Chapter 7: Forces in Beams and Cables

1 School of Mechanical EngineeringChapter 7: Forces in Beams and Cables of Mechanical EngineeringContents7-2 IntroductionInternal Forces in MembersSample Problem Types of beam Loading and SupportShear and Bending Moment in a BeamSample Problem Problem Among Load, Shear, and Bending MomentSample Problem Problem With Concentrated LoadsCables With Distributed LoadsParabolic CableSample Problem of Mechanical EngineeringIntroduction7-3 Preceding chapters dealt with:a)determining external Forces acting on a structure and b)determining Forces which hold together the various members of a structure. The current Chapter is concerned with determining the internalforces( , tension/compression, shear, and bending) which hold together the various parts of a given member.

2 Focus is on two important types of engineering structures:a) Beams -usually long, straight, prismatic members designed to support loads applied at various points along the ) Cables -flexible members capable of withstanding only tension, designed to support concentrated or distributed of Mechanical EngineeringInternal Forces in Members7-4 Straight two- force member ABis in equilibrium under application of Fand -F. Internal forcesequivalent to Fand -Fare required for equilibrium of free-bodies ACand CB. Multiforce member ABCDis in equili-brium under application of cable and member contact Forces . Internal Forces equivalent to a force -couple system are necessary for equili-brium of free-bodies JDand ABCJ.

3 An internal force -couple system is required for equilibrium of two- force members which are not of Mechanical EngineeringSample Problem the internal Forces (a)in member ACFat point Jand (b)in member BCDat : Compute reactions and Forces at connections for each member. Cut member ACFat J. The internal Forces at Jare represented by equivalent force -couple system which is determined by considering equilibrium of either part. Cut member BCDat K. Determine force -couple system equivalent to internal Forces at Kby applying equilibrium conditions to either of Mechanical EngineeringSample Problem :0= yF0N1800N2400=++-yENEy600=:0= xF0=xESOLUTION: Compute reactions and connection Forces .

4 :0= EM()()() +-FN1800=FConsider entire frame as a free-body:School of Mechanical EngineeringSample Problem member BCDas free-body::0= BM()()() +-yCN3600=yC:0= CM()()() +-yBN1200=yB:0= xF0=+-xxCBConsider member ABE as free-body::0= AM() =:0xF0=-xxAB0=xA =:0yF0N600=++-yyBAN1800=yAFrom member BCD,:0= xF0=+-xxCB0=xCSchool of Mechanical EngineeringSample Problem Cut member ACFat J. The internal Forces at Jare represented by equivalent force -couple free-body AJ::0= JM()() +-MmN2160 =M:0= xF() -FN1344=F:0= yF() +-VN1197=VSchool of Mechanical EngineeringSample Problem Cut member BCDat K. Determine a force -couple system equivalent to internal Forces at free-body BK::0= KM()() +MmN1800 -=M:0= xF0=F:0= yF0N1200=--VN1200-=VSchool of Mechanical EngineeringVarious Types of beam Loading and Support7-10 beam -structural member designed to support loads applied at various points along its length.

5 beam designis two-step process:1)determine shearing Forces and bending moments produced by applied loads2)select cross-section best suited to resist shearing Forces and bending moments beam can be subjected to concentratedloads or distributedloads or combination of of Mechanical EngineeringVarious Types of beam Loading and Support7-11 Beams are classified according to way in which they are supported. Reactions at beam supports are determinate if they involve only three unknowns. Otherwise, they are statically of Mechanical EngineeringShear and Bending Moment in a Beam7-12 Wish to determine bending moment and shearing force at any point in a beam subjected to concentrated and distributed loads.

6 Determine reactions at supports by treating whole beam as free-body. Cut beam at Cand draw free-body diagrams for ACand CB. By definition, positive sense for internal force -couple systems are as shown. From equilibrium considerations, determine M and Vor M and V .School of Mechanical EngineeringShear and Bending Moment Diagrams7-13 Variation of shear and bending moment along beam may be plotted. Determine reactions at supports. Cut beam at Cand consider member AC,22 PxMPV+=+= Cut beam at Eand consider member EB,()22xLPMPV-+=-= For a beam subjected to concentrated loads, shear is constant between loading points and moment varies of Mechanical EngineeringSample Problem the shear and bending moment diagrams for the beam and loading : Taking entire beam as a free-body, calculate reactions at Band D.

7 Find equivalent internal force -couple systems for free-bodies formed by cutting beam on either side of load application points. Plot of Mechanical EngineeringSample Problem : Taking entire beam as a free-body, calculate reactions at Band D. Find equivalent internal force -couple systems at sections on either side of load application points. =:0yF0kN201=--VkN201-=V:02= M()()0m0kN201=+M01=M0kN14mkN28kN14mkN28k N26mkN50kN2666554433=-= +=-= +== -==MVMVMVMVS imilarly,School of Mechanical EngineeringSample Problem Plot that shear is of constant value between concentrated loads and bending moment varies of Mechanical EngineeringSample Problem the shear and bending moment diagrams for the beam AB.

8 The distributed load of 40 N/m extends over m of the beam , from Ato C, and the 400-N load is applied at : Taking entire beam as free-body, calculate reactions at Aand B. Determine equivalent internal force -couple systems at sections cut within segments AC, CD, and DB. Plot of Mechanical EngineeringSample Problem : Taking entire beam as a free-body, calculate reactions at Aand B.:0= AM()()()()() :0= BM()()()()() +AN135=A:0= xF0=xB Note: The 400 N load at Emay be replaced by a 400 N force and 1600 Nm couple at of Mechanical EngineeringSample Problem :01= M()04013521=+--Mxxx220135xxM-=:02= M() +-+-Mxx()mN +=xMFrom Cto D: =:0yF012135=--VN 123=V Evaluate equivalent internal force -couple systems at sections cut within segments AC, CD, and A to C: =:0yF040135=--VxxV40515-=School of Mechanical EngineeringSample Problem :02= M()() +-+--+-Mxxx()cmN -=xM Evaluate equivalent internal force -couple systems at sections cut within segments AC, CD, and Dto B: =:0yF0400m12135=---VN 277-=VSchool of Mechanical EngineeringSample Problem Plot A to C.

9 XV40135-=220135xxM-=From Cto D:N 123=V()mN +=xMFrom Dto B:N 277-=V()mN -=xMSchool of Mechanical EngineeringRelations Among Load, Shear, and Bending Moment7-22 Relations between load and shear:()wxVdxdVxwVVVx-=DD==D-D+- D0lim0()curve loadunder area-=-=- DCxxCDdxwVV Relations between shear and bending moment:()()VxwVxMdxdMxxwxVMMMxx=D-=DD==D D+D--D+ D D2100limlim02()curveshear under area==- DCxxCDdxVMMS chool of Mechanical EngineeringRelations Among Load, Shear, and Bending Moment7-23 Reactions at supports,2wLRRBA== Shear curve, -=-=-=-=-=- xLwwxwLwxVVwxdxwVVAxA220 Moment curve,() ===-= -==- 0at 8222max200 VdxdMMwLMxxLwdxxLwMVdxMMxxASchool of Mechanical EngineeringSample Problem the shear and bending-moment diagrams for the beam and loading : Taking entire beam as a free-body, determine reactions at supports.

10 With uniform loading between Dand E, the shear variation is linear. Between concentrated load application points, and shear is Between concentrated load application points, The change in moment between load application points is equal to area under shear curve between With a linear shear variation between Dand E, the bending moment diagram is a of Mechanical EngineeringSample Problem Between concentrated load application points, and shear is With uniform loading between Dand E, the shear variation is : Taking entire beam as a free-body, determine reactions at supports. =:0AM()()()()()()()0m.