EXAMPLE 1 - ETU

1 Fig. 1 4bFig. 1 the resultant internal loadings acting on the cross sectionat Cof the beam shown in Fig. 1 N/m540 N2 m4 mVCMCNC(b)BC(a)ABC3 m6 m270 N/mFig. 1 4 SolutionSupport problem can be solved in the most directmanner by considering segment CBof the beam, since then thesupport reactions at Ado not have to be an imaginary section perpendicular to thelongitudinal axis of the beam yields the free-body diagram of segmentCBshown in Fig. 1 4b. It is important to keep the distributed loadingexactly where it is on the segment until afterthe section is made. Onlythen should this loading be replaced by a single resultant force. Noticethat the intensity of the distributed loading at Cis found by proportion, , from 4a,Themagnitude of the resultant of the distributed load is equal to the areaunder the loading curve (triangle) and acts throughthe centroid of ,which acts from Cas shown in Fig.

2 1 of the equations of equilibriumwe negative sign indicates that acts in the opposite direction tothat shown on the free-body diagram. Try solving this problem usingsegment AC, by first obtaining the support reactions at A, which aregiven in Fig. 1 N#m-MC-540 N12 m2=0d+ MC=0;VC=540 NVC-540 N=0+q Fy=0;NC=0-NC=0:+ Fx=0;1>316 m2=2 mF=121180 N>m216 m2=540 N,w=180 N> >6 m=1270 N>m2>9m, m1 m180 N/m90 N/m540 N135 NVCMCNC(c)1215 N3645 N mCAFig. 1 5cFig. 1 the resultant internal loadings acting on the cross sectionat Cof the machine shaft shown in Fig. 1 5a. The shaft is supportedby bearings at Aand B, which exert only vertical forces on the NCD200 mm100 mm100 mm50 mm50 mm800 N/mB(a)AFig. 1 m(800 N/m)( m) = 120 m225 NAyBy(b)BSolutionWe will solve this problem using segment ACof the free-body diagram of the entire shaft is shownin Fig.

3 1 5b. Since segment ACis to be considered, only the reactionat Ahas to be determined. Why?The negative sign for indicates that acts in the opposite sensetothat shown on the free-body an imaginary section perpendicular tothe axis of the shaft through Cyields the free-body diagram of segmentACshown in Fig. 1 of do the negative signs for and indicate? As an exercise,calculate the reaction at Band try to obtain the same results usingsegment CBDof the N#mMC+40 m2+ m2=0d+ MC=0;VC= N-40 N-VC=0+q Fy=0;NC=0:+ Fx=0;AyAyAy= m2+120 m2-225 m2=0d+ MB=0;(c)40 mMCVCCANCFig. 1 hoist in Fig. 1 6aconsists of the beam ABand attached pulleys,the cable, and the motor. Determine the resultant internal loadingsacting on the cross section at Cif the motor is lifting the 2000 N ( 200 kg) load Wwith constant velocity.

4 Neglect the weight of thepulleys and beam.(b) mC125 m2000 N2000 NVCNCMCAFig. 1 61 m125 mm125 mmACDBW(a)SolutionThe most direct way to solve this problem is to section both the cableand the beam at Cand then consider the entire left Fig. 1 of an exercise, try obtaining these same results by considering justthe beam segment AC, , remove the pulley at Afrom the beam andshow the 2000-N force components of the pulley acting on the beamsegment AC. Also, this problem can be worked by first finding thereactions at B, (Bx 0, By 4000 N, MB 7000 N m) and thenconsidering segment lb#ft500 lb ft2-500 lb ft2+MC=0d+ MC=0;-500 lb-VC=0 VC=-500 lb+q Fy=0;500 lb+NC=0 NC=-500 lb:+ Fx=0;2000 N 2000 N 2000 N 2000 N 2000 N m2000 N( m) 2000 N( m) 1 MC 0 Fig.

5 1 7dFig. 1 7cFig. 1 7b6200 N= 4650 N= 7750 NFBAFBD345(c)B(d)MGNGVG1 m3457750 N1500 the resultant internal loadings acting on the cross sectionat Gof the wooden beam shown in Fig. 1 7a. Assume the joints at A,B, C, D, and Eare pin connected.(a)600 N/m1 m1 m3 m1500 mEFig. 1 m3 m(3 m) = 2 m(3 m)(600 N/m) = 900 N121500 N= 2400 N= 6200 N= 6200 NEyExFBC(b)23 SolutionSupport we will consider segment AGfor theanalysis. A free-body diagram of the entirestructure is shown in Fig. 1 7b. Verify the computed reactions at Eand C. In particular,note that BCis a two-force membersince only two forces act on this reason, the reaction at Cmust be horizontal as BAand BDare also two-force members, the free-bodydiagram of joint Bis shown in Fig.

6 1 7c. Again, verify the magnitudesof the computed forces and Free-Body the result for the left section AGof the beam is shown in Fig. 1 of the equations of equilibriumto segment AG, we an exercise, compute these same results using segment lb#ftMG-17750 lb2A35B 12 ft2+1500 lb 12 ft2=0d+ MG=0;VG=3150 lb-1500 lb+7750 lbA35B-VG=0+q Fy=0;7750 lbA45B+NG=0 NG=-6200 lb:+ Fx=0;FBA, N 1500 N7750 N 3150 N(7750 N) 6200 N(1 m)(1500 N)(1 m) 03150 N mFig. 1 the resultant internal loadings acting on the cross sectionat Bof the pipe shown in Fig. 1 8a. The pipe has a mass of 2 kg/mand is subjected to both a vertical force of 50 N and a couple momentof at its end A. It is fixed to the wall at problem can be solved by considering segment AB, which doesnotinvolve the support reactions at x, y, zaxes are established at Band thefree-body diagram of segment ABis shown in Fig.

7 1 8b. The resultantforce and moment components at the section are assumed to act inthe positive coordinate directions and to pass through the centroidofthe cross-sectional area at B. The weight of each segment of pipe iscalculated as follows:These forces act through the center of gravity of each of the six scalar equations ofequilibrium, we have* do the negative signs for and indicate? Note that the normal force whereas the shear force is Also, the torsional moment is and the bending moment is*The magnitudeof each moment about an axis is equal to the magnitude of each forcetimes the perpendicular distance from the axis to the line of action of the force.

8 Thedirectionof each moment is determined using the right-hand rule, with positivemoments (thumb) directed along the positive coordinate N# +102=1MB2y= N#mTB=21022+ ,1MB2y1MB2x1MB2z=0 1MB2z=0;1MB2y= N#m1MB2y+ N m2+50 N m2=0 1MB2y=0;1MB2x= N# N m2=01MB2x+70 N#m-50 N N m2 1MB2x=0;1FB2z= N-50 N=0 Fz=0;1FB2y=0 Fy=0;1FB2x=0 Fx=0;WAD=12 kg> N>kg2= NWBD=12 kg> N>kg2= N70 N# m70 N m(b) mA50 NB(MB)z(MB)y(MB)x(FB)x(FB)y(FB)zFig. 1 m50 mCD70 N m(a)Fig. 1 16a(d)30 MPa35 mm10 mmFig. 1 bar in Fig. 1 16ahas a constant width of 35 mm and a thicknessof 10 mm. Determine the maximum average normal stress in the barwhen it is subjected to the loading shown.(b)9 kN9 kN12 kN12 kN= 12 kNPAB= 30 kNPBC22 kN= 22 kNPCDP(kN)x122230(c)12 kN22 kN9 kN9 kN4 kN4 kN35 mmADBC(a)SolutionInternal inspection, the internal axial forces in regions AB,BC, and CDare all constant yet have different magnitudes.

9 Using themethod of sections, these loadings are determined in Fig. 1 16b; and thenormal force diagram which represents these results graphically is shownin Fig. 1 16c. By inspection, the largest loading is in region BC, whereSince the cross-sectional area of the bar is constant,thelargest average normal stress also occurs within this region of the Normal Eq. 1 6, we stress distribution acting on an arbitrary cross section of thebar within region BCis shown in Fig. 1 16d. Graphically the volume(or block ) represented by this distribution of stress is equivalent tothe load of 30 kN; that is, 30 kN= MPa2135 mm2110 m2= MPaPBC=30 1 17cFig. 1 80-kg lamp is supported by two rods ABand BCas shown inFig.

10 1 17a. If ABhas a diameter of 10 mm and BChas a diameterof 8 mm, determine the average normal stress in each rod.(b)60 FBAFBCyx80( ) = NB345A60 BC345(a)Fig. 1 17 SolutionInternal must first determine the axial force in eachrod. A free-body diagram of the lamp is shown in Fig. 1 17b. Applyingthe equations of force equilibrium yieldsBy Newton s third law of action, equal but opposite reaction, theseforces subject the rods to tension throughout their Normal Eq. 1 6, we average normal stress distribution acting over a cross sectionof rod ABis shown in Fig. 1 17c, and at a point on this cross section,an element of material is stressed as shown in Fig. 1 m22= MPasBC=FBCABC= m22= MPaFBC= N, FBA= NFBCA35B+FBA sin 60 N=0+q Fy=0;FBCA45B-FBA cos 60 =0:+ Fx=0; MPa(c)(d)Fig.