E 160 Lb 2 Notes: Sher nd oment Digrms for ems Sher nd moment digrms re plots of how the internl bending moment nd sher vry long the length of the bem. Sign onvention for nd onsider the rbitrrily loded simple bem: ut the bem loose from its pin support t the left, roller support t the right nd through the bem t section - Free ody Digrms (F.. D.) of portion of bem to the left of - nd portion of the bem to the right of - (note tht there is no xil force in this bem): Rective forces re developed t ech support nd equl nd opposite internl sher force () nd bending moment () re present to keep ech bem segment in equilibrium. The bove senses for nd re the usul ivil Engineering convention for positive sher nd positive bending moment. It is good prctice to show the sign convention next to sher nd moment digrms to mke the user cler of the convention for positive sher nd moment. 1 ukzich E 160 Lb 2 Notes [L2]
Some common icons for showing sign convention Exmple Problem onstruct the Sher nd ending oment Digrm for the following bem: 10 ft 10 ft Find support rections t nd F..D. of entire bem cut loose from its supports x y 10 ft 10 ft y Need to replce distributed lod s n equivlent point lod (equivlent point lod = re of distribution plced t the centroid of the distribution) 2 ukzich E 160 Lb 2 Notes [L2]
()*(10 ft) = 20 k x y 5 ft 5 ft 10 ft y pply equtions of equilibrium to find unknown rections oment Equilibrium! = 0 + ounterclockwise moments bout point positive 20 k 5 ft + 10 ft +! 20 ft = 0 y = ( y is positive so y cts upwrd s ssumed) Force Equilibrium + F! = 0 Upwrd forces positive! 20k +! = 0 y = 20 k ( y is positive so y cts upwrd s ssumed) Force Equilibrium + F! = 0 forces positive to the right! = 0 x = 0 (no xil force in bem) 3 ukzich E 160 Lb 2 Notes [L2]
F..D of bem with known support rections shown 10 ft 10 ft b b 20 k Find internl sher nd moment from to ke cut t section -, n rbitrry point between points nd F..D of bem to left of cut, show unknown internl forces nd in their positive sense 20 k x Replce distributed lod s n equivlent point lod ()*(x) = 2x 20 k x/2 x/2 4 ukzich E 160 Lb 2 Notes [L2]
oment Equilibrium of segment! = 0 + ounterclockwise moments bout point re positive 20 x + 2x x/2 + = 0 = 20x x 2 (units in k nd ft) Force Equilibrium + F! = 0 Upwrd forces positive 20 2x = 0 = 20 2x (units in k nd ft) Plot nd between points nd t point Evlute t x = 0! = 20 2 0 = 20 k! = 20 0 0! = 0 k ft t point - (just to the left of point ) Evlute t x = 10ft! = 20 2 10 = 0 k! = 20 10 10! = 100 k ft is liner in x between nd is prbolic in x between nd 5 ukzich E 160 Lb 2 Notes [L2]
Find internl sher nd moment from to ke cut t section b-b, n rbitrry point between points nd F..D of bem to right of cut, show unknown internl forces nd in their positive sense b b x oment Equilibrium of segment! = 0 + ounterclockwise moments bout point b re positive 10 x = 0 = 10x! (units in k nd ft) Force Equilibrium + F! = 0 Upwrd forces positive 10 + = 0 = 10 (units in k) Plot nd between points nd t point + (just to the right of point ) Evlute t x = 10ft! =! = 10 10 = 100 k ft 6 ukzich E 160 Lb 2 Notes [L2]
t point Evlute t x = 0! =! = 10 0 = 0 k ft is constnt between nd is liner in x between nd Plot nd digrms under the F..D of the bem 10 ft 10 ft 20 k x x 20 k = 20 2x = 10 - = 20x x 2 100 k-ft = 10x' 0 0 7 ukzich E 160 Lb 2 Notes [L2]
For review, we will use equilibrium to find the vrition of nd cross the bem in this lb. However, we should note the differentil nd integrl reltionships between,, nd w tht re lso useful. Note lso tht these reltionships re dependent on the sign convention chosen. d dx = w d dx = slope of tngent to digrm t point = (distributed lod intensity t tht point) slope of tngent to digrm t point = vlue of t tht point = - (re of distributed lod between points nd ) = re of digrm between points nd 8 ukzich E 160 Lb 2 Notes [L2]