F y ) 0.5 = 90.5 Therefore, λ < λ p Compact section.  compact web  OK! This example demonstrates the method for designing beams and accounting for C b > 1.0 0
21 Example.5 Design the beam shown below. The concentrated live loads acting on the beam are shown in the Figure. The beam is laterally supported at the load and reaction points. w sw = 0.1 kips/ft. 30 kips 30 kips 1 ft. 8 ft. 10 ft. Lateral support / bracing 30 ft. Step I. Assume a selfweight and determine the factored design loads Let, w sw = 100 lbs/ft. = 0.1 kips/ft. P L = 30 kips P u = 1.6 P L = 48 kips w u = 1. x w sw = 0.1 kips/ft. The reactions and bending moment diagram for the beam are shown below. 48 kips 48 kips w sw = 0.1 kips/ft. A B C D 1 ft. 8 ft. 10 ft kips 53 kips A B C D kipft. 54 kipft. Step II. Determine L b, C b, M u, and M u /C b for all spans. 1
22 Span L b C b M u M u /C b (ft.) (kipft.) (kipft.) AB BC (assume) CD It is important to note that it is possible to have different L b and C b values for different laterally unsupported spans of the same beam. Step III. Design the beam and check all laterally unsupported spans Assume that span BC is the controlling span because it has the largest M u /C b although the corresponding L b is the smallest. From the AISCLRFD manual select W1 x 68 made from 50 ksi steel (page ) Check the selected section for spans AB, BC, and CD Span L b (ft.) φ b M n for C b = 1.0 from C b φ b M n for C b value col. 3 x col. 4 φ b M p limit AB kipft BC CD kipft. Thus, for span AB, φ b M n = 600 kipft. > M u for span BC, φ b M n = 57.0 kipft. > M u For span CD, φ b M n = 600 kipft. > M u  OK! OK! OK!
23 Step IV. Check for local buckling λ = b f / t f = 6.0; Corresponding λ p = 0.38 (E/Fy) 0.5 = 9.19 Therefore, λ < λ p  compact flange λ = h/t w = 43.6; Corresponding λ p = 3.76 (E/F y ) 0.5 = Therefore, λ < λ p  compact web Compact section.  OK! This example demonstrates the method for designing beams with several laterally unsupported spans with different L b and C b values. Example.6 Design the simplysupported beam shown below. The uniformly distributed dead load is equal to 1 kips/ft. and the uniformly distributed live load is equal to kips/ft. A concentrated live load equal to 10 kips acts at the midspan. Lateral supports are provided at the end reactions and at the midspan. w D = 1.0 kips/ft. w L =.0 kips/ft. 10 kips A B C 1 ft. 1 ft. Step I. Assume the selfweight and calculate the factored design loads. Let, w sw = 100 lbs/ft. = 0.1 kips/ft. w D = = 1.1 kips/ft. w L =.0 kips/ft. w u = 1. w D w L = 4.5 kips/ft. P u = 1.6 x 10 = 16.0 kips 3
24 The reactions and the bending moment diagram for the factored loads are shown below. w u = 4.5 kips/ft. 16 kips B 1 ft. 1 ft. 6.4 kips 6.4 kips x M(x) = 6.4 x x / Step II. Calculate L b and C b for the laterally unsupported spans. Since this is a symmetric problem, need to consider only span AB L b = 1 ft.; C b =.5 M max 1.5 M max + 3 M + 4 M A B + 3 M c M(x) = 6.4 x 4.5 x / Therefore, M A = M(x = 3 ft.) = kipft. M B = M(x = 6 ft.) = 9.08 kipft. M C = M(x = 9ft.) = kipft M max = M(x = 1 ft.) = kipft.  quarterpoint along L b = 1 ft.  halfpoint along L b = 1 ft. threequarter point along L b = 1 ft.  maximum moment along L b =1ft. Therefore, C b = 1.37 Step III. Design the beam section M u = M max = kipft. L b = 1.0 ft.; C b = 1.37 M u /C b = 41.44/1.37 = kipft.  Select W1 x 48 made from 50 ksi with φ b M n = 3 kipft. for L b = 1.0 ft. and C b =1.0  For C b = 1.37, φ b M n = _ kft., but must be < or = φ b M p = 398 kft. 4
25  Therefore, for C b =1.37, φ b M n = 398 kft. < M u Step IV. Redesign the section  Select the next section with greater capacity than W1 x 48  Select W18 x 55 with φ b M n = 345 kft. for L b = 1 ft. and C b = 1.0 For C b = 1.37, φ b M n = 345 x 1.37 = kft. but must be φ b M p = 40 kft. Therefore, for C b = 1.37, φ b M n = 40 kft., which is < M u (41.44 kft), (NOT OK!)  Select W 1 x 55 with φ b M n = 388 kft. for L b = 1 ft. and C b = 1.0 For C b 1.37, φ b M n = 388 x 1.37 = kft., but must be φ b M p = 473 kft. Therefore, for C b = 1.37, φ b M n = 473 kft, which is > M u (41.44 kft). (OK!) Step V. Check for local buckling. λ = b f / t f = 7.87; Corresponding λ p = 0.38 (E/F y ) 0.5 = 9.19 Therefore, λ < λ p  compact flange λ = h/t w = 50.0; Corresponding λ p = 3.76 (E/F y ) 0.5 = Therefore, λ < λ p  compact web Compact section.  OK! This example demonstrates the calculation of C b and the iterative design method. 5
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