Transcription of Bending Stresses in Beams
1 University of Michigan, TCAUP Structures I Slide 1 of 19 Architecture 314 Structures I Bending Stresses in Beams Elastic Bending Stress Equation Section Modulus Flexure CapacityUniversity of Michigan, TCAUP Structures I Slide 2 of 19 Elastic BendingFlexure results in internal tension and compression forces, the resultants of which form a couple which resists the applied the initial unloaded state, all transverse sections are application of load causes the member to bend in a curve.
2 This means the initial parallel plane sections, while remaining plane, now follow the radii of the that by the geometry of the curved member the top edge is shortened and the bottom edge is lengthened. Only the neutral axis remains its original of Michigan, TCAUP Structures I Slide 3 of 19 Elastic BendingThe change in lengths, top and bottom, results in the material straining. For a simple span with downward loading, the top is compressed and the bottom stretched. The change in length is linear and proportional to the distance from the Neutral material strains result in corresponding Stresses .
3 By Hooke s Law, these Stresses are proportional to the strains which are proportional to the change in length of the radial arcs of the beam fibers . This assumes that the Modulus of Elasticity is constant across the of Michigan, TCAUP Structures I Slide 4 of 19 Elastic BendingThe applied moment at any point on the beam is equal to the resisting moment which is formed by the internal force couple, Rcand of the external and internal momentsBalance of the internal force coupleExpressions of the internal resisting momentUniversity of Michigan, TCAUP Structures I Slide 5 of 19 Elastic BendingThe internal moment, Mr.
4 Can be expressed as the result of the couple Rcand turn, the forces Rcand Rt, can be written as the resultants of the stress volumes acting through the centroids of those volumes. The average unit stress, s = fc/2 and so the resultant R is the area times s:Using similar triangles, s can be expressed as:andSubstituting these values back into the moment equation gives:University of Michigan, TCAUP Structures I Slide 6 of 19 Elastic BendingBy definition:And for homogeneous materials with Ec=EtOr using the Ifor the whole section:And so,So, at extreme fibers:With c = h/2 at extreme fibersof a symmetric :The Section Modulus is.
5 University of Michigan, TCAUP Structures I Slide 7 of 19 beam CapacityAllowable Capacity (ASD):for steel: Fb= ( to ) Fyksifor wood: Fb= 1000 to 600 psiApplied Load:(uniform load)University of Michigan, TCAUP Structures I Slide 8 of 19 beam Capacity Analysis- procedure1. Determine section properties. (from table)2. Choose safe allowable stress. (from ASD code)3. Calculate allowable moment Set equal to applied moment and find of Michigan, TCAUP Structures I Slide 9 of 19 beam Capacity Analysis -exampleGiven: beam = W27x178Sx = 502 in3Fy = 50 ksiFb =.
6 6Fy = 30 ksiFind:Floor capacityUniversity of Michigan, TCAUP Structures I Slide 10 of 19 beam Capacity AnalysisGiven: beam = W27x178Sx = 502 in3Fy = 50 ksiFb = .6Fy = 30 ksiFind:Floor capacityUniversity of Michigan, TCAUP Structures I Slide 11 of 19 Section PropertiesSection Modulus TableSorted by Sx for design selectionwith:S = I/cfbis actual stressFbis allowable stressFyis the yield stressSo the design equations is:S = Mapplied/FbUniversity of Michigan, TCAUP Structures I Slide 12 of 19 beam Design- procedure1.
7 Choose a steel grade and allowable Determine the applied moment ( moment diagram)3. Calculate the section modulus, Sx4. Choose a safe section. (from Sxtable)University of Michigan, TCAUP Structures I Slide 13 of 19 beam Design - steelUsing Steel W section:1. Choose a steel grade: Using Fy= 50 ksi Fb= Fy2. Determine the applied momentUniversity of Michigan, TCAUP Structures I Slide 14 of 19 beam Design steelUsing Steel W section:2. Calculate section modulus, SxUniversity of Michigan, TCAUP Structures I Slide 15 of 19 beam Design steelUsing Steel W section:3.
8 Choose a safe section. (from Sxtable)Sx in3 University of Michigan, TCAUP Structures I Slide 16 of 19 beam Design GlulamUsing glulam Timber:Fb= 1250 psi ( DF grade L3)University of Michigan, TCAUP Structures I Slide 17 of 19 Section PropertiesUsing glulam Timber: glulam Timbers 8 wideSxrequired = in3 University of Michigan, TCAUP Structures I Slide 18 of 19 Section PropertiesSawn LumberUniversity of Michigan, TCAUP Structures I Slide 19 of 19 Modes of FailureStrength Tension rupture Compression crushingStability Column buckling beam lateral torsional bucklingServiceability beam deflection Building story drift cracking
