 Ch 4. Beam Stresses Multimedia Engineering Mechanics Bending Strainand Stress BeamDesign Shear Stress Built-upBeams
 Chapter 1. Stress/Strain 2. Torsion 3. Beam Shr/Moment 4. Beam Stresses 5. Beam Deflections 6. Beam-Advanced 7. Stress Analysis 8. Strain Analysis 9. Columns Appendix Basic Math Units Basic Equations Sections Material Properties Structural Shapes Beam Equations Search eBooks Dynamics Fluids Math Mechanics Statics Thermodynamics Author(s): Kurt Gramoll ©Kurt Gramoll MECHANICS - CASE STUDY SOLUTION Free Body Diagram ofTypical Box Beam The roof of a new room addition supports 80 lb/ft2 roof load. The 20 ft long box beams are evenly spaced every 4 ft. Thus the total load on a single beam (not end beams) is      w = (80 lb/ft2) (4 ft) = 320 lb/ft The wall reaction forces are equal due to symmetry.      RA = RB = (320 lb/ft)(20 ft)/2 = 3,200 lb = 3.2 kip Shear and Moment Diagrams Moment and Shear Diagrams A shear and moment diagram of the channel will help identify the maximum shear and moment. Both of these values are needed to determine the maximum shear and bending stress. Only one cut is needed due to the continuous loading.      ΣF = 0      V = 3.2 - 3.2x  kip      ΣM = 0      M + (0.32x)(x/2) - 3.2x = 0      M = 3.2x - 0.16x2 The shear load is maximum at each end of the beam (this is one reason old wood beams crack near the center at each end). To design the screw spacing, it is best to take the worst load, 3,200 lb, even though it changes along the beam. The maximum moment, 16.0 kip-ft, occurs at the beam mid-point. Shear Stress Shear Stress on Horizontal Surfaces (A) Shear Stress on Vertical Surfaces (B) There are two possible location to shear the screws. The first location is the horizontal joints and involves the full top section (label Area A in first diagram). The left side is part of the left 2x6 and it is assumed the wood will not shear. The second location is the vertical joints and involves only part of the top section (label Area B in the second diagram). In both cases, there are two surfaces that will carry the shear load. First determine Q for both cases.      QA = AAyA = (7.0)(1.5)(3.5 - 0.75) = 28.875 in3      QB = AByB = (4.0)(1.5)(3.5 - 0.75) = 16.5 in3 Next, the moment of inertia for the full cross sections needs to be known.      I = 7(7)3/12 - 4(4)3/12 = 178.75 in4 Using the shear stress equation gives, This shear stress needs to be carried by the screws. Each screw can carry 800 lb. The minimum spacing (sA and sB) of the screws can be found using      FScrew = τ A For each surface, this gives the spacing as      800 lb = (172.3 lb/in2) [(1.5 in) sA]      sA = 3.095 in      800 lb = (98.46 lb/in2) [(1.5 in) sB]      sB = 5.417 in Thus, the screws should be every 3 inches on the horizontal joints and every 5 inches on the vertical surfaces. Bending Stress The bending stress should also be checked. Unlike shear stress, the maximum bending stress occurs at the point furthest from the neutral axis. This is at the top or bottom surface. Using the bending stress equation gives,      σ = -My/I         = -(16 kip-ft)(12 in/ft)(-3.5 in) / (178.75 in4)         = 3.759 ksi This is lower than the allowable bending stress for wood so it should be safe.

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