STRESSES IN BEAMS. 



Impact. The stresses due to moving loads arc greater than the stresses due to loads at rest. 



Tin- in. ir. iso in stress of the moving load over tin lo.nl at rest is called impact. For a discussion 

 of impact stresses in railway bridges see page 161, Chapter IV. 



STRESSES IN BEAMS. When a straight lieam or bar is supported near the ends and 

 c.i i rics loads or forces applied transverse to the length of the axis of the beam or bar, the axis 

 of the member assumes a curve. The transverse loads or forces are carried by flexure, which is a 

 comliiii.it ion of the three simple stresses of tension, compression and shear. For example, a simple 

 l>e.mi renting horizontally on supports carries a concentrated load. The fibers on the lower or 

 convex side of the beam will be elongated and are therefore in tension, while the fibers on the 

 upper or concave side are shortened and arc therefore in compression. Shear is taking place 

 between each vertical plane of the beam and the plane adjoining between the load and each 

 support. Since the longitudinal stresses in a simple beam vary from a maximum rump re onion 

 on the concave side to a maximum tension on the convex side, the stresses will pass through 

 zero on some plane, called the neutral plane or axis. Also since the fibers on each side of the 

 neutral axis carry different amounts of stress, they will lengthen or shorten different amounts, 

 and there will therefore be horizontal shearing stresses as well as vertical shearing stresses. 



Neutral Surface and Neutral Axis. Under flexure a beam' is curved, and the fibers on the 

 concave side are in compression while the fibers on the convex side are in tension. The neutral 

 surface is a surface on which the fibers have zero stress, and the neutral axis is the trace of this 

 plane on any longitudinal section of the beam. In a simple horizontal beam carrying vertical 

 loads the neutral axis passes through the center of gravity of the cross section of the beam, for a 

 rectangular beam the neutral axis is at half the height of the beam. Where a beam carries loads 

 that are not at right angles to the neutral axis of the beam, the beam is in equilibrium under 

 flexure and direct stress, and the neutral axis or line of zero stress will not pass through the center 

 of gravity of the cross section of the beam, and may fall entirely outside the beam. A bar carrying 

 simple tension or compression may be considered as a beam in which the neutral axis is at an 

 infinite distance from the center of gravity of the cross section of the beam. 



Reactions. For any structure to be in equilibrium, (i) the sum of the horizontal components 

 of all forces acting on the beam must equal zero, (2) the sum of the vertical components of all 

 forces acting on the beam must equal zero, and (3) the sum of the moments about any point of 

 all forces acting on the beam must be equal to zero. Having the loads given the reactions can 

 be calculated by applying the three conditions of equilibrium. 



Vertical Shear. The vertical shear in a beam is equal to the algebraic sum of the forces 

 (reaction minus the loads) on the left of the section considered. 



. Bending Moment. The bending moment at any section of a beam is equal to the algebraic 

 sum of the moments of the reaction and the loads on the left of the section. 



Relations between Shear and Bending Moment. In a simple beam carrying vertical loads 

 the shear is a maximum at the supports and passes through zero at some intermediate point in 

 the beam. The bending moment is zero at the supports and is a maximum at some intermediate 

 point in the beam. The shear is the algebraic sum of all the forces on the left of a section, while 

 the bending moment may be defined as the algebraic sum of all the shearing stresses on the left 

 of the section. The definite integral of the loads to the left of the section equals the shear at the 

 section, and the definite integral of the shear to the left of the section is equal to the bending 

 , moment at the section. From the above it will be seen that maximum bending moment will 

 come at the point of zero shear. 



Formulas for Flexure. Applying the conditions for static equilibrium to any cross section 

 of a beam we have, (i) Sum of Tensile Stresses = Sum of Compressive Stresses; (2) Resisting 

 Shear = Vertical Shear; (3) Resisting Moment = Bending Moment. 



Resisting Shear. If the shearing stresses are uniformly distributed the shearing stress 

 will be 



/. = VIA. d) 



35 



