week 4---bmd+sfd+afd

INTERNAL FORCESINTERNAL FORCESINTERNAL FORCESINTERNAL FORCES

AXIAL FORCE DIAGRAMAXIAL FORCE DIAGRAMSHEAR FORCE DIAGRAMSHEAR FORCE DIAGRAM

BENDING MOMENT DIAGRAMBENDING MOMENT DIAGRAM

INTERNAL FORCESINTERNAL FORCES

T d T d Todays Todays ObjectiveObjective::

Students will Students will be able to use be able to use the method of the method of sections fosections fosections for sections for determining determining internal forcesinternal forcesinternal forces internal forces in 2in 2--D load D load cases.cases.

Chapter 10 OutlineChapter 10 OutlineChapter 10 OutlineChapter 10 Outline

BBBeams:Beams:

Axial Force Shear Force &Axial Force Shear Force & Axial Force, Shear Force & Axial Force, Shear Force & Bending MomentBending Momentgg

Shear Force & Bending Moment Shear Force & Bending Moment DiDiDiagramsDiagrams

Relations between Distributed Load, ShearRelations between Distributed Load, ShearRelations between Distributed Load, Shear Relations between Distributed Load, Shear Force & Bending MomentForce & Bending Moment

W k 4 T t i lW k 4 T t i lWeek 4 Tutorial Week 4 Tutorial P blP blProblems:Problems:

Chapter 10Chapter 10Chapter 10Chapter 10

AXIAL FORCE, SHEAR FORCE AXIAL FORCE, SHEAR FORCE & BENDING MOMENTS& BENDING MOMENTS Determining Determining the forces & momentsthe forces & moments within a beam within a beam

subjected to an external load & reactions:subjected to an external load & reactions: Cut the beam by a plane at an arbitrary cross section & Cut the beam by a plane at an arbitrary cross section &

isolate the part of the beam to the left of the planeisolate the part of the beam to the left of the plane

AXIAL FORCE, SHEAR FORCE AXIAL FORCE, SHEAR FORCE & BENDING MOMENTS& BENDING MOMENTS

The isolated part cannot be in equilibrium The isolated part cannot be in equilibrium unless it is subjected to some system of unless it is subjected to some system of forces & moments at the plane where it forces & moments at the plane where it joins the other part of the beam joins the other part of the beam i t l f & ti t l f & tinternal forces & momentsinternal forces & moments

Since the system of Since the system of external loadsexternal loads & & reactionsreactions on the beam is 2on the beam is 2--D we canD we canreactionsreactions on the beam is 2on the beam is 2--D, we can D, we can represent the represent the internal forces & momentsinternal forces & momentsby an equivalent system consisting of 2by an equivalent system consisting of 2by an equivalent system consisting of 2 by an equivalent system consisting of 2 components of force & a couplecomponents of force & a couple

AXIAL FORCE, SHEAR FORCE AXIAL FORCE, SHEAR FORCE

Th tTh t PP ll l t thll l t th& BENDING MOMENTS& BENDING MOMENTS

The component The component PP parallel to the parallel to the beams axis is called beams axis is called

the axial forcethe axial forcethe axial forcethe axial forceThe component The component VV normal to the normal to the

beams axis is calledbeams axis is calledbeam s axis is called beam s axis is called

the shear forcethe shear forceMMThe couple The couple MM is called is called

the bending momentthe bending momentNotice that the Notice that the axial force, shear force & axial force, shear force &

bending momentbending moment on the part of the beam to theon the part of the beam to thebending momentbending moment on the part of the beam to the on the part of the beam to the right of the cutting plane are equal in magnitude but right of the cutting plane are equal in magnitude but opposite in direction to that on the leftopposite in direction to that on the left


The directions of theThe directions of the axial force, shear axial force, shear force & bending momentforce & bending moment in the freein the free--body diagrams are the established definitions of body diagrams are the established definitions of y gy gthe positive directions of these quantitiesthe positive directions of these quantities

A positive axial force A positive axial force PP subjects the beam to subjects the beam to pp jjtensiontension

A positive shear forceA positive shear force VV tends to rotate the axis tends to rotate the axis ppof the beam clockwiseof the beam clockwise

A positive bending momentA positive bending moment MM tends to causetends to causeA positive bending moment A positive bending moment MM tends to cause tends to cause upward curvature of the beams axisupward curvature of the beams axis


Notice that a positive bending moment subjects the Notice that a positive bending moment subjects the upper part of the beam to upper part of the beam to compressioncompression, , shortening the beam in the direction parallel to its axisshortening the beam in the direction parallel to its axisshortening the beam in the direction parallel to its axis shortening the beam in the direction parallel to its axis & subjects the lower part of the beam to tension, & subjects the lower part of the beam to tension, lengthening the beam in the direction parallel to its lengthening the beam in the direction parallel to its axisaxisaxisaxis

AXIAL FORCE, SHEAR FORCE AXIAL FORCE, SHEAR FORCE

i t l f & ti t l f & t& BENDING MOMENTS& BENDING MOMENTS

Determining the Determining the internal forces & momentinternal forces & moment at a at a particular cross section of a beam typically involves 3 steps:particular cross section of a beam typically involves 3 steps:

1.Determine the external forces & moments 1.Determine the external forces & moments draw draw the the freefree--body diagrambody diagram of the beam & determine the of the beam & determine the reactions at its supports If the beam is a member of areactions at its supports If the beam is a member of areactions at its supports. If the beam is a member of a reactions at its supports. If the beam is a member of a structure, you must analyze the structure.structure, you must analyze the structure.

2. Draw the 2. Draw the freefree--body diagrambody diagram of part of the beam of part of the beam cut cut y gy g ppthe beam at the point at which you want to determine the the beam at the point at which you want to determine the internal forces & moment & draw the freeinternal forces & moment & draw the free--body diagram of 1 body diagram of 1 of the resulting parts. You can choose the part with the of the resulting parts. You can choose the part with the simplest freesimplest free--body diagram If your cut divides a distributedbody diagram If your cut divides a distributedsimplest freesimplest free--body diagram. If your cut divides a distributed body diagram. If your cut divides a distributed load, dont represent the distributed load by an equivalent load, dont represent the distributed load by an equivalent force until after you have obtained your freeforce until after you have obtained your free--body diagrambody diagram

3. Apply 3. Apply the equilibrium equationsthe equilibrium equations use the equilibrium use the equilibrium equations to determine equations to determine PP, , VV & & MM..

EXAMPLEEXAMPLEFor the beam shown below, For the beam shown below,

EXAMPLEEXAMPLE

determine determine the internal forces & moment at the internal forces & moment at CC..

EXAMPLEEXAMPLEEXAMPLEEXAMPLE

StrategyStrategy

After determining After determining the reactionsthe reactions at the at the supports, cut the beam by a plane at supports, cut the beam by a plane at pp , y ppp , y ppoint point CC & draw & draw the freethe free--body diagrambody diagram to to the left of the plane. the left of the plane.

Then useThen use the equilibrium equationsthe equilibrium equations totoThen use Then use the equilibrium equationsthe equilibrium equations to to determine the internal forces & determine the internal forces & moments at moments at CC..

EXAMPLEEXAMPLESolutionSolution

EXAMPLEEXAMPLESolutionSolutionDetermine the External Forces & MomentsDetermine the External Forces & Moments::Begin by drawing the freeBegin by drawing the free--body diagram of the beam &body diagram of the beam &Begin by drawing the freeBegin by drawing the free--body diagram of the beam & body diagram of the beam &

determining the reactions at its supports:determining the reactions at its supports:


EXAMPLEEXAMPLESolutionSolutionDraw the FreeDraw the Free--Body Diagram of Part of the Beam:Body Diagram of Part of the Beam:Cut the beam at Cut the beam at CC & draw the free& draw the free--body diagram of body diagram of y gy gthe left part, including the the left part, including the internal forces & momentinternal forces & moment

in their defined positive directions:in their defined positive directions:


EXAMPLEEXAMPLE

Apply the Equilibrium Equations:Apply the Equilibrium Equations:From the equilibrium equations:From the equilibrium equations:From the equilibrium equations:From the equilibrium equations:

we obtainwe obtain

EXAMPLEEXAMPLE We can check our results with We can check our results with

EXAMPLEEXAMPLE

the freethe free--body diagrambody diagram of the part of of the part of beam to the right ofbeam to the right of CC::beam to the right of beam to the right of CC::


The equilibrium equations are:The equilibrium equations are:

Confirming thatConfirming thatgg


For the beam in Figure below, determine the For the beam in Figure below, determine the internal forces & momentinternal forces & moment atat BBinternal forces & momentinternal forces & moment at at BB..


StrategyStrategy

To determineTo determine the reactionsthe reactions at the supportsat the supportsTo determine To determine the reactionsthe reactions at the supports, at the supports, represent the triangular distributed load by represent the triangular distributed load by an equivalent force Then determine thean equivalent force Then determine thean equivalent force. Then determine the an equivalent force. Then determine the internal forces & momentinternal forces & moment at at BB by cutting by cutting th b b l tth b b l t BB & d i th f& d i th fthe beam by a plane at the beam by a plane at BB & drawing the free& drawing the free--body diagram of the part of the beam to the body diagram of the part of the beam to the l ft f th ll ft f th l i l di th t f thi l di th t f thleft of the plane, left of the plane, including the part of the including the part of the distributed load to the left of the planedistributed load to the left of the plane..


EXAMPLEEXAMPLE

Determine the Determine the External Forces & MomentsExternal Forces & Moments::Draw the freeDraw the free--body diagrambody diagram of the beam & represent of the beam & represent

th di t ib t d l d b i l t fth di t ib t d l d b i l t fthe distributed load by an equivalent force:the distributed load by an equivalent force:


EXAMPLEEXAMPLESolutionSolutionThe equilibrium equations are:The equilibrium equations are:

Solving them, we obtain:Solving them, we obtain:

EXAMPLEEXAMPLE

SolutionFrom the equilibrium equations are:From the equilibrium equations are:

bt iwe obtain:

EXAMPLEEXAMPLE

h i lh i l

EXAMPLEEXAMPLE

If you attempt to determine If you attempt to determine the internal the internal forces & moment at forces & moment at BB by cutting the by cutting the ff b d di f h bb d di f h b ddfreefree--body diagram of the beam at body diagram of the beam at BB, you do , you do not not obtain the correct resultsobtain the correct results

fi h h l i ffi h h l i f You can confirm that the resulting freeYou can confirm that the resulting free--body diagram of the part of the beam to body diagram of the part of the beam to the left ofthe left of BB givesgivesthe left of the left of BB gives gives

PPBB = 0, = 0, VVBB = 120 [N] & = 120 [N] & MMBB = 360 [Nm]= 360 [Nm] The effect of the distributed load is not The effect of the distributed load is not

properly accounted for on your freeproperly accounted for on your free--body body dididiagramdiagram


You must wait until You must wait until afterafteryou have isolated part ofyou have isolated part ofyou have isolated part of you have isolated part of the beam before the beam before representing distributed representing distributed l d ti th t t bl d ti th t t bloads acting on that part by loads acting on that part by equivalent forcesequivalent forcesequivalent forcesequivalent forces

EXAMPLEEXAMPLE

C idC id a simply supported beama simply supported beam l d dl d d

EXAMPLEEXAMPLE

Consider Consider a simply supported beama simply supported beam loaded loaded by a force:by a force:

Cut the beam at an arbitrary positionCut the beam at an arbitrary position xx betweenbetween Cut the beam at an arbitrary position Cut the beam at an arbitrary position xx between between the left end of the beam & the load the left end of the beam & the load FF::


EXAMPLEEXAMPLESolutionSolutionDraw the Draw the FreeFree--Body Body DiagramDiagram of Part of theof Part of theDiagramDiagram of Part of the of Part of the Beam:Beam:Cut the beam atCut the beam at BB & obtain& obtainCut the beam at Cut the beam at BB & obtain & obtain the freethe free--body diagram.body diagram.Because point Because point BB is at the is at the midpoint of the triangular midpoint of the triangular distributed load, the value distributed load, the value of the distributed load atof the distributed load at BBof the distributed load at of the distributed load at BBis 30 [N/m]. By is 30 [N/m]. By representing the distributed representing the distributed load by an equivalent forceload by an equivalent forceload by an equivalent forceload by an equivalent force

EXAMPLEEXAMPLEApplying the equilibrium equations to thisApplying the equilibrium equations to this freefree--

EXAMPLEEXAMPLEApplying the equilibrium equations to this Applying the equilibrium equations to this freefree

body diagrambody diagram, we obtain:, we obtain:

EXAMPLEEXAMPLE To determine theTo determine the internal forces & momentinternal forces & moment

EXAMPLEEXAMPLETo determine the To determine the internal forces & momentinternal forces & momentfor values of for values of xx greater than greater than LL, we obtain a , we obtain a freefree--body diagram by cutting the beam at an body diagram by cutting the beam at an y g y gy g y garbitrary position arbitrary position xx between the load between the load FF & the & the right end of the beam:right end of the beam:


The results are:The results are:


The The shear force & bending shear force & bending moment diagramsmoment diagrams are are ggsimply the graphs of simply the graphs of VV & & MMrespectively, as functions of respectively, as functions of xx:: They permit you to see the They permit you to see the

changes in the shear force & changes in the shear force & b di t th tb di t th tbending moment that occur bending moment that occur along the beams length as along the beams length as well as their maximum (leastwell as their maximum (leastwell as their maximum (least well as their maximum (least upper bound) & minimum upper bound) & minimum (greatest lower bound) values(greatest lower bound) values

Shear Force & Bending Shear Force & Bending iiMoment DiagramsMoment Diagrams

Thus we can determine the distributions of the Thus we can determine the distributions of the internal forces & momentinternal forces & moment in a beam by in a beam by yyconsidering a plane at an arbitrary distance considering a plane at an arbitrary distance xx from from the end of the beam & solving for the end of the beam & solving for PP, , VV & & MM as as functions offunctions of xxfunctions of functions of xx

Depending on the complexity of the loading, it Depending on the complexity of the loading, it may be necessarymay be necessary to draw several freeto draw several free--may be necessary may be necessary to draw several freeto draw several freebody diagramsbody diagrams to determine the distributions to determine the distributions over the entire length of the beamover the entire length of the beamover the entire length of the beamover the entire length of the beam

The resulting equations for The resulting equations for VV & & MM allow us to draw allow us to draw the shear force & bending moment the shear force & bending moment t e s ea o ce & be d g o e tt e s ea o ce & be d g o e tdiagramsdiagrams

EXAMPLEEXAMPLE

D t i thD t i th h f & b dih f & b di

EXAMPLEEXAMPLE

Determine the Determine the shear force & bending shear force & bending moment diagramsmoment diagrams for the beam below.for the beam below.


StrategyStrategy

After determining the reactions at the supports, After determining the reactions at the supports, cut the beam at an arbitrary position betweencut the beam at an arbitrary position betweencut the beam at an arbitrary position between cut the beam at an arbitrary position between AA & & BB to determine the internal forces & to determine the internal forces & moment for 0



Begin by drawing Begin by drawing the freethe free--body diagrambody diagram of of th ti b & ti th di t ib t dth ti b & ti th di t ib t dthe entire beam & representing the distributed the entire beam & representing the distributed force by an equivalent force:force by an equivalent force:


SolutionSolutionFrom the equilibrium equations are:From the equilibrium equations are:From the equilibrium equations are:From the equilibrium equations are:

bt i th tibt i th tiwe obtain the reactions:we obtain the reactions:


EXAMPLEEXAMPLE

Cut the beam at an arbitrary position between Cut the beam at an arbitrary position between AA & & BB& obtain & obtain the freethe free--body diagrambody diagram::y gy g


EXAMPLEEXAMPLE

From the equilibrium equations:From the equilibrium equations:

we obtain:we obtain:


SolutionSolution

Cut the beam at an Cut the beam at an arbitrary position between arbitrary position between BB & & CC & draw the free& draw the free--b d di f hb d di f hbody diagram of the part body diagram of the part of the beam to the right of of the beam to the right of the cutting plane:the cutting plane:the cutting plane:the cutting plane:


EXAMPLEEXAMPLESolutionSolutionFrom the equilibrium equations:From the equilibrium equations:

we obtain:we obtain:


SolutionSolutionThe shear force &The shear force &The shear force & The shear force & bending moment bending moment diagrams are diagrams are ggobtained by plotting obtained by plotting the equations for the equations for VV & & MM f h 2 ff h 2 fMM for the 2 ranges offor the 2 ranges ofxx::

EXAMPLEEXAMPLE When you obtain equations for the shear forceWhen you obtain equations for the shear force

EXAMPLEEXAMPLE When you obtain equations for the shear force When you obtain equations for the shear force

& bending moment in a beam that apply to & bending moment in a beam that apply to different parts of the beam as we did in thisdifferent parts of the beam as we did in thisdifferent parts of the beam, as we did in this different parts of the beam, as we did in this example, there are 2 conditions you can often example, there are 2 conditions you can often use to check your results:use to check your results:use to check your results:use to check your results:

The The shear force diagramshear force diagram of a beam is of a beam is continuous except at points where the continuous except at points where the beam is subjected to a point forcebeam is subjected to a point force

The The bending moment diagrambending moment diagram of a beam of a beam is continuous except at points where theis continuous except at points where theis continuous except at points where the is continuous except at points where the beam is subjected to a point couplebeam is subjected to a point couple

EXAMPLEEXAMPLEIn this example, the equations weIn this example, the equations we

EXAMPLEEXAMPLEIn this example, the equations we In this example, the equations we

obtained fir the bending moment obtained fir the bending moment MMforforfor for

0 < 0 < xx < 2 [m] & for 2 < < 2 [m] & for 2 < xx < 4 [m] < 4 [m] 2 [ ]2 [ ]must agree at must agree at xx =2 [m]=2 [m]

Checking, we have:Checking, we have:100(2)100(2) 20(2)2 [kNm] 60(420(2)2 [kNm] 60(4 2) [kNm]2) [kNm]100(2) 100(2) 20(2)2 [kNm] = 60(4 20(2)2 [kNm] = 60(4 2) [kNm]:2) [kNm]:120 [kNm] = 120 [kNm]120 [kNm] = 120 [kNm]

and we confirm that they agree.and we confirm that they agree.

QUIZQUIZ1. In a multiforce member, the member is generally 1. In a multiforce member, the member is generally

subjected to an internal .subjected to an internal .j _________j _________A) normal force B) shear forceA) normal force B) shear forceC) bending moment D) All of the above.C) bending moment D) All of the above.) g )) g )

2. In mechanics, the force component V acting tangent 2. In mechanics, the force component V acting tangent to, or along the face of, the section is called the to, or along the face of, the section is called the _________ ._________ .

ffA) axial force A) axial force B) shear forceB) shear forceC) l fC) l fC) normal force C) normal force D) bending momentD) bending moment

APPLICATIONSAPPLICATIONS

Th bTh bThese beams are These beams are used to support the used to support the

f f thif f thiroof of this gas roof of this gas station.station.

Why are the beams Why are the beams yytapered? Is it because tapered? Is it because of the internal forces? of the internal forces?

If so, what are these If so, what are these forces and how do weforces and how do weforces and how do we forces and how do we determine them?determine them?

APPLICATIONSAPPLICATIONSA fixed column supports A fixed column supports this rectangular billboard. this rectangular billboard.

Usually such columns are Usually such columns are wider at the bottom than wider at the bottom than at the top. Why? at the top. Why?

i b f hi b f hIs it because of the Is it because of the internal forces? internal forces?

If so, what are they and If so, what are they and h d d t ih d d t ihow do we determine how do we determine them?them?

APPLICATIONSAPPLICATIONS

The concrete supporting aThe concrete supporting abridge has fractured.bridge has fractured.What might have caused What might have caused h d hi ?h d hi ?the concrete to do this?the concrete to do this?

How can we analyze or How can we analyze or design these structures todesign these structures todesign these structures to design these structures to make them safer?make them safer?

INTERNAL FORCESINTERNAL FORCESThe design of any structural member The design of any structural member requires finding the forces acting within requires finding the forces acting within q g gq g gthe member to make sure the material the member to make sure the material can resist those loads.can resist those loads.

For example, we want to For example, we want to determine the internal forces determine the internal forces acting on the cross section at C. acting on the cross section at C. First, we need to determine the First, we need to determine the support reactionssupport reactionssupport reactions.support reactions.

Then we need to cut the beam at C and Then we need to cut the beam at C and draw a FBD of one of the halves of the draw a FBD of one of the halves of the beam. This FBD will include beam. This FBD will include the the internal forces acting at Cinternal forces acting at C FinallFinallinternal forces acting at C.internal forces acting at C. Finally, Finally, we need to solve for these unknowns we need to solve for these unknowns using the EofE.using the EofE.

INTERNAL FORCESINTERNAL FORCESIn twoIn two--dimensional cases, typical internal dimensional cases, typical internal

loads areloads are normalnormal oror axial forcesaxial forces (N(Nloads are loads are normal normal or or axial forcesaxial forces (N, (N, acting acting perpendicularperpendicular to the section), to the section),

shear forcesshear forces (V, acting along the surface), (V, acting along the surface), ( , g g ),( , g g ),and the and the bending momentbending moment (M).(M).

The loads on the left and right sides of the section at C are equal in The loads on the left and right sides of the section at C are equal in

iimagnitude but magnitude but opposite opposite in direction. This is because when the in direction. This is because when the two sides are reconnected, the net loads are zero at the section.two sides are reconnected, the net loads are zero at the section.

STEPS FOR DETERMINING STEPS FOR DETERMINING INTERNAL FORCESINTERNAL FORCES1. Take an imaginary cut at the place where you need to 1. Take an imaginary cut at the place where you need to

determine the internal forces. Then, decide which determine the internal forces. Then, decide which resulting section or piece will be easier to analyze.resulting section or piece will be easier to analyze.

2. If necessary, determine2. If necessary, determine any support reactionsany support reactions oror2. If necessary, determine 2. If necessary, determine any support reactionsany support reactions or or joint forces you need by drawing a FBD of the entire joint forces you need by drawing a FBD of the entire structure and solving for the unknown reactions.structure and solving for the unknown reactions.

3. Draw a FBD of the piece of the structure youve decided 3. Draw a FBD of the piece of the structure youve decided to analyze. Remember to show theto analyze. Remember to show the N, V, and MN, V, and M loadsloadsto analyze. Remember to show the to analyze. Remember to show the N, V, and MN, V, and M loads loads at the cut surface.at the cut surface.

E fE t th FBDE fE t th FBD4. Apply the 4. Apply the EofE to the FBDEofE to the FBD (drawn in step 3) and solve (drawn in step 3) and solve for the unknown internal loads.for the unknown internal loads.

EXAMPLEEXAMPLEGiven:Given:The loading on the beamThe loading on the beamThe loading on the beam.The loading on the beam.

Find:Find:Find:Find:The internal forces at point C.The internal forces at point C.

PlanPlan::Follow the procedure!!Follow the procedure!!

Solution:Solution:

1. Plan on taking the imaginary cut at C. It will be 1. Plan on taking the imaginary cut at C. It will be easier to work with the left section (point A to the easier to work with the left section (point A to the s o o s o (po os o o s o (po ocut at C) since the geometry is simpler.cut at C) since the geometry is simpler.

2. We need to determine A2. We need to determine Axx and Aand Ayy using a FBD of using a FBD of the entire frame.the entire frame.

FBD: 400 N1.2 m

FBD:

ByAx Ay

3 m 2 m

EXAMPLEEXAMPLE3.3. Now draw a FBD of the left section. Now draw a FBD of the left section.

Assume directions for VAssume directions for VCC, , NNCC and Mand MCC..

MC1.5 m400 N

V

NCA

400 N

C

4. Applying the 4. Applying the EofEEofE to this FBD, we getto this FBD, we get

96 N VC

pp y gpp y g , g, g

QUIZQUIZQQ1.1. A column is loaded with a vertical 100 N force. A column is loaded with a vertical 100 N force. At which sections are the internal loads the same?At which sections are the internal loads the same? PAt which sections are the internal loads the same?At which sections are the internal loads the same?

A) P, Q, and R A) P, Q, and R

B) P and QB) P and Q

P

Q 100 N) a d Q) a d QC) Q and R C) Q and R

D) None of the above.D) None of the above.

R

))

2. A column is loaded with a horizontal 100 N 2. A column is loaded with a horizontal 100 N force. At which section are the internal force. At which section are the internal loads largest?loads largest?

A) PA) P

P

Q 100NA) PA) P

B) QB) Q

C) RC) R

QR

SC) R C) R

D) SD) S

QUIZQUIZ1. Determine the magnitude of the internal 1. Determine the magnitude of the internal

loads (loads (normal, shear, and bendingnormal, shear, and bending

QQ100 N

loads (loads (normal, shear, and bending normal, shear, and bending momentmoment) at point C.) at point C.

A) (100 N 80 N 80 NA) (100 N 80 N 80 Nm)m) 0.5m

1 m

80 N

A) (100 N, 80 N, 80 NA) (100 N, 80 N, 80 N m) m) B) (100 N, 80 N, 40 Nm)B) (100 N, 80 N, 40 Nm)C) (80 N, 100 N, 40 Nm)C) (80 N, 100 N, 40 Nm)D) (80 N 100 N 0 N )D) (80 N 100 N 0 N )

C

D) (80 N, 100 N, 0 Nm )D) (80 N, 100 N, 0 Nm )

2. A column is loaded with a horizontal 1002. A column is loaded with a horizontal 1002. A column is loaded with a horizontal 100 2. A column is loaded with a horizontal 100 N force. At which section are the internal N force. At which section are the internal loads the lowest?loads the lowest?

P

Q 100N

A) P A) P B) QB) QC) RC) R

R

SC) RC) RD) SD) S

S

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