Department of Building Engineering An-Najah National University

Department of Building Engineering An-Najah National University
68402: Structural Design of Buildings II 61420: Design of Steel Structures 62323: Architectural Structures II Design of Beams for Flexure Monther Dwaikat Assistant Professor Department of Building Engineering An-Najah National University 68402

Design of Beams for Flexure
Introduction Moment Curvature Response Sectional Properties Serviceability Requirements (Deflections) Compact, Non-compact and Slender Sections Lateral Torsional Buckling Design of Beams 68402

Beams under Flexure Members subjected principally to transverse gravity loading Girders (important floor beams, wide spacing) Joists (less important beams, closely spaced) Purlins (roof beams, spanning between trusses) Stringers (longitudinal bridge beams) Lintels (short beams above window/door openings) 68402

Design for Flexure Limit states considered Yielding
Lateral-Torsional Buckling Local Buckling Compact Non-compact Slender 68402

Design for Flexure – LRFD Spec.
Commonly Used Sections: I – shaped members (singly- and doubly-symmetric) Square and Rectangular or round HSS Tees and Double Angles Rounds and Rectangular Bars Single Angles Unsymmetrical Shapes Will not be covered in this course 68402

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 SF and BM in the beams as shown in Fig. 1 Fig. 1- SF & BM in a SS Beam 68402

These internal SF & BM cause longitudinal axial stresses and shear stresses in the cross-section as shown in the Fig. 2 Curvature =  = 2/d (Planes remain plane) Fig. 2 - Longitudinal axial stresses caused by internal BM 68402

Steel material follows a typical stress-strain behavior as shown in Fig 3 below. E = 200 GPa Fig 3 - Typical steel stress-strain behavior. 68402

If the steel stress-strain curve is approximated as a bilinear elasto-plastic curve with yield stress equal to y, then the section Moment - Curvature (M-) response for monotonically increasing moment is given by Fig. 4. In Fig. 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 - the shape factor f for the section. For a rectangular section, f = 1.5. For a wide-flange section, f ≈ 1.1. 68402

Moment-Curvature (NEW)
Beam curvature  is related to its strain and thus to the applied moment e y  (1) (2) (3) (4) 68402

Moment-Curvature (NEW)
When the section is within elastic range Where S is the elastic section modulus When the moment exceeds the yield moment My Then Where Z is the plastic section modulus S 68402

Fig. 4 - M- response of a beam section
Section Force-Deformation Response & Plastic Moment (MP) Fig. 4 - M- response of a beam section 68402

Calculation of MP: Cross-section subjected to either +y or - y at the plastic limit. See Figure 5 below. (a) General cross-section (c) Force distribution (b) Stress distribution Figure 5. Plastic centroid and MP for general cross-section. 68402

Moment-Curvature When the whole section is yielding a plastic hinge will be formed plastic hinge Structural analysis by assuming collapsing mechanisms of a structure is known as “Plastic analysis” The plastic moment Mp is therefore the moment needed at the section to form a plastic hinge 68402

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. As shown below, the c.g. is defined as the axis about which A1y1 = A2y2. 68402

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 = y 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. 68402

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. bMp = 0.90 Z Fy b - strength reduction factor 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. 68402

Ex. 4.1 – Sectional Properties
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 A992 steel. 300 mm 15 mm 400 mm 10 mm 25 mm 400 mm 68402

Ag = 300 x 15 + ( ) x x 25 = mm2 Af1 = 300 x 15 = 4500 mm2 Af2 = 400 x 25 = mm2 Aw = 10 x ( ) = 3600 mm2 distance of elastic centroid from bottom = Ix = 400x253/ ( )2 + 10x3603/ ( ) x153/ ( )2 = 503.7x106 mm4 Sx = 503.7x106 / ( ) = x103 mm3 My-x = Fy Sx = kN-m. Sx - elastic section modulus 68402

distance of plastic centroid from bottom = y1 = centroid of top half-area about plastic centroid = mm y2 = centroid of bottom half-area about plas. cent. = mm Zx = A/2 x (y1 + y2) = 9050 x ( ) = mm3 Zx - plastic section modulus 68402

Mp-x = Zx Fy = x 344/106 = kN.m Design strength according to AISC Spec. F1.1= bMp= 0.9 x = kN.m Check = Mp  1.5 My Therefore, kN.m < 1.5 x = kN.m - OK! 68402

Flexural Deflection of Beams - Serviceability
Steel beams are designed for the factored design loads. The moment capacity, i.e., the factored moment strength (bMn) should be greater than the moment (Mu) 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 steel 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 service-level loads. The deflection due to service-level loads must be less than the specified values. 68402

Serviceability Requirements
Steel beams need to satisfy SLS in addition to ULS Serviceability limit states are usually checked using non-factored loads. Deflection under live loads shall be limited to L/360 Dead load deflections can be compensated by cambering beams. SLS might also include limiting stresses in bottom or top flanges if fatigue is a concern in design (Will be further discussed with plate girders). Standard equations to calculate deflection for different load cases: 68402

The AISC Specification gives little guidance other than a statement, “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 max. allowable deflections. LL DL+LL Plastered floor construction L/360 L/240 Unplastered floor construction L/240 L/180 Unplastered roof construction L/180 L/120 DL deflection – normally not considered for steel beams 68402

In the following examples, we will assume that local buckling and lateral-torsional buckling are not controlling limit states, i.e, the beam section is compact and laterally supported along the length. 68402

Ex Deflections Design a 9 m long simply supported beam subjected to UDL of 6 kN/m dead load and a UDL of 8 kN/m live load. The dead load does not include the self-weight of the beam. Step I. Calculate the factored design loads (without self-weight). wu = 1.2 wD wL = 20 kN/m Mu = wu L2 / 8 = 20 x 92 / 8 = kN.m (SS beam) Step II. Select the lightest section from the AISC Manual design tables. Zx = Mu/(bFy) = 202.5x106/(0.9x344) = 654x103 select W16 x 26 made from A992 steel with bMp = 224 kN.m 68402

Ex Deflections Step III. Add self-weight of designed section and check design wsw = 0.38 kN/m Therefore, wD = 6.38 kN/m wu = 1.2 x x 6 = kN/m Therefore, Mu = x 92 / 8 = kN.m < bMp of W16 x 26. OK! Step IV. Check deflection at service loads. w = 8 kN/m D = 5 w L4 / (384 E Ix) = 5 x (8) x103 x (9)4 / (384 x 200x125) D = 27.3 mm > L/ for plastered floor construction Step V. Redesign with service-load deflection as design criteria L /360 = 25 mm > 5 w L4/(384 E Ix) Therefore, Ix > 136.7x106 mm4 68402

Ex Deflections Select the section from the moment of inertia selection from Section Property Tables – select W16 x 31 with Ix = 156x106 mm4 Note that the serviceability design criteria controlled the design and the section hjhkjkjolklktytvtvguu 68402

Ex. 4.3 – Beam Design Design the beam shown below. The unfactored dead and live loads are shown in Fig. 6 below. Step I. Calculate the factored design loads (without self-weight). wu = 1.2 wD wL = 1.2 x x 11 = 29.6 kN/m Pu = 1.2 PD PL = 1.2 x x 40 = 64 kN Mu = wU L2 / 8 + PU L / 4 = = kN.m 40 kN 10 kN/m 11 kN/m 4.5 m hjhkjkjolklktytvtvguu 9 m 68402

Ex. 4.3 – Beam Design Distributed load = w = 11 kN/m
Step II. Select W21 x 44 Zx = 1563x103 mm3 bMp = 0.9x1563x103x344/ = kN.m Self-weight = wsw = 0.64 kN/m. Step III. Add self-weight of designed section and check design wD = = kN/m wu = 1.2 x x 11 = 30.4 kN/m Therefore, Mu = 30.4 x 92/ = < bMp of W21 x 44. OK! Step IV. Check deflection at service live loads. Service loads Distributed load = w = 11 kN/m Concentrated load = P = L = 40 kN 68402

Ex. 4.3 – Beam Design Deflection due to uniform distributed load = d = 5 w L4 / (384 EI) Deflection due to concentrated load = c = P L3 / (48 EI) Therefore, service-load deflection =  = d + c  = 5x11x94x109/(384x351x106x200) +40x93x109/(48x351x106x200)  = = 22.1 mm L = 9 m. Assuming unplastered floor construction, max = L/240 = 9000/240 = 37.5 mm Therefore,  < max OK! 68402

Local Buckling of Beam Section – Compact and Non-compact
Mp, the plastic moment capacity for the steel shape, is calculated by assuming a plastic stress distribution (+ or - y) over the cross-section. The development of a plastic stress distribution over the cross-section can be hindered by two different length effects: Local buckling of the individual plates (flanges and webs) of the cross-section before they develop the compressive yield stress y. Lateral-torsional buckling of the unsupported length of the beam / member before the cross-section develops the plastic moment Mp. 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 cross-sections. 68402

Figure 7. Local buckling of flange due to compressive stress (s) 68402

Steel sections are classified as compact, non-compact, or slender depending upon the slenderness (l) ratio of the individual plates of the cross-section. Compact section if all elements of cross-section have   p Non-compact sections if any one element of the cross-section has p    r Slender section if any element of the cross-section has r   It is important to note that: If   p, then the individual plate element can develop and sustain y for large values of e before local buckling occurs. If p    r, then the individual plate element can develop y at some locations but not in the entire cross section before local buckling occurs. If r  , then elastic local buckling of the individual plate element occurs. 68402

Classification of Sections
Classifications of bending elements are based on limits of local buckling The dimensional ratio l represents Two limits exist p and r p represents the upper limit for compact sections r represents the upper limit for non-compact sections 68402

Thus, slender sections cannot develop Mp due to elastic local buckling. Non-compact sections can develop My but not Mp before local buckling occurs. Only compact sections can develop the plastic moment Mp. Figure 8. Stress-strain response of plates subjected to axial compression and local buckling. 68402

All rolled wide-flange shapes are compact with the following exceptions, which are non-compact. W21x48, W40x174, W14x99, W14x90, W12x65, W10x12, W8x10, W6x15 (made from A992) 68402

Classification of Sections
The limits are Flange Web 68402

Lateral-Torsional Buckling (LTB)
The laterally unsupported length of a beam-member can undergo LTB due to the applied flexural loading (BM). Figure 9. Lateral-torsional buckling of a wide-flange beam subjected to constant moment. 68402

Lateral-Torsional Buckling (LTB)
LTB is fundamentally similar to the flexural buckling or flexural-torsional buckling of a column subjected to axial loading. The similarity is that it is also a bifurcation-buckling type phenomenon. The differences are that lateral-torsional 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 LTB causing bending moment M(x) varies along the unbraced length. The worst situation is for beams subjected to uniform BM along the unbraced length. Why? 68402

Lateral-Torsional Buckling (LTB) –Uniform BM
Consider a beam that is simply-supported at the ends and subjected to four-point loading as shown below. The beam center-span is subjected to uniform BM (M). Assume that lateral supports are provided at the load points. Laterally unsupported length = Lb. If the laterally unbraced length Lb is less than or equal to a plastic length LP then lateral torsional buckling is not a problem and the beam will develop its plastic strength MP. 68402

Lateral-Torsional Buckling (LTB) – Uniform BM
Lp = 1.76 ry x for I members & channels If Lb is greater than LP then lateral torsional buckling will occur and the moment capacity of the beam will be reduced below the plastic strength MP as shown in Fig. 10. As shown in Fig. 10, the lateral-torsional buckling moment (Mn = Mcr) is a function of the laterally unbraced length Lb and can be calculated using the equation: 68402

Mn = Mcr = Mn - moment capacity Lb - laterally unsupported length. Mcr - critical lateral-torsional buckling moment. E – 200 GPa; G – 77 GPa Iy - moment of inertia about minor or y-axis (mm4) J - torsional constant (mm4) from the Section Property Tables. Cw - warping constant (mm6) from the Section Property Tables. This Eq. is valid for ELASTIC LTB only (like the Euler equation). This means it will work only as long as the cross-section is elastic and no portion of the cross-section has yielded. 68402

(0.7Fy) Plastic No Instability Inelastic LTB Elastic LTB Fig. 10 Lateral Torsional Buckling (Uniform Bending) 68402

As soon as any portion of the cross-section reaches the yield stress Fy, the elastic LTB equation cannot be used. Lr is the unbraced length that corresponds to a LTB moment Mr = Sx (0.7Fy). Mr will cause yielding of the cross-section due to residual stresses. When the unbraced length is less than Lr, then the elastic LTB Eq. cannot be used. When the unbraced length (Lb) is less than Lr but more than the plastic length Lp, then the LTB Mn is given by the Eq. below: 68402

Lateral-Torsional Buckling – Uniform BM
If Lp  Lb  Lr, then This is linear interpolation between (Lp, Mp) and (Lr, Mr) See Fig. 10 again. For a doubly symmetric I-shape: c = 1 h0 = distance between the flange centroids (mm) 68402

Moment Capacity of Beams Subjected to Non-uniform BM
As mentioned previously, the case with uniform bending moment is worst for lateral torsional buckling. For cases with non-uniform bending moment, the LTB moment is greater than that for the case with uniform moment. The AISC specification says that: The LTB moment for non-uniform bending moment case Cb x lateral torsional buckling moment for uniform moment case. 68402

Cb is always greater than 1.0 for non-uniform bending moment. Cb is equal to 1.0 for uniform bending moment. Sometimes, if you cannot calculate or figure out Cb, then it can be conservatively assumed as 1.0. for doubly and singly symmetric sections Mmax - magnitude of maximum bending moment in Lb MA - magnitude of bending moment at quarter point of Lb MB - magnitude of bending moment at half point of Lb MC - magnitude of bending moment at three-quarter point of Lb Use absolute values of M 68402

Flexural Strength of Compact Sections
Mmax MA MB MC @ quarter @ mid @ three-quarter Moments determined between bracing points 68402

Values of Cb 3-1 68402

Moment Capacity of Beams Subjected to Non-uniform Bending Moments
The moment capacity Mn for the case of non-uniform bending moment Mn = Cb x {Mn for the case of uniform bending moment}  Mp Important to note that the increased moment capacity for the non-uniform moment case cannot possibly be more than Mp. Therefore, if the calculated values is greater than Mp, then you have to reduce it to Mp 68402

Figure 11. Moment capacity versus Lb for non-uniform moment case Cb = 1.0 means uniform BM 68402

Structural Design of Beams
Steps for adequate design of beams: 1) Compute the factored loads, factored moment and shear 2) Determine unsupported length Lb and Cb 3) Select a WF shape and choose Zx assuming it is a compact section with full lateral support 4) Check the section dimension for compactness and determine bMn 5) Use service loads to check deflection requirements 68402

Ex. 4.4 – Beam Design Use Grade 50 steel to design the beam shown below. The unfactored uniformly distributed live load is equal to 40 kN/m. There is no dead load. Lateral support is provided at the end reactions. Select W16 section. wL = 40 kN/m 7.5 m 68402

Ex. 4.4 – Beam Design Step I. Calculate the factored loads assuming a reasonable self-weight. Assume self-weight = wsw = 1.46 kN/m. Dead load = wD = = 1.46 kN/m. Live load = wL = 40 kN/m. Ultimate load = wu = 1.2 wD wL = 65.8 kN/m. Factored ultimate moment = Mu = wu L2/8 = kN-m. Is BM uniform?? Yes Cb =1.0 No Go to Step II Step II. Determine unsupported length Lb and Cb There is only one unsupported span with Lb = 7.5 m Cb = 1.14 for the parabolic bending moment diagram, See values of Cb shown in Table 3-1. 68402

Ex. 4.4 – Beam Design Step III. Select a wide-flange shape
Compute Zx = 462.3*106/(0.9*344) = 1493x106 mm3. Select W16 x 50 steel section Zx = 1508x103 mm3 Sx = 1327x103 mm ry = 40.4 mm Cw = 610x109 mm6 Iy = 15.5x106 mm J = 0.63x106 mm4 68402

Ex. 4.4 – Beam Design h0 = D - TF = 414 – 16 = 398 mm Lb > Lr 68402

Ex. 4.4 – Beam Design Step IV. Check if section is adequate
Mu > Mn Not OK Step V. Try a larger section. After few trials select W16 x 67 Mn = > Mu OK Step VI. Check for local buckling.  = Bf / 2Tf = 7.7; Corresponding p = 0.38 (E/Fy)0.5 = 9.19 Therefore,  <  p - compact flange  = H/Tw = 35.9; Corresponding p = 3.76 (E/Fy)0.5 = 90.5 Therefore,  <  p - compact web Compact section. - OK! This example demonstrates the method for designing beams and accounting for Cb > 1.0) Values for Lr and Lp can be obtained from Tables too 68402

Ex. 4.5 – Beam Design 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. Use Grade 50 steel. 135 KN 135 KN 1.5 KN/m 3.6 m 2.4 m 3.0 m 9.0 m 68402

Ex. 4.5 – Beam Design Step I. Assume a self-weight and determine the factored design loads Let, wsw = 1.5 kN/m PL = 135 kN Pu = 1.6 PL = 216 kN wu = 1.2 x wsw = 1.8 kN/m The reactions and bending moment diagram for the beam are shown below. 68402

Ex. 4.5 – Beam Design 216 KN 216 KN 1.8 KN/m 3.6 m 2.4 m 3.0 m
68402

Ex. 4.5 – Beam Design Step II. Determine Lb, Cb, Mu, and Mu/Cb for all spans. Span Lb (m) Cb Mu (kN-m) Mu/Cb AB 3.6 1.67 754.9 452.8 BC 2.4 1.0 (assume) CD 3.0 715.5 429.2 It is important to note that it is possible to have different Lb and Cb values for different laterally unsupported spans of the same beam. Cb – Table 3-1 68402

Ex. 4.5 – Beam Design Step III. Design the beam and check all laterally unsupported spans Assume that span BC is the controlling span because it has the largest Mu/Cb although the corresponding Lb is the smallest. Required Zx = 754.9*106/(0.9*344) = 2438x103 mm3 After few trials select W21 x 68 from section property Table. Lp = 1.94 m Lr = 5.73 m (From Tables) For all members Lp < Lb < Lr Check the selected section for spans AB, BC, and CD Span Lb (m) Cb fbMn for Cb value fbMp limit AB 3.6 1.67 kN.m 811.8 kN.m BC 2.4 1.0 773.6 CD 3 kN.m 68402

Ex. 4.5 – Beam Design For span BC, fbMn = 773.6 kN.m > Mu - OK!
Thus, for span AB, fbMn = kN.m > Mu - OK! For span BC, fbMn = kN.m > Mu - OK! For span CD, fbMn = kN.m > Mu - OK! Step IV. Check for local buckling Bf / 2Tf = 6.04; Corresponding p = 0.38 (E/Fy)0.5 = 9.19 Therefore compact flange H/Tw = 43.6; Corresponding p = 3.76 (E/Fy)0.5 = 90.55 Therefore compact web Compact section OK! This example demonstrates the method for designing beams with several laterally unsupported spans with different Lb and Cb values. 68402

Ex. 4.6 – Beam Design Design the simply-supported beam shown below. The uniformly distributed dead load is equal to 15 kN/m and the uniformly distributed live load is equal to 30 kN/m. A concentrated live load equal to 40 kN acts at the mid-span. Lateral supports are provided at the end reactions and at the mid-span. Use Grade 50 steel. 40 kN 15 kN/m 30 kN/m 3.6 m 3.6 m 68402

Ex. 4.6 – Beam Design Step I. Assume the self-weight and calculate the factored design loads. Let, wsw = 1.5 kN/m wD = = 16.5 kN/m wL = 30 kN/m wu = 1.2 wD wL = 67.8 kN/m Pu = 1.6 x 40 = 64 kN The reactions and the bending moment diagram for the factored loads are shown below. 64 kN 67.8 kN/m 3.6 m 3.6 m 276.1 kN 276.1 kN M(x) = 276.1(x) (x)2/2 68402

Ex. 4.6 – Beam Design Step II. Calculate Lb and Cb for the laterally unsupported spans. Since this is a symmetric problem, need to consider only span AB Lb =3.6 m, M(x) = x – 67.8 x2/2 Therefore, MA = M(x = 0.9 m) = 221 kN.m - quarter-point along Lb = 3.6 m MB = M(x = 1.8 m) = 387 kN.m - half-point along Lb = 3.6 m MC = M(x = 2.7 m) = 498 kN.m - three-quarter point along Lb= 3.6 m Mmax = M(x = 3.6 m) = kN.m - maximum moment along Lb =3.6 m Therefore, Cb = 1.36 68402

Ex. 4.6 – Beam Design Step III. Design the beam section
Mu = Mmax = kN.m Lb = 3.6 m, Cb = 1.36 Required Zx = 554.6*106/(0.9*344) = 1791x103 mm3 After few trials, select W21 x 57 steel section Lp = 1.46 m Lr = 4.37 m Lp < Lb < Lr bMn = 699 kN.m > bMp = kN.m bMn= >Mu OK 68402

Ex. 4.6 – Beam Design Step V. Check for local buckling. Bf / 2Tf = 7.87; Corresponding p = 0.38 (E/Fy)0.5 = 9.192 Therefore, compact flange l = h/tw = 50.0; Corresponding p = 3.76 (E/Fy)0.5 = 90.55 Therefore, compact web Compact section OK! This example demonstrates the calculation of Cb and the iterative design method. 68402

Shear Capacity The shear capacity of the beam is
For I-shaped sections, the factors Cv and v are functions of the shear buckling of the web and thus the ration h/tw representing the case of no web instability. Aw = dtw 68402

Shear Capacity For all other doubly and singly sym. sections and channels except round HSS The first case represents the case of no web instability. The second case represents inelastic web buckling The last case represents elastic web buckling 68402

Shear Capacity Aw = dtw The web plate buckling coefficient, kv, is given For unstiffened webs with h/tw <260, kv = 5 except for the stem of tee shapes, kv =1.2 For stiffened webs a = clear distance between transverse stiffeners h = for rolled shapes, the clear distance between flanges less the fillet of corner radii. 68402

AISC Specifications: Chapter K
Beam Bearing Plates Design of a beam bearing plate would require checking: 1- Web Yielding and Web crippling to determine N 2- Bearing capacity to determine B 3- Plate moment capacity to determine t AISC Specifications: Chapter K 68402

Beam Bearing Plates When a bearing plate is used at beam end, two limit states shall be considered 1. Web Yielding This represents yield of the web at the vicinity of the flange due to excessive loading CASE 1: At Support CASE 2: Interior Load 68402

Beam Bearing Plates 2. Web Crippling
Web crippling represent the possible buckling of the web CASE 1: At Support CASE 2: Interior Load 68402

Ex. 4.7 – Beam Design Check the beam shown in the figure below for:
Shear capacity. Web yielding. Web crippling. Assume the width of the bearing plate is 100 mm. Use Grade 50 steel. 40 kN 10 kN/m 25 kN/m W16x26 2 m 2 m 68402

Ex. 4.7 – Beam Design Step I. The section used from Example 4.6 is W21x57. The self-weight wsw = 0.83 kN/m wD = = kN/m wL = 25 kN/m wu = 1.2 wD wL = 52.5 kN/m Pu = 1.6 x 40 = 64 kN The reactions and the bending moment diagram for the factored loads are shown below. 64 kN 52.5 kN/m 2 m 2 m 137 kN 137 kN Vu = 137 kN 68402

Ex. 4.7 – Beam Design Step II. h/tw = 56.8
Assume unstiffened web, h/tw <260, kv = 5 Vn = 0.9*(0.6Fy)*d*tw*Cv Vn = 0.9*(0.6x344)*399*6.4x10-3 = kN> Vu 68402

Ex. 4.7 – Beam Design Step III. Web yielding critical is support
k = 19 mm R = (2.5k + N)*Fy*tw R = 1x(2.5x )x344x6.4/1000 = kN R > reaction = 137 kN OK Step IV. Web crippling critical is support d = 399 mm tw = 6.4 mm tf = 8.8 mm N/d = 100/399 = 0.25 > 0.2 68402

Ex. 4.7 – Beam Design R = 179 kN > reaction = 137 kN OK 68402

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