 Ch 1. Stress and Strain Multimedia Engineering Mechanics NormalStress Shear andBearing Stress NormalStrain Hooke'sLaw ThermalEffects IndeterminateStructures
 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 The objective for this case is to determine the shearing and bearing stress at joint A or B due to a 50 lb/ft2 snow load on the roof of a small shed. Truss Loading Wall Reaction Forces Snow Loading acting Through Top Purlins Free Body Diagram of Roof Truss with Snow Load Transfers to Truss through Purlins at A, B and C. Since the structure and loading is symmetrical, joints A and B will be the same. This means, the reaction force from the wall at both joints will be equal, or      FA = FB = (50 lb/ft2) (2 ft) (6 ft) / 2 = 300 lb Note that the truss spacing is 2 ft, so each truss must support a 2 ft wide section of roof loading. Also, only the vertical load is considered (50 lb/ft2) over the span of 6 ft. Next, the snow loading needs to be analyzed and replaced by equivalent force loads at the joints. The snow load on the plywood is supported by two purlins on each side of the roof. These purlins (one at the top and the bottom of each side) hold the trusses in place and transfer the roof load to the truss joints. The top purlins on each side of the roof will transfer only half of the roof load. The two bottom purlins will transfer the other half of the snow load. The load at the top center is      FC = (50 lb/ft2)(2 ft)(1.5 ft + 1.5 ft) = 300 lb Half the load on each side of the roof is taken by the center purlins, so the center 3 ft width section of the roof is supported by the center joint. The remaining snow load on the roof is transferred to joints A and B. The roof angle is      θ = tan-1 (1/3) = 18.43o Member Loads Free Body Diagram of Joint C Before the bearing and shear stress at joint B can be determined, the force in member CB must be determined. This can be done by analyzing the top joint, C using static equilibrium equations,      ΣFx = 0      FCA = FCB      ΣFy = 0      2 FCB sin18.43 - 300 = 0      FCB = 474.3 lb Shear Stress in Dowel from Member CB Shear Stress in Dowel The shear stress in the dowel pin at joint B is determined from the shear stress equation,      τ = V/A         = 474 / ((1/4)2π) = 474 lb/0.1963 in2         = 2,414 PSI The load in member FAB ( = 474 cos18.4 = 450 lb) will also cause a shear stress in the dowel, but since the load is less than the load in member FCB, the shear stress will also be lower. Bearing Stress in Member AB Bearing Stress in Member CB at Joint B The load in member AB is transferred into joint B through the dowel. This causes a bearing stress on the surface of the dowel against the member. The bearing stress is found by      σb = P/Ab         = 474 / ((3/4)(1/2)) = 474 lb/0.375 in2         = 1,265 PSI

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