ENGR 1990 Engineering athematics pplications of Derivatives E 560, E 570 Eample #1 Consier a long slener beam of length with a concentrate loa acting at istance a from the left en. Due to this loa, the beam eperiences an internal bening moment ( ) an internal shearing force V( ). s presente in earlier notes, the bening moment is zero at both ens of the beam an rises linearly from there to a maimum value at a. The shearing force is the erivative of the bening moment. ( ) V ( ) ( ) Y a V V b B Given: 100 (lbs), 5 (ft), a 3.5 (ft) an ma ab ma in: (a) ( ) for 0 ; (b) V( ) for 0 ; an (c) plot the functions. a Solution: ma ab (3.5)(1.5)100 5 105 (ft-lb). (a) or 0 a, the slope is ( ) 30 (ft-lb) or a form m 105 0 3.5 0 30 (ft-lb/ft)., the slope is m 0 105 5 3.5 70 (ft-lb/ft). Using the point-slope 105 70 3.5 ( ) 350 70 (ft-lb) Kamman ENGR 1990 Engineering athematics page: 1/6
(ft-lb) or V (lb) (b) or 0 a (c) ( ) ( ) 30 30 (lb), V, V or a 110 90 70 50 30 10 ( ) ( ) 350 70 350 70 70 (lb) -10-30 -50-70 oment (ft-lb) V() (lb) (0<<a) V() (lb) (a<<) 0 0.5 1 1.5.5 3 3.5 4 4.5 5 (ft) Question: What is the value of () at 3.5 (ft)? Eample : Given: 10 (ft), w 100 (lb/ft), an 0 ( ) 500 50 (ft-lb) in: (a) the shearing force V( ); (b) the maimum bening moment an its location; an (c) plot ( ) an V( ). Solution: (a) or 0 V ( ) ( ) 500 50 500 50 500 100 (lb) Y V B Kamman ENGR 1990 Engineering athematics page: /6
oment (ft-lb) or Shear (lb) (b) Because the shearing force is continuous, the bening moment is a maimum (or minimum) either at an en of the beam or where the shear zero. (c) V ( ) ( ) 500 100 0 500 100 5 (ft) an (0) ( ) 0 ( 5) 5005 505 150 (ft-lb) To verify that it is a maimum, check the sign of () : ( ) 500 100 100 0 (it is a maimum) 1300 1100 900 700 500 300 100-100 -300-500 oment (ft-lb) Shear (lb) 0 1 3 4 5 6 7 8 9 10 (ft) ma Eample 3: Y Consier a cantilevere beam with a uniformly istribute loa of. If the beam is cut at a istance from the wall, we epose the internal shearing force V an bening moment. Given: 10 (ft), w 100 (lb/ft), an w w w 1 1 ( ) (ft-lb) w V w Kamman ENGR 1990 Engineering athematics page: 3/6
oment (ft-lb) or Shear (lb) in: (a) the shearing force V( ); (b) the maimum bening moment an its location; an (c) plot ( ) an V( ). Solution: Using the values for an w, ( ) 50 1000 5000 (ft-lb) V ( ) ( ) 50 1000 5000 1000 100 (lb) (a) (b) gain, the shearing force is continuous, so the bening moment is a maimum (or minimum) either at an en of the beam or where the shear zero. V ( ) ( ) 1000 100 0 1000 100 10 (ft) (at the en) (0) 5000 (ft-lb) (10) 0 (ft-lb) 5000 (ft-lb) ma In this case, the maimum occurs at the en of the beam, an not where ( ) 0, because our concern is with the absolute value of the bening moment. We must esign the beam to withstan 5000 (ft-lb) of bening moment, not zero. (c) 1000 0-1000 -000 oment (ft-lb) Shear (lb) -3000-4000 -5000 0 1 3 4 5 6 7 8 9 10 (ft) Kamman ENGR 1990 Engineering athematics page: 4/6
Eample 4: Consier a bar with rectangular cross-sectional area an applie force as shown. Because is perpenicular (or normal) to the irection of, the material eperiences normal stress only an is efine as rea, rea, Now consier a plane at an angle to the vertical. Since this plane is not normal to, the material along this plane eperiences both normal stress an shear stress. The normal stress is efine (as before) by the area an the normal force. The shear stress is efine by the area an the tangential force. cos( ) cos( ) cos( ) cos ( ) n cos( ) sin( ) cos( ) sin( )cos( ) t In a simple tension test, such as that escribe above, brittle materials ten to fail ue to ecessive normal stress, an uctile material ten to fail ue to ecessive shear stress. By thinking of the normal an shear stresses as functions of the angle, we can fin which planes eperience the highest normal an shear stresses. We can fin maima an minima by setting 0 an 0 an solving for the angle. Using the prouct an chain rules gives cos ( ) cos( ) sin( ) sin( )cos( ) sin( )cos( ) sin ( ) cos ( ) sin( )cos( ) cos ( ) sin ( ) cos( ) rea, cos( ) t n n t Kamman ENGR 1990 Engineering athematics page: 5/6
cos( ) sin( ) sin( ) Setting the erivatives to zero an consiering the range of is 0, we get the following results. Stress ngle, 1 st Derivative n Derivative Type 0 0 negative maimum o 4 (ra) 45 0 negative maimum So, brittle materials will be more likely to fail on a plane normal to the loa, an uctile materials will be more likely to fail on a 45 o plane. Kamman ENGR 1990 Engineering athematics page: 6/6