,Click to edit Master title style,Click to edit Master text styles,Second level,Third level,Fourth level,Fifth level,*,Click to edit Master title style,Click to edit Master text styles,Second level,Third level,Fourth level,Fifth level,*,1,BEAMS,SHEAR AND MOMENT,2,Beam Shear,Shear and Moment Diagrams,Vertical shear:,tendency for one part of a beam to move vertically with respect to an adjacent part,3,Beam Shear,Magnitude(V),=sum of vertical forces on either side of the section,can be determined at any section along the length of the beam,Upward forces(reactions)=positive,Downward forces(loads)=negative,Vertical Shear,=reactions loads,(to the left of the section),4,Beam Shear,Why?,necessary to know the maximum value of the shear,necessary to locate where the shear changes from positive to negative,where the shear passes through zero,Use of shear diagrams give a graphical representation of vertical shear throughout the length of a beam,5,Beam Shear,Simple beam,Span=20 feet,2 concentrated loads,Construct shear diagram,6,Beam Shear Example 1,Determine the reactions,Solving equation(3):,Solving equation(2):,Figure,6.7a=,7,Beam Shear Example 1(pg.64),Determine the shear at various points along the beam,8,Beam Shear Example 1,Conclusions,max.vertical shear=5,840 lb.,max.vertical shear occurs at greater reaction and equals the greater reaction(for simple spans),shear changes sign under 8,000 lb.load,where max.bending occurs,9,Beam Shear Example 2,Simple beam,Span=20 feet,1 concentrated load,1 uniformly distr.load,Construct shear diagram,designate maximum shear,locate where shear passes through zero,10,Beam Shear Example 2,Determine the reactions,Solving equation(3):,Solving equation(2):,11,Shear and Moment Diagrams,12,Beam Shear Example 2,Determine the shear at various points along the beam,13,Beam Shear Example 2,Conclusions,max.vertical shear=11,000 lb.,at left reaction,shear passes through zero at some point between the left end and the end of the distributed load,x=,exact location from R,1,at this location,V=0,14,Beam Shear Example 3,Simple beam with overhanging ends,Span=32 feet,3 concentrated loads,1 uniformly distr.load acting over the entire beam,Construct shear diagram,designate maximum shear,locate where shear passes through zero,15,Beam Shear Example 3,16,Determine the reactions,Solving equation(3):,Solving equation(4):,17,Beam Shear Example 3,Determine the shear at various points along the beam,18,Beam Shear Example 3,Conclusions,max.vertical shear=12,800 lb.,disregard+/-notations,shear passes through zero at three points,R,1,R,2,and under the 12,000lb.load,19,Bending Moment,Bending moment:,tendency of a beam to bend due to forces acting on it,Magnitude(M),=sum of moments of forces on either side of the section,can be determined at any section along the length of the beam,Bending Moment,=moments of reactions moments of loads,(to the left of the section),20,Bending Moment,21,Bending Moment Example 1,Simple beam,span=20 feet,2 concentrated loads,shear diagram from earlier,Construct moment diagram,22,Bending Moment Example 1,Compute moments at critical locations,under 8,000 lb.load&1,200 lb.load,23,Bending Moment Example 2,Simple beam,Span=20 feet,1 concentrated load,1 uniformly distr.Load,Shear diagram,Construct moment diagram,24,Bending Moment Example 2,Compute moments at critical locations,When x=11 ft.and under 6,000 lb.load,25,Negative Bending Moment,Previously,simple beams subjected to positive bending moments only,moment diagrams on one side of the base line,concave upward(compression on top),Overhanging ends create negative moments,concave downward(compression on bottom),26,Negative Bending Moment,deflected shape has inflection point,bending moment=0,See example,27,Negative Bending Moment-Example,Simple beam with overhanging end on right side,Span=20,Overhang=6,Uniformly distributed load acting over entire span,Construct the shear and moment diagram,Figure 6.12,28,Negative Bending Moment-Example,Determine the reactions,Solving equation(3):,Solving equation(4):,29,Negative Bending Moment-Example,2)Determine the shear at various points along the beam and draw the shear diagram,30,Negative Bending Moment-Example,3)Determine where the shear is at a maximum and where it crosses zero,max shear occurs at the right reaction=6,540 lb.,31,Negative Bending Moment-Example,4)Determine the moments that the critical shear points found in step 3)and draw the moment diagram,32,Negative Bending Moment-Example,4)Find the location of the inflection point(zero moment)and max.bending moment,since x cannot=0,then we use x=18.2,Max.bending moment=24,843 lb.-ft.,33,Rules of Thumb/Review,shear is dependent on the loads and reactions,when a reaction occurs;the shear“jumps up by the amount of the reaction,when a load occurs;the shear“jumps down by the amount of the load,point loads create straight lines on shear diagrams,uniformly distributed loads create sloping lines of shear diagrams,34,Rules of Thumb/Review,moment is depen