6.2 GRAPHICAL METHOD FOR CONSTRUCTING SHEAR AND MOMENT EXAMPLE 6.8 6.2 269 GRAPHICAL METHOD FOR CONSTRUCTING SHEAR AND MOMENT DIAGRAMS Draw the shear and moment diagrams for the cantilever beam in Fig. 6–15a. 2 kN ! EXAMPLE 6.8 Ölçü ve yükleme durumu şekilde verilen konsol kirişin kesme kuvveti ve moment diyagramlarını çiziniz. MB " 11 kN#m Draw the shear and moment diagrams for the cantilever beam 2 kN 2 kN in Fig. 6–15a. 1.5 kN/m 1.5 kN/m 2m OD FOR 269 CONSTRUCTING SHEAR AND MOMENT2 D IAGRAMS kN Çözüm: 2 kN B By " 5 kN 2m w"0 slope " 0 (b) 2V6(kN) 9 w " negative constant slope " negative constant B CONSTRUCTING SHEAR 6.2 (a) AND MOMENT DIAGRAMS 11RAPHICAL kN#m METHOD FOR MB "G 1.5 kN/m 2m w"0 slope " 0 2m 2 EXAMPLE 6.8 B 2m 2m 2m A ntilever beam (b) 1.5 kN/m A V (kN) (a) !2 SOLUTION MB " 11 kN#m Draw the shear and moment diagrams for the cantilever beam 2 kN 2 4 By " 5 kN x (m) Support Reactions. The support reactions at the1.5 fixed kN/msupport in Fig. 6–15a. 2m 2m (c) B are shown in Fig. 6–15b. !2 SOLUTION (b) Shear Diagram. The shear at end is M !2 kN. D This value is G RAPHICAL METHOD FOR CONSTRUCTING SHEARAAND OMENT IAGRAMS 269 Support Reactions. The 6.2 support reactions at the fixed support w "2 0kN w " negative constant plotted at x " 0, Fig. 6–15c. Notice how the shear diagram is (c) 1.5 kN/m B are shown in0Fig. 6–15b. slope " By " !5 5VkN slope " negative constant " negative constant constructed by following the slopes defined by the 2loading w. 2m m slope " negative constant Shear Diagram. The shear at end A is !2 kN. This value is The shear at x = 4 m is !5 kN, the reaction on the beam. This EXAMPLE 6.8 6 V (kN) plotted at A x " 0, Fig. 6–15c. Notice theverified shear diagram is the area under (b) the distributed value how can be by finding M (k N#m ) V " negative constant constructed following slopes defined by the loading MB " 11 kN#m B Draw the shearbyand momentthe diagrams for the beam w. 2slope kN " negative loading, Eq.cantilever 6–3. V " negative increasing w " 0 constant w " negative constant 2 4 shear at x = 4 m is !5 kN, the reaction on the beam. This 1.5 kN/m in The Fig. 6–15a. x (m) slope " negative increasing 2 slope " 0 slope " 2m 2m negative constant 0 value can be verified by finding the area under the distributed M (kN#m) V ƒ x = 4 m = V ƒ x = 2 m + ¢V = -2 VkN !2 loading, Eq. 6–3. (kN)- (1.5 kN>m)(2 m) = -5 kN he fixed support 2 kN (c) (a) 1.5 kN/m 2 0 4 x (m) Moment Diagram. of zero at x2 = 0 is plotted4 in By " 5 kN V ƒ x = 4 m = V ƒ x = 2 m + ¢V = - 2!5 kN - (1.5 kN>m)(2 m) The = - 5moment kN 2m 2m x (m) Fig. 6–15d. Notice how the moment diagram is constructed based !4 N. This value is 6 its slope, which is equal to the shear on knowing (b) at each point. A SOLUTION 2 kN Moment Diagram. The moment of zero at x = 0 is plotted !2 in hear diagram is The change of moment from to m is determined !11 x = 0 x = 2 B VFig. " negative constant 6–15d. Notice The how support the moment diagram is constructed based w " 0 (d) w " negative constant the loading w. Support Reactions. reactions at the fixed support slope " negative constant V " negative increasing from themarea under the shear diagram. Hence, moment slope(c) " 0 the 2m slope " negativeat constant knowing its6–15b. slope, which is equal to2the shear at each point. n the beam. This B areon shown in Fig. slope " negative increasing !5 2 kN m is x = 2 change of moment from x = 0 to x = 2 m is determined the distributed MThe (kN#m ) Shear Diagram. shear end A is !2Hence, kN. This is at V (kN) (a) from the area The under the at shear diagram. thevalue moment 6 V " 2 kN plotted 6–15c. Notice how is = 0 + [-2 kN(2 m)] = -4 kN # m 2m M ƒ xthe = M ƒ xdiagram is 0, Fig. x =at2 xm " = 2 m shear = 0 + ¢M 2 4 M " 4 kN#m 0 x (m) constructed by following the slopes defined by the loading w. V " negative constant 2 4 (e) (2 m) = - 5 kNThe shear at x = 4 m is !5 kN, the reaction on the beam. This# slope " negative constant V " negative increasing x (m) slope "of negative increasing value determined from the2 mmethod sections, M ƒ x = 2 m = M ƒ x = 0 + ¢M = 0 This + [ -same 2 kN(2 m)] can = -be 4 kN m !4 by finding the area under the distributed value can be verified M !2 (kN#m) Fig. 6–15e. SOLUTION (e) loading, Eq. 6–3. = 0 is plotted in This same value can The be determined from the sections, Support Reactions. support reactions at method the fixedofsupport 2 (c) 4 !11 onstructed based B are (d) 6–15b. Fig.shown 6–15e. in Fig. 0 x (m) Fig. 6–15 !5 ar at each point.V ƒ x = 4 m = V ƒ x = 2 m + ¢V = - 2 kN - (1.5 kN>m)(2 m) = - 5 kN 2 kN m is determined Shear Diagram. The shear at end A is !2 kN. This value is 6 !4 Fig. 6–15c. Notice how the shear diagram is the moment at plotted at x " 0,V " 2 kN V " negative constant Moment Diagram. The moment of zero at x =by0 the is plotted constructed by following the slopes defined loadingin w. slope " negative constant V " negative increasing !11 M "!5 4 kN#m Fig. 6–15d. Notice the is constructed based (d) The shear at x how ismoment kN, diagram the reaction on the beam. This = 4m slope " negative increasing on knowing its slope, which is equal to the shear at each point. value can be verified by finding the area under the distributed 2M kN(kN#m) 2m = -4 kN # m The change of6–3. moment from x = 0 to x = 2 m is determined loading, Eq. from the area (e) under the shear diagram. Hence, the moment at 4 V "22 kN 0 x (m) thod of sections, x = 2 m is V ƒ x = 4 m = V ƒ x =Fig. + ¢V = - 2 kN - (1.5 kN>m)(2 m) = - 5 kN M " 4 kN#m 2 m 6–15 M ƒ x = 2 m = M ƒ x = 0 + ¢M = 0 + [ - 2 kN(2 m)] = - 4 kN # m Moment Diagram. The moment of zero at x = 0 is plotted in Fig. 6–15d. Notice how the moment diagram is constructed based This same value can be determined from the method of sections, on knowing its slope, which is equal to the shear at each point. Fig. 6–15e. The change of moment from x = 0 to x = 2 m is determined from the area under the shear diagram. Hence, the moment at x = 2 m is M ƒ x = 2 m = M ƒ x = 0 + ¢M = 0 + [ - 2 kN(2 m)] = - 4 kN # m PAGE "73 This same value can be determined from the method of sections, Fig. 6–15e. 2m (e) !4 !11 (d) 2 kN Fig. 6–15 V " 2 kN M " 4 kN#m 2m (e) Fig. 6–15 !4 (d) V" Fig. 270 DING ! CHAPTER 6 EXAMPLE 6.9 BENDING Ölçü ve yükleme durumu şekilde verilen çıkmalı kirişin kesme kuvveti ve Draw the shear and moment diagrams for the overhang beam in moment çiziniz.for the Draw the shear diyagramlarını and moment diagrams Fig.overhang 6–16a. beam in Fig. 6–16a. 4 kN/m 4 kN/m A 270 A 270 CHAPTER 6 CHAPTER 6 " 10 kN B B BENDING 2m (a) (a) slope " 0 w " negative constant slope " negative constant gative constant NDING egative constant 2m 4m 2m 4m A BENDING 4m EXAMPLE (b)6.9 Ay " 2 kN B " 10 kN EXAMPLE w " 0 6.9 y 2m 4 kN/m 4 kN/m Draw the shear and moment diagrams for the overhang beam in Fig. 6–16a. Draw the shear and moment diagrams for the overhang beam in SOLUTION Fig. 6–16a. Support Reactions. The support reactions are shown in V (kN) SOLUTION 4 kN/m 8 Support Reactions. The support4 kN/m reactions are Fig. shown 6–16b. in 4 kN/m 4 kN/m Fig. 6–16b. Shear Diagram. The shear of !2 kN at end A of the beam A Shear Diagram. The shear of !2 kN at end Aisofplotted the beam at x " 0, Fig. 6–16c. The slopes are determined A B A x (m) Draw the moment diagrams for the overhang beamloading in 0shear and is plotted at x " 0, Fig. 6–16c. The slopes are determined from the and from this the shear diagram is 4 6 A B x (m) Fig. 6–16a. from the loading4 m and !2 from this2 m the shearconstructed, diagram isas indicated in the figure. In particular, notice 6 2 m 4m 6 the positive jump of 10 kN at m due to the force By, as x = 4 constructed, as indicated in the figure. In particular, notice 2m 4 kN/m 4m 2m 4m (b) Ay "jump 2 kN of(c) indicated 4 kN/m the positive 10 kN at x =By4"m10due force Byin , asthe figure. kN to the (a) w " 0 (b) indicatedAin the figure. 2 kN y" By " 10 kN (a) slope "0" 0 negative V decreasing w" V " negative constant w " negative constant Moment Diagram. The moment of zero at x " 0 is A slope " negative decreasing plotted, Fig. 6–16d. Then following the behavior of the slope slope " 0 slope " negative constant SOLUTION Moment Diagram. The moment of zero at x " 0 is slope negative constant negative decreasing B w ""negative constant V (kN) " negative decreasing plotted, Fig. found from the shear diagram, the moment diagram is 6–16d. Then following the behavior of the slope SOLUTION slope " negative constant Support Reactions. The support reactions are shown in M (kN#mV) (kN) constructed. The moment at x " 4 m is found from the area found from the shear diagram, the moment diagram is Fig. 6–16b. Reactions. The support reactions are shown in Support 4 m 8 slope " 0 2 m 2m under thearea shear diagram. constructed. The moment at x " the 8 4 m is found from Fig. 6–16b. 4 slope " 0 Shear Diagram. The shear of !2 kN at end A# of the beam under the shear diagram. 0 By " 10 kN (a) x (m) M ƒ x = 4 is = M ƒ x = 0at+ x¢M = 0Fig. + [6–16c. - 2 kN(4 m)]slopes = - 8 kN m plotted " 0, areA m determined 6 Shear Diagram. The shear of The !2 kN at end of the beam x (m) # x (m) M = kN(4 m)] = 8 kN m 0 M ƒ x = 0 + ¢M = 0 + [-2 ƒ x = 4 m from the loading and from this the shear diagram is We canisalso obtain by using method sections, plotted atthis x "value 0, Fig. 6–16c.the The slopesofare determined 4 6 6 constant negative x (m) SOLUTION 0 " negative constant constructed, as indicated in the figure. In particular, notice !24 as of shown in Fig. from the6–16e. loading and from this the shear diagram is We the6 method sections, !8 by using 6 can also obtain this value the positive of 10 kN in at x the force B 4 m due constructed, as indicated the= figure. Intoparticular, notice The support reactions are shown in jump !2 y, as as6Support shown in Reactions. Fig. 6–16e. (d) indicated in the figure. the positive jump of 10 kN at x = 4 m due to the force By, as Fig. 6–16b. (c) indicated in the figure. V " 2 kN (c) Moment Shear Diagram. The shearVVof !2 kN decreasing at end A of the beam Diagram. The moment of zero at x " 0 is " negative V" negative constant " 2 kN M " 8 the kN#m slope "The negative decreasing plotted, Fig.Diagram. 6–16d. ThenThe following behavior of xthe"slope A Moment moment of zero at 0 is slope " negative constant is plotted at x " 0, Fig. 6–16c. slopes are determined V " negative decreasing V " negative constant x (m) slope " negative decreasing found from the shear diagram, the moment diagram is M " 8 k N#m plotted, Fig. 6–16d. Then following the behavior of the slope from the loading and from this the shear diagram is slope " negative constant 4 6 A 4m M (kN#m) constructed. at x " 4 m from the areais found themoment shear diagram, theis found moment diagram constructed, as indicated in the figure. In particular, noticefromThe 2 M (kN#m) 4 m under constructed. Thediagram. moment at x " 4 m is found from the area the positive jump of 10 kN at x = 4 mslope due "to0 the force By,the as shear slope "0 4 2 kN the shear diagram. under indicated in the figure. x (m) 0 M ƒ x = 4 m = M ƒ x = 0 + ¢M = 0 + [ - 2 kN(4 m)] = - 8 kN # m 4 6 (e) 20kN x (m) Moment Diagram. The moment of zero at x M "ƒ x =04 mis = M ƒ x = 0 + ¢M = 0 + [ - 2 kN(4 m)] = - 8 kN # m " negative decreasing 6 We can also obtain this value by using the method of sections, (e) " negative decreasing plotted, Fig. 6–16d. Then Fig. 6–16 following the behavior of the slope as shown in Fig. 6–16e. We can also obtain this value by using the method of sections, !8 found from the shear the moment diagram is Fig. diagram, 6–16 as shown in Fig. 6–16e. (d) at x!8 constructed. The moment " 4 m is found from the area slope " 0 under the shear diagram.(d) V " 2 kN 8 6 x (m) M ƒ x = 4 m = M ƒ x = 0 + ¢M = 0 + [ - 2 kN(4 m)] = - 8 kN # m A We can also obtain this value by using the method of sections,A as shown in Fig. 6–16e. 2 kN 2 kN V " 2 kN M " 8 kN#m A 4m 2 kN (e) Fig. 6–16 PAGE "74 V " 2 kN M " 8 kN#m M " 8 kN#m 4m 4m (e) (e) Fig. 6–16 Fig. 6–16 6.2 EXAMPLE 6.10 ! EXAMPLE 6.10 Ölçü ve yükleme durumu şekilde verilen şaft için kesme kuvveti ve The shaft in çiziniz. Fig. 6–17a supported byyatay a thrust bearing at A and a moment diyagramlarını A ismesnedinde hareket The shaft injournal Fig. 6–17a is supported by athe thrust at A and aengellenmiş, B mesnedi ise bearing at B. Draw shearbearing and moment diagrams. kayıcı mesnettir. journal bearing at B. Draw the shear and moment diagrams. 120 lb/ft A A D FOR CAL CONSTRUCTING SHEAR AND MOMENT DIAGRAMS 120 lb/ft A B 271 METHOD FOR CONSTRUCTING SHEAR AND MOMENT DIAGRAMS 12 ft Çözüm: 2 6.2 GRAPHICAL METHOD FOR CONSTRUCTING SHEAR AND MOMENT DIAGRAMS GRAPHICAL METHOD FOR CONSTRUCTING SHEAR AND MOMENT DIAGRAMS 271 B 12 ft 271 (a) (a) 120 lb/ft A B 120 lb/ft B 12 ft 12 ft (b) Ay = 240 lb (b)w # negative increasing By # 480 lb Ay = 240 lb # negative increasing w # negativeslope increasing By # 480 lb V (lb) slope # negative increasing V (lb) 240 SOLUTION 12 6.93 240 aring at A and a 120 lb/ft x (f SOLUTIONSupport Reactions. 0 12 The support reactions are shown in 6.93 diagrams. thrust bearing at A and a x IAGRAMS (ft) 120 lb/ft 6.2 G RAPHICAL M ETHOD FOR C ONSTRUCTING S HEAR AND M OMENT D 0 Support Reactions. Fig. 6–17b. The support reactions are shown in (c) moment diagrams. /ft Fig. 6–17b. Shear Diagram. (c) As shown in Fig. 6–17c, the shear at is x = 0 6.2 B GRAPHICAL METHOD FOR CONSTRUCTING SHEAR AND MOMENT IAGRAMS 271 V #D positive decreasing A 120 lb/ft " 480 Shear Diagram. As shown the in Fig. 6–17c, the shear at xloading, = 0 is the Following slope defined by the shear slope # positive decreasing V # positive decreasing A !240. EXAMPLE 6.10 B V # negative " 480 increasing !240. Following the slope defined by the loading, the shear slope # positive decreasing diagram is constructed, where at B its value is "480 lb. Since the 12 ft # negative increasing B V # negativeslope increasing diagram is constructed, where at Bpoint its value is "480 shear changes sign, the6–17a bethe located. doand M V = by 0lb.must EXAMPLE 6.10 (b)The 12 ft shaft in Fig. iswhere supported aSince thrust bearing To at A a (lb$ft) 120 lb slope # negative increasing V#0 B Ashear lb y = 240 changes thissign, we will use the method free-body of the point where must beGThe located. To do diagram VDraw =of0 sections. M (lb$ft) 6.2shear RAPHICAL METHOD FOR CONSTRUCTING SHEARslope AND M DIAGRAMS # OMENT 0 (b) journal bearing at lb B. the and moment diagrams. w # negative increasing B # 480 y V#0 AyFig. = the 2406–17a lb left of the shaft, sectioned at at andiagram arbitrary this we will use the segment method sections. The free-body The shaft in is supported by a thrust bearing A and a ofposition x, is slope 120 lb/ft slope # negative #0 w # increasing negative increasing By # 480 lb V (lb) 120 lb/ft 1109 6 shown in Fig. 6–17e. Notice that the intensity of the distributed the left segment of the shaft, sectioned at an arbitrary position x, is journal bearing at B. # Draw the shear and moment diagrams. slope negative increasing A 6.2 GRAPHICAL METHOD FOR CONSTRUCTING SHEAR AND MOMENT DIAGRAMS 271 (lb) load at x is w # 10x, which has been found by proportional 1109 shown inV Fig. 6–17e. Notice that the intensity of the distributed EXAMPLE 6.10 120 lb/ft by proportional i.e.,which 120!12has # w!x. load at x istriangles, w # 10x, been found 240 A B 12 ft x( 0 A 12 6–17a Thus, V #in0,Fig. B 6.93 Thefor shaft is supported by a thrust bearing at A and a triangles, i.e., 120!12 # w!x. 240 6.93 12 120 lb x (ft) (b) EXAMPLE 6.10 0 are shown in x (ft) 0 12 6.93 at B. Draw journal bearing the A = 240 lb Thus, for V # 0, 1 shear and moment diagrams. 12 ft y 6.93w # negative (ft)2 (10x)x = 0 240 lbx = 0; (d) 12 increasing B # 0 + c ©F 12 ft A y eactions are shown in y B (c) The+shaft Fig. 6–17a is supported thrust bearing at A and a 120 lb/ft 120 lb/ft slope # negative increasing (b)(d) c ©Fin 240 lb by - 12a(10x)x = 0 y = 0; V (lb) x = 6.93 ft 1 A = 240 lb (c) A hear at x = 0 is journal y [10 x ] x V # positive decreasing bearing at B. Draw the shear and moment diagrams. w # negative increasing By # 480 lb2 12 ft x " 480 7c, the shear at xslope = 0#ispositive x = 6.93 ft (a) ding, the shear 1 [10 x ] x V #decreasing positive decreasing slope # negative increasing Moment Diagram. diagram starts at 0 V since 3 10 x 2 (lb) there V # negative increasingThe moment "120 480lb/ft decreasing 12 ft 80the lb. loading, Since thethe shear slope # positive 240 Bx A it is constructed based on Athe is slope no SOLUTION moment atincreasing A; then (a) # negative Moment Diagram. The diagram starts at 0 since there B slope as 3 10 (b) V #moment negative increasing 12 alue is "480 the e located. To lb. doSince 6.93 V x M (lb$ft) determined from the shear diagram. The maximum moment Ay0= 240 lb slope # negative increasing is no at A; then it is constructed based on the slope as V # 0 Support Reactions. The support reactions are shown in A be located. To do moment 0 must ody diagram of V M (lb$ft) w # negative increasing By # 12 ft is equal to 240 slope #0 M occurs at the shear zero, since 12 ft = 6.93diagram. ft, whereThe A Fig. V6–17b. #x0shear determined from the slope #12negative The free-body of 6 Bmaximum moment ary position x, diagram is SOLUTION (c) increasing A(b) 6.93 slope # 0 V (lb) Fig. 6–17d, dM>dx = V = 0, x (ft) occurs where the shear is equal zero, since x = 6.93 ft,The 0 lbat x = 0 is = 240 x M 6to(a) an position x, is atReactions. 1109 thearbitrary distributed Support support reactions shown in Shear Diagram. As shown inare Fig. 6–17c, theAyshear V # positive decreasing w # negative increasing By # 480 lb 12 ft Fig. 6–17d, dM>dx = V = 0, 1109 ensity of the distributed d + ©M = 0; by proportional Aypositive # 240 lbdecreasing Fig. 6–17b. " !240. Following the slope defined by the loading, shear(c) slope x# slopethe # negative increasing 1 1 V (lb) 240 n found by proportional M + [(10)(6.93)] 6.93 (6.93) 240(6.93) = 0 V # negative increa A B diagram is constructed, where at B its value is "480 lb. Since the (e) max SOLUTION 2 3 d+ ©M = 0; A # 240 lb Shear Diagram. As shown in Fig. (a) 6–17c, the shear at x = 0 is y 12 V # positive decreasing slope #6.93 negative incre x (ft) 0 1 changes sign, 1 point where V = 0 must be located. To do shear the M0 (lb$ft) # Mmax +6.93 [(10)(6.93)] 6.93 (6.93) 240(6.93) = 0 slope A B "6–17 480 !240. Following the slope defined by the loading, the shear # positive decreasing (e) Support Reactions. The support reactions are shown in 12 M = 1109 lb ft 2 3 Fig. max x (ft) 0 V#0 240 this we will at use the method of sections. Thethe free-body diagram ofV # negative increasing 6.93 12 diagram is constructed, where B its value is "480 lb. Since slope #(c) 0 SOLUTION # (d)Fig. 6–17b. M = 1109 lb ft 12 6.93 Fig. 6–17 max Finally, notice how integration, first of the loading w which is linear, slope # negative increasing the left segment of the shaft, sectioned atTo an do arbitrary position x, is x (ft) x = 0 shear changes sign, the point where must be located. V = 0 0 (d) M (lb$ft) Shear Diagram. As shown in Fig. 6–17c, the shear at is x = 0 Support Reactions. The support reactions shown and in then a moment V # positive decreasing 1109 produces shear diagram which isare parabolic, ft 1a[10 shown inxof 6–17e. Notice that intensity of the distributed V#0 notice how integration, first of the loading w the which linear, ] Fig. x sections. thisFinally, we will use the method free-body diagram of loading, " !240. theThe slope defined by isthe the slope shear slope # positive decreasing Fig. 6–17b. 2 Following # 0 (c) x = 6.93 ft diagram which is cubic. 1 load at xxsectioned is w[10is #x]parabolic, which has been found by proportional x10x, produces a shear diagram which and then avalue moment 6 the left segment of the shaft, at an arbitrary position x, is V # negative increas 2 diagram is constructed, where at B its is "480 lb. Since the at 0 since there Shear Diagram. As shown 3in shear at x = 0 is V # positive decreasing x # the 10 x 6–17c, triangles, i.e.,Fig. 120!12 w!x. diagram which is shear cubic. slope # negative incre NOTE: Having studied these examples, test= yourself bylocated. coveringTo do 1109 shown in Fig. 6–17e. that the of thethe distributed am the starts at 0 as since there changes sign, the point where be V 0 must 3intensity M0(lb$ft) " 480 on slope !240. Following the Notice slope defined by the loading, shear slope # positive decreasing 10 x V# 0, Thus, for V 6.93 the shear and moment diagrams in Examples 6–1 through 6–4 V#0 12 load at isx constructed, isHaving wover # this 10x, which has been found by ed based on the slope as we will the method of sections. # negative increasing imum moment NOTE: studied these yourself byThe covering diagram atuse Bexamples, its value "480 lb.proportional Since the free-body diagramVof V istest slope # 0 A and see ifwhere 1 concepts discussed here. you can construct them using the c slope # negative increasing triangles, i.e., 120!12 # w!x. 240 lb + ©F = 0; (10x)x = 0 The maximum moment the left segment of the shaft, sectioned at an arbitrary position x, is (d) M= 0 to zero, since shear over the shear diagrams inmust Examples 6–1 2 through changes sign,and theAmoment pointy where be located. To do 6–4 V M (lb$ft) x (ft)1109 0 distributed M V#0 forifuse Vyou #the 0, method is equal to zero, since shown in Fig. 6–17e. Notice that diagram the intensity of the and can construct them using the concepts discussed here. thisThus, we see will of sections. The free-body of 6.93 12 x = 6.93 ft x slope # 0 1 [10 x ] x load at x is w # 10x, which has been found by proportional 6 2 the of the shaft, sectioned at an arbitrary position x, is x - 1 (10x)x 240 lb + cleft ©Fsegment = 0 x (d) y =A0; # 240 lb y i.e., 120!12 # w!x. 1109 shown in Fig. 6–17e.triangles, Notice that the2 intensity of the distributed Moment 3 10 A # 240 lb Diagram. The moment diagram starts at 0 since there 240(6.93) = 0 y(e) 0 forhas V #at 0,A; then x = 6.93 1 [10 x ] x load= at been found proportionalbased on the slope as is Thus, nowhich moment it by is ft constructed 6.93 12 6.93) B - 240(6.93) 0 x is w # 10x, V (e) 2 1 triangles, i.e., 120!12 # w!x. from the shear diagram. The maximum moment x Fig.determined 6–17 c 240 lb - 2(10x)x = 0 + ©Fy = 0; (d) A Moment TheFig. moment diagram starts the at 0 shear since there 3 10x x(ft) 0 to zero, since t Thus, forDiagram. V # 0, occurs at 6–17 is equal x = 6.93 ft, where 6.93 12 w which is linear, is no moment at A; then it is constructed based on the slope as x = 6.93 ft 1 V dM>dx = V = 1 0, Fig. 6–17d, loading w which is linear, c ©Fy = 0; from the x 2 [10 x ] x 240 lb +determined (10x)x = 0 then a moment (d) 2 shear diagram. The maximum moment x bolic, and then a moment Theequal moment diagram starts at 0 sinceAthere dft, + ©M = Diagram. 0;the shear 3 10 Ay # 240 M lb occurs at x = 6.93Moment where x 1 =is 6.93 ft to zero,1 since 1 [10 x ] x is no moment at A; it is constructed based the slope Mmax + then 6.93 A 3(6.93) 240(6.93) = 0as 2 B - on (e) V 2 [(10)(6.93)] self by covering dM>dx = V = 0, Fig. 6–17d, x x determined from the shear diagram. The maximum moment Moment Diagram. The moment diagram starts at 0 since there 3 test yourself by covering # A 10 x 6–1 through 6–4 d+ ©M = 0; Mmax = 1109 lb ft Fig. 6–17 Ay # 240 lb whereonthe since x = 6.93 ft, based is no6–4 moment at A; occurs then it at is constructed the shear slope is as equal to zero, Examples through V s discussed6–1 here. 1 1 Mmax the +dM>dx 6.93 240(6.93) = 0 AThe B - first 6–17d, = diagram. V =how 0, Fig. Finally, integration, ofmoment the loading w which is linear, (e) determined from shearnotice maximum 2 [(10)(6.93)] 3 (6.93) e concepts discussed here. x A produces a shear diagram which is parabolic, and then a moment M occurs at x = 6.93 ft, where the shear is equal to zero, since # Mmax= =0;1109 lb ft d + ©M Ay # 240 lb Fig. 6–17 diagram is cubic. 1 6–17d, which dM>dx = V = 0, Fig. Mmax + 12[(10)(6.93)] 6.93 (6.93) B - 240(6.93) = 0 x A (e) PAGE " 7 5 3 Finally, notice how integration, first of the loading w which is linear, NOTE: Having studied these examples, test yourself by covering d+ ©M = 0; A # 240 lb # ft y Mmax = then 1109 albmoment produces a shear diagram which is parabolic, and Fig. 6–17 1 the shear6.93 andA 13moment 6–1 through 6–4 (e) M + over (6.93) B -diagrams 240(6.93)in=Examples 0 2 [(10)(6.93)] diagram whichmax is cubic. Finally, howconstruct first of the w which is linear, and seenotice if you can them using theloading concepts discussed here. # integration, is applied about an axis perpendicular to this axis of symmetry as shown in Fig. 6–18. The behavior of members that have unsymmetrical 6.3 orBare ENDING DEFORMATION OF A STRAIGHT MEMBER 281 cross sections, made from several different materials, is based on similar observations and will be discussed separately in later sections of this chapter. DeformationBy using a highly deformable material such as rubber, we can illustrate what happens when a straight prismatic member is subjected to a bending 6.10 Prizmatik bir kirişin eğilme deformasyonu ight Member moment. Consider, for example, the undeformed bar in Fig. 6–19a, which has a square cross section and is marked with longitudinal and transverse discuss the deformations that occur when a of grid lines. When a bending moment is Axis applied, it tends to distort these y symmetry made of homogeneous material, is subjected to Fig. 6–19b. lines into the pattern shown in Notice that the longitudinal will be limited to beams havingcurved a cross-sectional lines become and the vertical transverse lines remain straight and with respect to an yet axis,undergo and thea bending rotation. moment M is perpendicular toThe thisbending axis ofmoment symmetry z within the bottom portion of causesasthe material the bar stretch and the material within the top portion to compress. behavior of members thattohave unsymmetrical Neutral x Consequently, between these on two regions there must be a surface, called de from several different materials, is based surface the neutral surface, in which longitudinal fibers of the material will not d will be discussed separately in later sections of Longitudinal undergo a change in length, Fig. 6–18. axis rmable material such as rubber, we can illustrate aight prismatic member is subjected to a bending xample, the undeformed bar in Fig. 6–19a, which n and is marked with longitudinal and transverse ing moment is applied, it tends to distort these own in Fig. 6–19b. Notice that the longitudinal d the vertical transverse lines remain straight and causes the material within the bottom portion of he material within the top portion to compress. hese two regions there must be a surface, called hich longitudinal fibers of the material will not gth, Fig. 6–18. rmation M z Neutral surface x Longitudinal axis Fig. 6–18 6 Fig. 6–18 M 6 Horizontal lines become curved M Vertical lines remain straight, yet rotate Before deformation M After deformation (a) (b) Şekildeki moment kirişin alt tarafında çekme üst tarafında basınç oluşturmaktadır. Bu iki Fig. 6–19 bölge arası geçiş sırasında bir “tarafsız yüzey” bulunacağı aşikardır. x ekseninin konumlandırıldığı boyuna eksende boy değişimi olmadığı görülmektedir. Tüm kiriş kesitlerinin deformasyon sırasında düzlem kaldığı ve boyuna eksene dik olduğu gözlemlenmektedir. 282 CHAPTER 6 BENDING From these observations we will make assumptions regarding the way the stress deform longitudinal axis x, which lies within the neutral Vertical lines remain not experience any change in length. Rather t straight, yet rotate deform the beam so that this line becomes a c After deformation plane of symmetry, Fig. 6–20b. Second, all cro (b) remain plane and perpendicular to the longi deformation. And third, any deformation of th Fig. 6–19 own plane, as noticed in Fig. 6–19b, will be neg z axis, lying in the plane of the cross section an section rotates, is called the neutral axis, Fig. 6–2 In order to show how this distortion will str isolate a small segment of the beam located a dis Note the distortion of the lines due to length and having an undeformed thickness ¢x, from the beam, is shown in profile view bending oföncesi this rubber bar. Thegösterilmiştir. top line taken Şekildeki lastik elemanın deformasyon ve sonrası Deformasyon sonrası Horizontal lines become curved M theorta bottom line compresses, üst çizginin uzadığı alt çizgininstretches, kısaldığı çizginin ise aynı and uzunlukta kaldığı görülmektedir. the center line remains the same length. Furthermore the vertical lines rotate and yet remain straight. 6 z PAGE "76 Note the distortion of the lines due to bending of this rubber bar. The top line stretches, the bottom line compresses, and the center line remains the same length. Furthermore the vertical lines rotate and yet remain straight. y 6 z 6.3 283 BENDING DEFORMATION OF A STRAIGHT MEMBER 6.3 x BENDING DEFORMATION OF A STRAIGHT MEMBER 283 6.3 !xB BENDING 6.3 ENDING D DEFORMATION EFORMATION OF OF A AS STRAIGHT TRAIGHT M MEMBER EMBER O¿ x 6.3 O¿ BENDING DEFORMATION OF A STRAIGHT MEMBER O¿ O¿ y (a) O¿ r 6.3 z longitudinal axis !s " !x !x y !x d element M r r r O¿ !u !u !u !s¿ !x !x longitudinal !x !s¿ axis Undeformed element (b) y Deformed element !x Undeformed element Undeformed element (a) (b)(a) (a) neutral surface longitudinal 6.3 !u EFORMATION STRAIGHT MEM BENDING6.3 D!EFORMATION OF A STRAIGHTOF MAEMBER s¿ BENDING D longitudinal axis longitudinal y axis axis y y (b) !s¿ !x !xO¿ ! s¿ !x O¿ Fig. 6–20 Deformed element !x !s " !x ) Undeformed element y !xFig. 6–21 (a) r neutral axis y longitudinal longitudinal axis!x r ! s "!s¿ !x axis !s " !x ! s " ! x x y !x longitudinal y y longitudinal longitudinal y! x axis !u!x axis axis !x !s " !x z 283 BENDING DEFORMATION OF A STRAIGHT MEMBER !u gitudinal axis !x dinal isolate a small segment of the beam located a distance x along the beam’s length and having an undeformed thickness ¢x, Fig. 6–20a. This element, taken from the beam, is shown in profile view in the undeformed and Fig. 6–21 Fig. 6–21 Fig. 6–21 Fig. 6–21 !x Deformedrelement r Deformed Deformed elementelement (b) (b) (b) !u 6 !u 6 ndeformed element element ions in Fig. 6–21. Notice that any line segmentDeformed ¢x, located longitudinal longitudinal! s¿ ! s¿ !s " !x !s " !x surface, does not change its length, whereas any linein Fig. 6–21. axis deformed positions Notice any lineany segment ¢x, located y located y axis deformed in Fig. 6–21. that Notice that line segment ¢x, (b) positions ormed (a) positions in Fig. 6–21. Noticelongitudinal that any line segment located ¢x, ! x y !x y surface, longitudinal onthe the neutral surface, does not change its whereas line cated at the arbitrary distance y above neutral deformed positions in Fig. 6–21. Notice thatchange anylength, line located ¢x,any on the neutral surface, does not itssegment length, whereas any line he neutral surface, does not change its length, whereas any line 6–21 axis axis segment located the arbitrary distance y above neutral surface, ¢s, ! x arbitrary and become ¢s¿ afterFig. deformation. By definition, the !at xlocated segment atnot the distance y the above the neutral surface, ¢s, on the neutral surface, does change its length, whereas any line ment ¢s, located at the arbitrary distance y above the neutral surface, will contract and become after deformation. By definition, the the ¢s¿ long ¢s is determined from Eq. 2–2,segment namely, willlocated contract and become distance By definition, ¢s¿ afterydeformation. at the arbitrary above the neutral surface, ¢s, contract and become ¢s¿ after deformation. By definition, the normal strain along is determined from Eq. 2–2, namely, ¢salong normal strain is determined from Eq. 2–2, namely, 6 ¢s will contract and ¢s¿ - ¢s ! x after deformation. By definition, the ! xbecome ¢s¿ mal strain along is determined from Eq. 2–2, namely, ¢s P = lim ¢s¿ - ¢s ¢s¿ - ¢snamely, normal strain along from Eq. 2–2, ¢s is determined ¢s : 0 ¢s that ed positions in Fig. 6–21. Notice segment located ¢x, P = limP = lim ¢s¿any - line ¢s Undeformed element Deformed element Undeformed element Deformed element ¢s : 0 ¢s : 0 ¢s ¢s Pin=terms lim its length, ¢s¿ - ¢s surface, does not change whereas any line w neutral represent this strain ¢s : 0 of the ¢s location y of the (a) P = lim Weof will now represent strain in of terms the y(b)of the nt located the arbitrary y above theaxis neutral surface, We will now represent this ¢s strain in the of location y of the (a) (b) location :this 0 ¢sterms he¢s, radius of at curvature the longitudinal the r ofdistance We will and now become represent strain insegment terms of the location yofof the of curvature segment and the radius the longitudinal axis of the r of ntract after deformation. By definition, the ¢s¿this and the radius curvature of the longitudinal axis of the r e deformation, ¢s = ¢x, Fig. 6–21a. After deformation now represent this strain in ofFig. the6–21a. location y deformation of the Fig. 6–21 Fig. 6–21 element. Before After ¢sterms = ¢x, ment the¢sradius of curvature ofWe thewill longitudinal axis deformation, of the relement. strainand along is determined from Eq. 2–2, namely, Before deformation, Fig. 6–21a. After deformation ¢s = ¢x, us of curvature r, with center of curvature at point segment and theaO¿, radius of thecenter longitudinal axis at of point the O¿, r of has of curvature curvature with of curvature ment. Before deformation, ¢s Fig. 6–21a. After deformation ¢x, has a¢x radius ofradius curvature center of curvature at point ¢x r, with r, O¿, ¢s¿ =-the ¢ssides ce ¢u defines the angle between of the element, element. Before deformation, Fig. 6–21a. After deformation ¢s = ¢x, P = lim r, with center Fig. 6–21b. Since defines thebetween angle between theof sides of the element, ¢u the has a radius of curvature of curvature at point O¿, Fig. 6–21b. Since defines angle the sides the element, ¢u ¢s : 0 the¢s deformed length ¢u. In the same manner, a ¢x radius of¢s curvature with curvature at point ¢x r, manner, In the samecenter manner, the deformed length = of ¢sof =the r¢u. defines the angle between the element, ¢u In the same the ofdeformed length of O¿, ¢x has = ¢s =sides r¢u. ¢sof ¢s H6–21b. A P T E R Since 6 B END ING Substituting intointhe above equation, we¢u get 1r now - y2¢u. will represent this strain terms of6–21b. thepositions location y¢s¿ of the deformed positions in Fig. 6–21. Notice that any line segment located ¢x, Fig. Since defines the angle between the sides of the element, becomes Substituting into the above equation, we get = 1r y2¢u. deformed in Fig. 6–21. Notice that any line segment located ¢x, becomes into the above equation, we get ¢s¿ = 1rlength - y2¢u. the deformed ofSubstituting = ¢s = r¢u. In the same manner, ¢s nt and the radius of curvature r ofonthe longitudinal axis of the the surface, on the neutral does not change its length, whereas any line In same manner, the deformed length of ¢x = ¢s = r¢u. ¢s the neutral surface, does not change its length, whereas any line omes ¢s¿ = 1r -1ry2¢u. Substituting into the above equation, we get 1r - y2¢u - r¢u - y2¢u r¢u t. Before deformation, Fig. 6–21a.¢s, After deformation ¢s = -¢x, 1r -lim y2¢u r¢u segment located arbitrary distance yequation, abovesurface, thewe neutral ¢s, segment located arbitrary distance y-above the neutral becomes Substituting into the above get surface, ¢s¿ = 1rat -the y2¢u. Pat =Pthe # P = lim max P = lim ¢u : 0 r¢udeformation. a radius of¢ucurvature with center of curvature at point r,r¢u O¿, :0 become after By definition, the ¢s¿ 1r - will y2¢ucontract - r¢uwill andcontract becomeand By definition, the ¢s¿ ¢u : 0after deformation. r¢u y 1r -# y2¢u - 2–2, r¢u 21b. Since ¢u defines thelim angle between thestrain sides of the element, P = P is P or normal strain along determined from Eq. 2–2, namely, ¢s ! normal along is determined from Eq. namely, ¢s max c 0 r¢u c = lim ordeformed length ofP ¢s manner, the ¢s = r¢u. In the same¢u : r¢u y ¢u : 0 ¢s¿ y - ¢s ¢s= -¢s¿ y es ¢s¿ = 1r - y2¢u. Substituting into the above equation, we get (6–7) P =yPlim P = lim P = (6–7) or - ¢s : 0 r ¢s (6–7) ¢s : 0P = ¢s r r yr¢u 1r - y2¢u P = --We (6–7) y strain This important result indicates that of thein longitudinal normal We will nowthis represent terms of the location of the will now represent strain inthis terms the location y of thestrainy of P = lim " x Pthat r This = (6–7) nt result indicates that the longitudinal normal strain of important result indicates the longitudinal normal ofaxis r¢usegment ! ¢u : 0 any element within the beam depends on its the location the cross andofthe radius ofr curvature longitudinal of the r of andsegment the radius curvature of axisystrain ofonthe r the longitudinal within the beam depends on its location y onelement. theNormal cross anylongitudinal element thebirim beam on its location ydeformation on the cross deformasyon Before deformation, Fig. 6–21a. After deformation ¢s 6–21a. =dağılımı ¢x, his important result indicates that the normal strain ofdepends element. Beforewithin deformation, After ¢s = ¢x, Fig. Normal strain distribution This result indicates that ther,longitudinal strain of point O¿, a on radius of with center normal ofatcurvature ¢x has element within the beam depends its location the cross has a important radius of ycurvature with center of curvature point r,curvature O¿,at y ¢x on any element within the beam depends on its location y on the cross P = - Fig. (6–7) Fig. 6–21b. Since defines the angle between the sides of the ¢u Fig.angle 6–22 between the sides of 6–21b. Since ¢u defines the element, element, r In the same manner, the deformed length of ¢s ¢x = ¢s = r¢u. ¢x = ¢s = r¢u. In the same manner, the deformed length of ¢s PAGE " 7 7 important result indicates that thebecomes longitudinal normal strain of becomes Substituting into the above equation, we get ¢s¿ = 1r y2¢u. ¢s¿ = 1r - y2¢u. Substituting into the above equation, we get ement within the beam depends on its location y on the cross P = lim 1rr¢u - y2¢u - r¢u 1r - y2¢u P = lim ¢u : 0 r¢u 6 - P2 willoccurs occur atinthe fibers locatedfiber, above the neutral axis by1division, Here the then maximum strain outermost located a 1+y2, whereas elongation 1+P2 will6–22. occur in contraction fibers located below the 1 + y2, 1 + P2 whereas elongation will occur in fibers located below the distance of yat !the c from the neutral on and the radius of curvature of the beam’s longitudinal axis axis 1-y2. This variation in strain over the cross section is shown in Fig.axis. Using Eq. 6–7, since Pmax = c>r, 6 1 y2. axis This variation in strain over the cross section is shown in Fig. then 6–22. for Here thespecific maximum occurs thedivision, outermost fiber, located a .ature In other any cross section, theatby longitudinal y>r of thewords, beam’s longitudinal axis atstrain the P 6–22. Here thePaxis maximum fiber, located a section andoflinearly the radius of the curvature ofaxis. the beam’s longitudinal at the strain occurs =at-the a outermost b distance y ! c from neutral Using Eq. 6–7, al strain will vary with y from the neutral axis. A since max = c>r, ny specific point. cross section, the longitudinal Pmax c>r Pmax = c>r, distance of y ! c from the neutral axis. Using Eq. 6–7, since In other words, for any specific cross section, the longitudinal then by division, action will occur fibers located the neutral axis y>r P A nearly 1-P2 with y6from thein neutral axis. Aabovewith thenthe by division, normal strain will vary linearly y from neutral axis. = -a b , in 1+P2the whereas willneutral occur in fibers located below the So the that neutral fibers elongation located above axis Pmaxaxis c>r 1 - P2 contraction will deformasyonlar occur in section fibers located above Boyuna birim tarafsız eksenden gösterilen y mesafesine bağlı olarak doğrusal -y2. This variation in strain over the cross is shown in Fig. y>r in fibers located below the +P2 will occur in whereas fibers located below1 +the P will occur 1 + y2, P2 elongation y>r Here the maximum strain occurs at the outermost fiber, located a = a b P birsection değişim that train over the is gösterecektir. shownininstrain Fig. Pmax overSothe 1 - y2. axiscross This variation cross section is shown in Fig. c>r = P-=a - a by b P (6–8) Pmax = c>r, nce of y ! c from the neutral axis. Using Eq. 6–7, since c>r c max ain occurs at the Here outermost fiber, located 6–22. the maximum straina occurs at the outermost fiber, located a Pmax by division, y So that distance y !since c from the=neutral Pmax c>r, axis. Using Eq. 6–7, since Pmax =P c>r, utral axis. Using Eq.of6–7, = - a bPmax (6–8) So that c then by division, This normal strain depends only on the assumptions made with regards y>r y P to the deformation. y is applied to the beam, therefore, it (6–8)When a moment = -a b P = - a bPmax cThis normal strain Pstress = - ain the b P longitudinal (6–8) P c>r y>r max will only cause a normal or x direction. All the depends only on the assumptions made with regards P y>r P c max y= - a b = -a b other components of normal and shear stress will be zero. It is this to the deformation. When a moment is applied to the beam, therefore, it Pmax c>r c>r max at uniaxial state of stress that causes the material to have the longitudinal will only cause a normal stress in the longitudinal or x direction. All the This normaletkisi: strain depends only on the assumptions made with regards Poisson y normal strain component depends only the with regards by Pon normal strain Eq. 6–8. Furthermore, components of normal anditshear stress willassumptions be by zero. Itmade is this Sotothat x, defined the deformation. When a momentother is applied toThis the beam, therefore, y to the deformation. When a moment is applied to the beam, therefore, it there must also be the associated strain components state(6–8) ofPoisson’s stress thatratio, causes the material to have longitudinal will only cause stress in theuniaxial longitudinal or x direction. All the P = a-normal a bPmax will only cause a normal stress in the longitudinal or x direction. All the c normal Pcomponent = nPx and = - nPxby , which the plane of by the cross-sectional normal strain Eq. deform 6–8. Furthermore, y y be other components of and shear stress will zero. ItPxis,Pzdefined this y y P = - a Poisson’s b Pmax other (6–8) components of normal and shear stress will be zero. It is this Such area, although here we have neglected these deformations. ratio, there must also be associated strain components state of stress that causes the c material to have the longitudinal P = - a bPuniaxial (6–8) max uniaxial state of stress that causes the material to have the longitudinal M Eq. c normal strain component Px, defined deformations will,deform causeofthe cross-sectional dimensions to Py = by -nP -nPx, which and Pz =Furthermore, the plane the cross-sectional x regards 6–8. byhowever, is normal strain depends only on the assumptions made with P , normal strain component defined by Eq. 6–8. Furthermore, by axis. z x become smaller below the neutral axis and larger above area, although here we have neglected these deformations. Such the neutral Poisson’s ratio, there must toalso be associated strain components x e deformation. When a moment is applied the beam, therefore, it ratio, This normal strain depends only on the assumptions made with regards Poisson’s there must also be associated strain components M For ifcause the beam has a square dimensions cross section, deformations will, however, the cross-sectional to it will actually Py = -nP = -nP , which deform the plane of theexample, cross-sectional x and z When nly atonormal stress in Pthe longitudinal or xisdirection. All the onlycause on the assumptions made with regards the deformation. ax moment appliedsmaller to the beam, therefore, it P = nP = nP , and P which deform the plane of the cross-sectional z y x z x deform as shown in Fig. 6–23. become below the neutral axis and larger above the neutral axis. Fig. 6–23 area, although here we have deformations. x neglected components of normal and shear stress zero. these Itarea, isorthis moment is applied to the beam, therefore, it will only cause a normal stress inwill the be longitudinal x direction.here AllSuch the have neglected these deformations. Such For example, if although the beam has a we square cross section, it will actually M deformations will, however, cause the cross-sectional dimensions ial state of stress that causes the material to have the longitudinal other components of normal and shear will be zero.will, It is however, thisto ss in the longitudinal or x direction. All the M stressdeformations cause the cross-sectional dimensions to deform as shown in Fig. 6–23. Fig. 6–23 becomestate smaller below the neutral axisFurthermore, and larger above thelongitudinal neutral axis. x component ,ofdefined al strain Eq. by smaller uniaxial the material to have the zstress l and shear stress willPxbe zero. that Itbyis causes this 6–8. become below the neutral axis and larger above the neutral axis. x For6.4 example, if thebe a square section, it will actually Thave HE Falso LEXURE Fbeam ORMULA 285 on’s ratio, there must associated strain components Phas normal strain component bycross Eq.For 6–8. Furthermore, by has a square cross section, it will actually auses the material to the longitudinal x, defined example, if the beam deform aswhich shown in Fig. 6–23. -nPx and PPoisson’s = -nP deform the plane theassociated cross-sectional ratio, there must alsoofbe strain components zby Eq. x,6–8. Furthermore, by deform as shown in Fig. 6–23. Fig. 6–23 x, defined nPx and = components - nPx, which Pz neglected the plane Such of the cross-sectional although here have thesedeform deformations. y = -we also be Passociated strain 6.4 THE FLEXURE FORMULA 6.4. Eğilme Formülü area, although here wecross-sectional have neglected these deformations. Such mations will, cause the dimensions to hich deform thehowever, plane of the cross-sectional M deformations will,axis however, cause thethe cross-sectional me belowthese the neutral and larger above neutral axis. dimensions to ave smaller neglected deformations. Such 6.4 THE FLEXURE FORMULA 285 y become smaller below the neutral axis and larger above the neutral axis. xample, if the beam has a dimensions square crosstosection, it will actually tes the stress x the cause cross-sectional Flexure Formula For example, if the beamPmax hasThe a square cross section, it will actually m as axis shown inlarger Fig. 6–23. moment acting utral and above the neutral axis. deform as shown in Fig. 6–23. at the material y as a square cross section, it In willthis actually section, we will develop an equation that relates the stress The Flexure Formula r variation of c 3. Pdistribution in a beam to the internal resultant bending moment acting Pmax y ar variation in x on the beam’s cross section. To do this we will assume that the material y this section, we will develop an equation that relates the stress ariation, sInwill behaves in a linear-elastic manner and therefore a linear variation of c m value, smax ,a distribution in a beam to the internal resultant bending moment acting Pmax P y normal strain, Fig. 6–24a, must then be the result of a linear variation in roportionality on the beam’s cross section. To do this we will assume that the material x normal stress, Fig. 6–24b. Hence, like the normal strain variation, s will Normal strain variation nd Eq. 6–8, we behaves in a linear-elastic manner and therefore a linear variation of c (profilezero view)at the member’s neutral axis to a maximum value, smaxP, a vary from y normal strain, Fig. 6–24a, must then be the result of a linear variation in distance c farthest from the neutral axis. Because of the proportionality x (a) like the normal strain variation, s will normal stress, Fig. 6–24b. Hence, Normal strain variation of triangles, Fig. 6–23b, or by using Hooke’s law, s = EP, and Eq. 6–8, we (profile view) y vary from zero at thecan member’s neutral axis to a maximum value, s , a max write 6.4 6.4 smaxfrom the neutral axis. Because of the proportionality distance c farthest (a) (6–9) Normal strain variation of triangles, Fig. 6–23b, or by using Hooke’s law, s = EP, and Eq. 6–8, we y (profile view) can write 2 8 6 C H A P T E R 6c B E N D I N G s M max y s y (a) s = - a b smax (6–9) x c y s c cross-sectional M s y smax s t. For positive y x = -variation a bsmax (6–9) smax Bendingsstress give negative y c 6 This(profile equation view) describes the stress distribution over the cross-sectional the negative x c area. The sign convention established here is significant. For positive s dAM ive or tensile y (b) z Bending stress variation M, which acts in the + z direction, positive values of xy give negative d at a specific (profile view) values forstress s, that is, a compressive since it acts in the negative x Fig. 6–24 This equation describes the distribution over thestress cross-sectional essive normal direction. Similarly, negative y values will give (b) s or tensile M positive dF area. ated at +y is The sign convention established here is significant. For positive values for s. If a volume element of material is selected at a specific Bending stress variationFig. 6–24 M, which acts in the +z direction, positive values of y give negative 6 point on the cross section, only these tensile normal (profile y c view) x or compressive oss section by for s, that is, a compressive stress since it acts in the negative values x stresses will act on it. For example, the element located at + y is d by the stress direction. Similarly,shown negative y values (b) in Fig. 6–24c. will give positive or tensile o zero. Noting values for s. If a volume element material selected at a axis specific We can locateofthe positionisof the neutral on the cross section by Fig. 6–24 A in Fig. 6–24c, stress variation point on the cross section, tensile compressive normal satisfyingonly the these condition thatorthe resultant force produced byBending the stress stresses will act ondistribution it. For example, element located at +y is over thethe cross-sectional area must be equal to zero. Noting(c) shown in Fig. 6–24c.that the force dF = s dA acts on the arbitrary element dA in Fig. 6–24c, Fig. 6–24 (cont.) We can locate the we position of the neutral axis on the cross section by require satisfying the condition that the resultant force produced by the stress PAGE "78 Noting distribution over the cross-sectional area must be equal to zero. Since smax>c is not equal to zero, then that the force dF = s dA acts on the arbitrary element dA in Fig. 6–24c, we require FR = ©Fx; 0 = dF = s dA y dA = 0 (6–10) smax s z ydA dF s s distributionBending about neutral axis. T stressthe variation M, which acts in the +z sdirection, valuesby of ythe givestress negative y positive max 6 shown incan Fig.s6–24c. s (profile view) We determine the stress in the beam from the requirement s y values for s, that is, a compressiveWe stress it the acts inmaxthe negative x Fig. 6–24c about the neutral isbydM = y dF. that Sin cansince locate position of the neutral axis on the crossaxis section s smax y dA direction. Similarly, negative y values will give positive or tensile the resultant internal moment M must be equal to the moment produced (b) z s satisfying the condition the resultant by thecross stress section, Eq. 6–9, haveforce for produced the entire dA that y values for s. If a volume element of material is the selected at a we specific zdistribution distribution over cross-sectional area must equal Fig. to axis. zero. Noting by about thebeneutral The moment of dF in dA the stress 6–24 zthese tensile or compressive normal point on the cross section, onlythat s the force dF = s dA acts on the arbitrary element dA in Fig. 6–24c, s M F Fig. 6–24c aboutlocated the neutral axis is dM = y dF. Since dF = s dA, using s ond it. stresses will act For example, the element at +y is we require s M dF smaxin Fig. Eq.s 6–9, we haveiçin forgerilmelerin the entirebileşkesi cross section, shown y 6–24c. dF Tarafsız ekseninM konumunu belirlemek (Resultant) y c x M = yolan dF = y1s dA2 = R2 z = ©M z; We can locate the position of the neutral axis on the1M cross section by c y dA kuvvetler için denge denkleminix yazarsak: LA LA satisfying the condition that the resultant force producedc by the stress z x M x y distribution over area must be equal to zero. Noting y c the cross-sectional 1M =dF = y dF 2x;z =element ©MzdA ; 0 in=M Rvariation Bending stress F = ©F s dA= R that the Fig. 6–24c, s force dF = s dA acts on the arbitrary dF L A L A L A or Bending stress variation c we yrequire (c) Bending stress variation y c y = -(c) a bsmax dA Bending stress variation Fig. 6–24 (cont.) c (c) or LA (c) Fig. 6–24 (cont.) ending stress variation Bending stress variation is not equal to zero, then s >c Fig.Since 6–24 (cont.) F =max ©Fx; 0 = dF = s dA (c) R (c) LA LA Fig. 6–24 (cont.) LA y1s dA2 = y y ¢ smax ≤ dA LA c M = smax y2 dA Thisc wood LAspecimen f -smax smax 2 to its fibers being crush y TdA M = (6–11) HE F LEXURE FORMULA y dA 2 8 7 apart at its bottom. c 6.4 LA c LA Since smax>c is not equal to zero, then Since is not equal to zero, then s >c max the moment 6–24 (cont.) y 0 of The integral inertia the cross-sectional area Fig. represents 6–24 (cont.) (6–10) yof dA = bs = a dA al to zero, then max Kesit ağırlık merkezinin konumu: about the neutral axis. We will symbolize its value as I. Hence, Eq. 6–11 *Recall that the centroid of the cross-sect LALA c y dA = 0 the location y for (6–10) can be solved for smax and written as L A = centroid y dA> dA. y dA = 0, the equation If 1 then This wood failed in bending due y = 0, and (6–10) y dAthat = 0the 1 1specimen -smax *Recall y fory the location of the cross-sectional area sthen not equal to zero, then the first moment 6.4 THE FLEXURE 2 8torn 7 is defined from Iny dA other of the L member’s cross-sectional area AdA to its fibers beingFORMULA crushed at its top and = y (6–10) = words, 0 reference (neutral) axis. See Appendix A. y = y dA> dA. y dA = 0, = 0, the equation If then y and so the centroid lies on the cThis at its bottom. area 1 be satisfied 1 the member’s LA LAabout the neutral axis must beInzero. condition can only if 1 apartcross-sectional other words, the first moment of Mc reference (neutral) axis. See Appendix A.area smoment = the (6–12) the neutralIn is=also thethe horizontal centroidal axis for the cross section.* about neutral must be zero. This condition can only be satisfied if (6–10) yaxis dA 0words, max other the first moment member’s cross-sectional represents the cross-sectional area (6–10) y dA = 0The integral Iof inertiaofofthe L A Consequently, once the centroid for the member’s cross-sectional area is st moment of the cross-sectional thesymbolize neutral also the horizontal centroidal axis for about the neutral axis mustarea beaxis zero. This condition canEq. only be satisfied if the cross section.* about the member’s neutral axis. We will itsisvalue as I. Hence, 6–11 Yani, tarafsız eksene göre alan momentinin gerekiyor. Bu şartarea ancak determined, location of the neutral axis is known must be zero. condition only be satisfied if atalet Consequently, once the centroid member’s cross-sectional is tarafsız thethe neutral axis is written also theashorizontal centroidal axis forsıfır the olması cross section.* scan can This be solved for and max ds, the first moment of the member’s cross-sectional area We can determine the stress in the beam from the requirement that he horizontal centroidal axis for the cross determined, the of the neutral axis ağırlık is known eksenin ağırlık merkezinden geçmesi sağlanabilir. Yani hesaplayarak Consequently, once thesection.* centroid forlocation theile member’s cross-sectional areamerkezini is t of the Here member’s cross-sectional area tral axis must be zero. This condition can only be satisfied the resultant internal moment M must be equal toifthe moment produced We can determine the stress in the beam from the requirement that centroid for the member’s cross-sectional area is determined, the location of the neutral axis is known tarafsız eksenin konumunu elde edebiliriz. ro. can only be satisfied if the cross s isThis alsocondition the horizontal centroidal axis for Mcsection.* by the stress distribution about the stress neutral axis. The moment ofrequirement dF the resultant internal moment M the must be in equal to the moment produced n of the neutral axis is known s = (6–12) We can determine the in the beam from that max = the s maximum normal stress in the member, which occurs at ntal centroidal section.* once the centroid forthe thecross member’s cross-sectional area is max axis for 6–24c about thethe neutral axis is dM using produced =IM ydistribution dF. Since dF =tosthe dA, by the stress about the neutral axis. The moment of dF in he stress Fig. in of the beam requirement that the from resultant internal moment must beaway equal moment aKesitte point on theiscross-sectional area farthest from the he the neutral axis known forlocation the member’s cross-sectional area is oluşacak moment, kesitteki gerilmelerin bileşkesi kuvvetlerin tarafsız eksene göre Eq. 6–9, we have for the entire cross section, Fig. 6–24c about the neutralaxis. axisThe is dM = y dF. oment M must be equal to the moment produced by the stress distribution about the neutral moment of Since dF indF = s dA, using neutral axis ermine the stress in the beam from the requirement that eutral axis is known momentleri toplamı olarak elde edilebilir. Eq. 6–9, we for the crossdF section, on about the neutral axis. moment of dF inhave Fig. 6–24c about neutral axis is dM Since =from yentire dF.the nternal moment must beThe equal tothe the moment produced n the beam from the requirement that M =M the resultant internal moment, determined method = s dA, using Here y utral axis is dM Since using cross = Eq. ythe dF. dF =for s the dA, distribution about neutral axis. The moment of dF in 6–9, we have entire section, the must be equal the moment produced M = equations 1MRto 2zof =sections ©M ; and y dF = of equilibrium, y1s dA2 = and y ¢calculated s ≤ dA 6.4 THE FLEXURE FORMULA 287 eutentire cross section, the neutral axis is dMz = y dF. Since dF = L s dA, using L y c max L A in Ain A which about the neutral axis of the cross section = the s maximum normal stress the member, at dA2 = the neutral axis. The moment of dF max 1MR2z = ©Mz ; M = y dFoccurs = y1s y ¢ smax ≤ dA ave for the entire cross section, point the cross-sectional areaneutral farthest away L A afrom LAy LA c is dM = y dF. dF = on s dA, c =Since theaperpendicular distance axis to pointthe 1M ©Musing ; M =fromythe dF = y1s dA2 = y ¢ smax ≤ dA R2the z = moment zy or The integral represents of inertia of the cross-sectional area neutral axis c oss section, away axis. This is A LAwhere smax acts LA = y dF = farthest y1s dA2 = from y ¢ the s neutral ≤ dA yL c=ormaxmoment, axis. We will symbolize itsssvalue Eq.the 6–11 max = neutral Mzabout ;L yA= dFthe = resultant y1s dA2 y¢ dA ≤ as A Mthe L LA 2 I. Hence, maxdetermined M internal from method M = y dA (6–11) IL = the moment of inertia of the cross-sectional area about the c A for s L A L A can be solved and written asequations smax c ofL A ofmax sections and the equilibrium, and M calculated or y neutral axis = y2 dA (6–11) F = y1s dA2 = about y ¢ the smax dA ≤ c neutral axis of the cross section s A L max 2 c LA L A M = y dA (6–11) smax 6 Mc(6–11) 2s distance fromstress the neutral to a point c the I=M = >cthe=yperpendicular -dA s>y, Eq. Since the normal at intermediate max LAaxis M = scmax 2 6–9, s = (6–12) max = y dA (6–11) c y *Recall that the location for the centroid of the cross-sectional area is defined from s farthest away from the neutral axis. This is where acts A L I c L distance y can be determined from an equation similar to Eq.max 6–12. A the equation y = 1 y dA> 1 dA. If 1 y dA = 0, then y = 0, and so the centroid lies on the *Recall thatcross-sectional the location y for area the centroid of the cross-sectional area is defined from have I = the moment of inertia of the about the sWe reference (neutral) axis. See Appendix A. max the equation y = 1 y dA> 1 dA. If 1 y dA = 0, then y = 0, and so the centroid lies on the M = y2 dA neutral (6–11) axis y for(neutral) *Recall that the location the centroid of the cross-sectional area is defined from c LA reference axis. See Appendix A. Here yefor the centroid of centroid thethe cross-sectional is defined y for the location of the y cross-sectional area is1defined = area y dA = from 0, then y = 0, and so the centroid lies on the equation Iffrom 1 y dA> 1 dA. My 6 dA.Since y dA 0, =the 0,centroid ys>y, and centroid lies onstress the dA. =If0,1then 0,then y ==>c and lies on the reference (neutral) axis. See Appendix A. sso =the (6–13) 1 yIfdA> 1 y1dA smax = -so Eq. 6–9, the normal at the intermediate I member, l) axis. See Appendix = the A. smax maximum normal stress in the which occurs at Appendix A. (Eğilme formülü) distance y can be determined from an equation similar to Eq. 6–12. a point onarea theiscross-sectional area farthest away from the ntroid of the cross-sectional defined from We have neutral axiscentroid lies on the A = 0, then y = 0, and so the Note thatresultant the negative is necessary sincefrom it agrees with the . M = the internalsign moment, determined the method My established x, y, z axes. By the right-hand rule, M is positive along of sections and the equations and calculated the s =ofs -equilibrium, (6–13) + z axis, y is positive upward, and therefore must be negative I about the neutral axis of the cross section (compressive) since it acts in the negative x direction, Fig. 6–24c. c = the perpendicular distance from the neutral axis to a point Either of the above two equations is often referred to as the flexure farthest away from the neutral axis. This is where smax acts formula. is used determinesign the is normal stress since in a straight member, NoteIt that theto negative necessary it agrees with the I = moment inertia ofthe theright-hand cross-sectional about having athe cross section isBy symmetrical with respect axis,the and the the established x, y, zofthat axes. rule, area Mto is an positive along neutral axispositive moment is applied perpendicular to this have be assumed + z axis, y is upward, andaxis. s Although thereforewemust negative that(compressive) the member issince prismatic, cannegative in most cases of engineering design it acts we in the x direction, Fig. 6–24c. 6 s >c flexure = the - s>y, Since Eq. two 6–9, the normalisthe stress at the intermediate also use the formula toequations determine normal stress members Either of above often referred toinas the flexure max distance y can beisdetermined from an equation similarin to Eq. 6–12. thatformula. have a slight taper.to For example, using a mathematical based It used determine the normal stress aanalysis straight member, Weon have the theory of elasticity, a member having with a rectangular section having a cross section that is symmetrical respect tocross an axis, and the andmoment a lengthis that is tapered 15° willtohave an actual maximum normal applied perpendicular this axis. Although we have assumed stress is about is5.4% less than that calculated using the flexure thatthat the member prismatic, we can in most cases of engineering design My s = to - determine the normal stress(6–13) formula. also use the flexure formula in members I that have a slight taper. For example, using a mathematical analysis based on the theory of elasticity, a member having a rectangular cross section and a length that is tapered 15° will have an actual maximum normal Note that the negative sign is necessary since it agrees with the PAGEthe "79 flexure stress that is about 5.4% less than that calculated using established x, y, z axes. By the right-hand rule, M is positive along the formula. + z axis, y is positive upward, and s therefore must be negative (compressive) since it acts in the negative x direction, Fig. 6–24c. = 6.11 6.4 6.4 THE FLEXURE FORMULA T6.4 HE FLEXURE FORMULA THE FLEXURE FORMULA 289 2 8 9 289 EXAMPLE 6.11 6.11 Dikdörtgen kesitli bir kiriş gösterilen gerilme dağılımına maruzdur. EXAMPLE Kesit momenti M değerini (a) Eğilme ile (b) TGerilme 6.4 HE FLEXURE FORMULA 6.4 formülü THE FLEXURE FORMULA 289 dağılımının bileşkesi ve temel prensipleri kullanarak hesaplayınız. beam has a rectangular cross section is subjected the stress A beamAhas a rectangular cross section and is and subjected to the to stress 289 6.4 THE M 289 THE FLEXURE FORMULA 289 distribution in6.11 Fig. 6–25a. Determine the internal moment at6.4 FORMULA distribution shown shown in Fig. 6–25a. Determine the internal moment M at FLEXURE 6 in. 6 in. EXAMPLE EXAMPLE 6.11 6.4 T HE F LEXURE F ORMULA 2 8 9 ectangular cross section and is subjected to the stress the section the distribution stress distribution (a) the using the flexure the section causedcaused by the by stress (a) using flexure wn in Fig. 6–25a. Determine internal Mof at the of 6 in.distribution formula, (b) by finding the resultant the distribution stress formula, (b) bythe finding themoment resultant stress using using EXAMPLE 6.11 ksi 2 ksi EXAMPLE 6.11 used by the stress distribution (a) using the flexurecross section and is subjected to the2 stress basic principles. A beam has a rectangular basic principles. A beam has a rectangular cross section and is subjected to the stress N finding EXAMPLE the resultant of 6.11 the stress distribution using distribution shown in Fig. 6–25a. Determine the internal moment M 6 in. N at distribution shown in Fig. 6–25a. Determine the internal moment M at 6 in. 2 ksi the section caused by the stress distribution (a) using the flexure A beam has a rectangular cross section and is subjected to the stress 6 in. 6 in. A SOLUTION beam has by a rectangular section (a) and using is subjected to the stress theSOLUTION section caused the stress cross distribution the flexure N of the stress distribution using formula, (b) by finding the resultant distribution shown in Fig. 6–25a. Determine the internal moment M at 6 in. shown in Fig.section 6–25a. the internal moment at in. A beamdistribution has abyrectangular cross and to the stress formula, (b) finding resultant ofDetermine theisisstress distribution using 26 ksi Part (a). Thethe flexure formula From Fig.6 M 6–25a, ssubjected max = Mc>I. basic principles. in. 2 ksi Partcaused (a). The flexure formula is smax Fig.M 6–25a, =using Mc>I. A the section by the stress distribution (a) theFrom flexure the section caused by the stress distribution (a) using the flexure distribution shown in Fig. 6–25a. Determine the internal moment at A 6 in. basic principles. and The neutral axis is defined as line NA, c = 6 in. s = 2 ksi. max N 6 in. and The neutral axis is defined as line NA, c = 6 in. s = 2 ksi. max formula, (b) by finding the resultant of the stress distribution using 6 in. formula, bystress finding the resultant the stress distribution using the section caused by the stress distribution (a) using the flexure the is zero along this of line. Since the cross section has Na flexure formula isbecause From Fig. 6–25a, sthe =(b) Mc>I. 2 ksi maxstress 2 ksi 6 in. because is zero along this line. Since the cross section has a A basic principles. basic principles. SOLUTION formula, (b) by finding the resultant of the stress distribution using 6 in. rectangular shape, the moment of inertia for the area about NA is The neutral axis is defined as line NA, smax = 2 ksi. 2 ksi 2 ksi 6 in. SOLUTION rectangular shape, the moment of inertia for the area about NA isN 2 ksi Part the (a). Thesection flexurehas formula is smax From front Fig. 6–25a,N = Mc>I. (a) from the formula for given on the inside ess is zero basic alongprinciples. thisdetermined line. Since cross aa rectangle A determined the formula for a rectangle given on the inside front Part (a). cover; Thefrom flexure formula is From Fig. 6–25a, s = Mc>I. 6 in. (a) A 6 in. N max and The neutral axis is defined as line NA, c = 6 in. s = 2 ksi. i.e., pe, the moment of inertia for the area about NA is max SOLUTION 6 in. SOLUTION 2 ksi and sbecause The neutral axis isthis defined as line c =cover; 6 in. i.e., = 2 ksi. max given the stress is zero along line. Since theNA, cross 6 in. (a) section has a m the formula for a rectangle on is thesinside SOLUTION Part (a). The formula From Fig. section 6–25a, = front Mc>I. Part (a). is The flexure formula is smax Fromhas Fig. =cross Mc>I. max line. because theflexure stress zero along this Since the a 6–25a, A rectangular shape, the moment of inertia for the area about NA is 1 1 2 ksi A 3 4 neutral is16 as NA, c = 6 Part in. and = flexure 2and ksi. 6 in. The axis isline defined line NA, cs=max 6shape, in. 2 ksi. (a). The formula is3 axis From 6–25a, s =defined Mc>I. I == bh =formula in.2112 in.2 =Fig. 864 inas 1max 1the 6 in. maxneutral rectangular thesThe moment inertia for the area about NA is 6 in. (a) 3 12 of 3 4 A determined from for a rectangle given on the inside front 2 ksi 12 6 in. I stress = 2 ksi. bh =line. 16 in.2112 in.2 = 864 because stress is szero along this Since the cross section has a section because the is zero along this line. thein cross and The neutral axis is Since defined as line NA, has a cdetermined =the 6 in. = max (a) 6 in. from the formula for a rectangle given on the inside front 12 12 cover; i.e., 1 1 shape, thein.2 moment ofininertia forof the area about NA 3zero along 4moment rectangular inertia the the stress isshape, this line. Since the for cross section has a NA is Çözüm: 2 ksi Therefore, 6area in. is about 2 ksi I = rectangular bh3because = 16 in.2112 = 864the cover; i.e., 12 rectangular 12from (a) determined the formula for a rectangle given on the inside front Therefore, (a) determined frommoment the formula for a rectangle given on the inside front shape, the of inertia for the area about NA is 2 ksi 1 Mc 3 2 1 M16 in.2 3 4 6 N 6 in. cover;determined i.e., I = bh = 16 in.2112 in.2 = 864 in cover; i.e., (a) from the formula for a rectangle given on the inside front s = ; 2 kip>in = M1612 max 1 3 1 (a) Mc 4 4 12 3in.2 6 "F 2 N I 4 in. 864 in 6 in. I = bh = 16 in.2112 in.2 = 864 in 6 in. cover; smaxi.e., = ; 2 kip>in = 12 1 12 "F IM16 4 in. 6 in. in.2 864# in4 4 N 1Therefore, 6 # ft in4 1 1 3kip 2 3 Ans.6 in. M3 = 288 in. in =in.2 243 kip 2 kip>in I= = bh4 = I = 16 in.2112 =in.2112 864 6 in. bh = in.2 16 = 864 4 in. "F A # # 1 kip12in. = 24 Therefore, 864 12 in 1 12 4 Ans. in. 6 in. M 288 3 12 3 kip ftM16 4 in.2 Mc 6 in. Is = =bh == 16 in.2112 in.2 = 2 864 in N 4 in. 6 in. A ; 12 2for kip>in = force two triangular stress 12The # Part # ftIresultant M16 in.2each of the "F Ans. M = 288 kipTherefore, in. = (b). 24max kip Therefore, Mc 4 in. 66 in. 864 in4 2 N 6 in. in. volume F distributions in 2Fig. 6–25b graphically equivalent to 4stress the smax = (b). ; The resultant kip>in = iseach A Part force for of the two triangular 4 "F Therefore, Icontained within each M16 4 in. 6 in. 864 in M16 #Thus, # ftvolume in.2 Mc =2 288 kip in.2 in. = each 24 stressM distribution. volume is6Nin. 6 Mc (b) FAns.N 2 distributions in Fig. 6–25b equivalent to kip the 6 resultant each two triangular stress 4 in. smax =force ;for = is 2graphically A sMc = of the ;2 kip>in kip>in = 4 max M16 in.2 "F # # 4 I "F 4 in. 864 in Ans. M = 288 kip in. = 24 kip ft 6 contained within each stress distribution. Thus, each volume is Iequivalent 4 in. 6 in. 6 in.(b) 2 volume 864 in N F Fig. 6–25b is graphically to the 6 in. smax = ; Part kip>in = 4 in.Fig. 6–25 (b). 2 The force each of the two triangular A 4 for 1resultant "F stress I 2 4 in. 864 in # # 6 in. n each stress distribution. Thus, each volume is (b) # # Ans. M = 288 kip in.=Fig. =288 24 kipkip>in ft F = 16 in.212 216 in.2 ft = equivalent 36 kip Ans. M kip in. = 24 kip F distributions in 6–25b is graphically to the volume 6 in. Fig. 6–25A 4 in. 4 in. Part (b). The resultant1 force2for# each 2of the #two triangular stress A Ans.volume is 288 kip in.stress = 24 kip FM = =16 in.212 kip>in 216 in.2ft= 36tokip contained within each distribution. Thus, each (b) Fig. 6–25 F distributions in Fig. 6–25b is graphically equivalent the volume 6 4in. in. 6 in. A 1 (b). The 2 Part resultant force for each of the two triangular stress 2(b). The resultant force for each of the two triangular stress Part F = 16 contained in.212 kip>in 216forces, in.2 36 kip These which form a couple, in volume the same as the each=is stress distribution. Thus,act each is direction (b) 6 in. Fig. 6–25 F distributions in within Fig. 6–25b graphically toequivalent the volume 2 Part (b). The force for is each of Fig. the two triangular F distributions in Fig. 6–25b graphically tostress the volume 1equivalent 2 Furthermore, they act stressesresultant within each distribution, 6–25b. F = 16 in.212 kip>in 216 in.2 = 36 kip These forces, which form a couple, act in the same direction as the contained within each stress distribution. Thus, each volume is (b) F distributions in Fig. is stress graphically equivalent to the volume contained within each distribution. Thus, volume (b) 2 volume, Fig. 6–25 through the6–25b centroid of each i.e., 23each 16 in.2 = 4 in.is from the 1same 2 the within each distribution, Fig. 6–25b. Furthermore, they act hich formcontained a stresses couple, within act inFeach the direction as stress distribution. Thus, each volume is (b) = 16 in.212 kip>in 216 in.2 = 36 kip neutral axis 2of the beam. Hence the distance between them is 8 in. as Fig. 6–25 2 Fig. 6–25 1These through centroid of each volume, i.e., from direction the =in4the in. same each distribution, Fig.the 6–25b. Furthermore, they act 1which 2the 3 16 in.2 forces, form a couple, act as the 2therefore shown. The moment of couple is F = 16 in.212 kip>in 216 in.2 = 36 kip 2 F = 16 in.212 kip>in 216 in.2 = 36 kip Fig. 6–25 (b) neutral axisi.e., of the the distance between themFurthermore, is 8 in. as ntroid of each volume, from the 161beam. in.2 = Hence 2stresses within each distribution, Fig. 6–25b. they act 24 in. 2 F 3=between 16 in.212 kip>in 216 in.2 = 36 kip 2 These forces, which form a couple, act in the volume, same direction as the= 4 in. from the shown. moment ofthe the couple is therefore he beam. Hence theThe distance them is 8 in. as # # 2M through centroid of each i.e., 16 in.2 = 36 kip 18 in.2 = 288 kip in. = 24 Ans. 3 kip ft within each awhich distribution, Fig. 6–25b. Furthermore, they act them mentThese of thestresses coupleThese is therefore forces, which form couple, act abeam. in the same direction as the neutral axis of the Hence thethe distance between forces, form couple, act in same direction as theis 8 in. as 2 # in. through the centroid of each volume, i.e., in.2 =24 4they in. # from stresses within each distribution, Fig. 6–25b. Furthermore, actas the M = 36 kip18 in.2 =of 288 kip = therefore kip ft 3 166–25b. shown. The the couple is These forces, which form amoment couple, act in the same direction the stresses within each distribution, Fig. Furthermore, they act NOTE: This result can also be obtained by choosing a Ans. horizontal 2 # # 2 neutral axis of the beam. Hence the distance between them is 8 in. as = 36 kip18 in.2 288 kip Ans. in. = 24 kip ft through the =centroid of each volume, i.e., from the 16 in.2 = 4 in. stresses within each distribution, Fig. 6–25b. Furthermore, they act through the centroid of each volume, i.e., from the 16 in.2 = 4 in. 3 strip of area dA ! (6 in.) dy and using 3 integration by applying 2 shown. moment the couple is # ftas Ans. neutral axis ofThe beam. Hence the is 8 #in. as NOTE: This result can also be obtained by=them choosing afrom horizontal Mdistance =Hence 36therefore kipbetween 18 in.2 288 kip in. = 24 through the centroid each volume, i.e., the =between 4 in. neutral axis ofofof the beam. the distance them iskip 8 in. Eq. 6–11. 3 16 in.2 esultshown. can neutral also be obtained by choosing a dy horizontal The moment of the couple is therefore strip of area ! (6 in.) and using integration axis of The thedA beam. Hence the distance them by is 8 applying in. as shown. moment of the couple is therefore # in.between M NOTE: = 36 in.2result = isapplying 288 kipalso Ans. = obtained 24 kip # ft by choosing A ! (6 in.) dy 6–11. and moment using integration by can be a horizontal Eq. shown. The ofkip18 theThis couple therefore # # M = 36 strip kip 18M in.2 = 288 kip in.=in.) =288 24dy kip ft using # in. # ft of dA18 !in.2 (6 and integration by applying =area 36 kip kip Ans. = 24Ans. kip NOTE: ThisMresult can also by choosing Eq. 6–11. = 36 kip 18 in.2be=obtained 288 kip # in. Ans. = 24 kip # afthorizontal NOTE: result can be choosing a horizontal stripThis ofNOTE: area dA !also (6 in.)obtained dy also and by using integration by applying This result can be obtained by choosing a horizontal strip of dA (6 in.) dyalso by applying Eq.area 6–11. NOTE: This! can obtained choosing a horizontal strip ofresult area dA !and (6beusing in.) dyintegration andbyusing integration by applying Eq. 6–11. strip ofEq. area dA ! (6 in.) dy and using integration by applying 6–11. Eq. 6–11. PAGE "80 6

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