THEORY OF STRUCTURES CHAPTER 5 : THREE PIN ARCH

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For updated version, please click onhttp://ocw.ump.edu.myTHEORY OF STRUCTURESCHAPTER 5 : THREE PINARCHbySaffuan Wan AhmadFaculty of Civil Engineering & Earth Resourcessaffuan@ump.edu.myby Saffuan Wan Ahmad

Chapter 5 – Three Pin Arch Aims– Determine internal forces, shear forces and bending moments in arch member Expected Outcomes :––––– Able to explain the function of archAble to describe the function of archAble to determine the reaction at support for three arch structureAble to determine the internal forces at any point at arch structureAble to draw shear force, axial force and bending moment diagramReferences–––––Mechanics of Materials, R.C. Hibbeler, 7th Edition, Prentice HallStructural Analysis, Hibbeler, 7th Edition, Prentice HallStructural Analysis, SI Edition by Aslam Kassimali,Cengage LearningStructural Analysis, Coates, Coatie and KongStructural Analysis - A Classical and Matrix Approach, Jack C. McCormac and James K.Nelson, Jr., 4th Edition, John Wileyby Saffuan Wan Ahmad

5.1 WHAT IS ARCH?crownhingeSpringinglineby Saffuan Wan Ahmad

5.1 WHAT IS ARCH?by Saffuan Wan Ahmad

5.2 FUNCTIONS OF ARCH Arches have been used for a very long time to spanlarge distance i.e bridges/building to carrytransverse loading efficiently. Arch carries most of the load axially with bendingmoment greatly reduced due to the curvature ofthe archby Saffuan Wan Ahmad

5.3 TYPES OF ARCH(a)(a)(a)Two-hinge archFixed archThree-hinge arch(a)Tied arch#Due to support systemby Saffuan Wan Ahmad

5.4 EQUATION OF PARABOLIC ARCHyC (L/2,h)P (x,y)h height ofthe archxA (0,0)B (L,0)Lby Saffuan Wan Ahmad

5.5 FORCES IN ARCHThe axial force, N shear force, Q and bending moment, Min the arch rib Shear force must be parallel to the cross sectionsurface, whilst the axial force must be perpendicularto the shear force. The positive were shown in figurebelow.by Saffuan Wan Ahmad

yc hyL/2hL/2xLy kx ( L- x ) Parabolic equationWhen x L/2 , y yc sub into enq.yc k (L/2) ( L- L/2 ) kL2/4Thereforek 4yc/L2 4h/L2then sub into eqn.y (4yc/L2 ) ( Lx- x2 )y (4h/L2 )( Lx- x2)by Saffuan Wan Ahmadsimplifyy (4h/L2 )x ( L- x )

At any point of the arch (parabolic)SlopebutSlopedydx4 hxy L 24 hxL L 24 hx Ldy dx4 h L4 h L 2 Where,by Saffuan Wan Ahmad ( L x )4 hx 2 L 24 hx 2 L 24 h2 ( 4 h ) x LL 28 hx L 2( L 2 x )tan-1slope

EXAMPLE 1 Calculate the reaction at support A and B asshown in figure below.by Saffuan Wan Ahmad

SolutionExample 1 Draw FBDConsider the whole structure MA 0130(5) 140(10) 150(15) VB (40) 0VB 107.5kN Fy 0VA 130 140 150 107.5 0VA 312.5kNby Saffuan Wan AhmadVB 107.5kNVA 312.5kN

SolutionExample 1 Now, consider segment CB MC 0107.5(20) H B (8) 0H B 268.75kNHence, H A 268.75kNby Saffuan Wan Ahmad

EXAMPLE 2 Determine the internal forces at the points D and E inthe three hinge parabolic arch as shown in figurebelow.16 kN/mCDE2.5 mAB2.5 mby Saffuan Wan Ahmad2.5 m3m2m

SolutionExample 2 Apply equation of equilibrium, consider the wholestructure then taking moment about A and B.V A 60kN , VB 20kNConsider RHS or LHS, take moment about C is equalto zeroH 40 kNby Saffuan Wan Ahmad

Consider segment CD4hx( L x)L24(2.5)(2.5) (10 2.5)10 2 1.875m4hSlope 2 ( L 2 x)L4(2.5) (10 2(2.5))10 2 0.5yD tan 1 (0.5) 26.57 oResolving in Q directionQ 40 sin( 26.57) 60 cos(26.57) 40 cos(26.57) 0Q 0.003kNby Saffuan Wan Ahmad

Consider segment CD4hx( L x)L24(2.5)(2.5) (10 2.5)10 2 1.875m4hSlope 2 ( L 2 x)L4(2.5) (10 2(2.5))10 2 0.5yD tan 1 (0.5) 26.57 oResolving in N directionN 40 cos( 26.57 ) 60 sin( 26.57 ) 40 sin( 26.57 ) 0N 44.72 kNby Saffuan Wan Ahmad

Consider segment CD4hx( L x)2L4(2.5)(2.5) (10 2.5)210 1.875m4hSlope 2 ( L 2 x)L4(2.5) (10 2( 2.5))210 0.5yD tan 1 (0.5) 26.57 oBending moment MD 02.5) 60( 2.5) 40(1.875) 02M D 25kNm M D 40(by Saffuan Wan Ahmad

EXAMPLE 3 Determine the bending moment at 25 m fromthe right hand support B and axial and shearforce at point D and E.10 kN/m8 kN10 m20 m10 mC5 kNE15 mDHA AVAB30 mby Saffuan Wan Ahmad30 mVBHB

SolutionExample 3 Calculate YE4(15)(10)yE (60 10)260 8.33m Consider whole structure MA 0 VB (60) 8(10) 10(20)(10 20) 5(8.33) 0VB 100.64kN Fy 0VA 100.64 8 10(20) 0VA 107.36kNby Saffuan Wan Ahmad

Consider RHS of the arch, taking moment at C is equal tozero.10 kN/m5 kN15 m8.33 m10 m MC 010 m10 mHB100.64 kN 100.64(30) 10(10)(5) 5(15 8.33) H B (15) 0H B 165.72kN Fx 0H A 165.72 5 0H A 170.72kNby Saffuan Wan Ahmad

Calculate the bending moment at 25 m from support , saypoint F4(15)(25)yF (60 25)260 14.58m MF 0M F 100.64(25) 165.72(14.58) 5(14.58 8.33) 10(5)(2.5) 0M F 56.45kNmby Saffuan Wan Ahmad

At point D (with point load)Resolving in Q directionQ 170.72 sin(33.67) 107.36 cos(33.67) 8 cos(33.67) 0Q 12.03kNResolving in N directionN 170.72 cos(33.67) 107.36 sin(33.67) 8 sin(33.67) 0N 197.16kNby Saffuan Wan Ahmad

At point F (with point load)Resolving in Q directionQ 100.64 cos(33.67) 165.72 sin(33.67) 5 sin(33.67) 0Q 10.96kNResolving in N directionN 100.64 sin(33.67) 165.72 cos(33.67) 5 cos(33.67) 0N 197.87kNby Saffuan Wan Ahmad

EXAMPLE 4Determine the reactions at supports and bending momentunder the load :by Saffuan Wan Ahmad

SolutionExample 4 l1r 1l2r220r 14016 r1 4my A 16 4 12mOr 4hx( L x)L24(16)(10)yA (40 10)40 2y A 12myA by Saffuan Wan Ahmad

Consider the whole structure. Taking moment at A isequal zero MA 080(5) 100(20) VB (30) H B (12) 030VB 12 H B 2400 . (i) Consider RHS, taking moment at C is equal to zero MC 0100(10) VB (20) H B (16) 020VB 16 H B 1000 .(ii )by Saffuan Wan Ahmad

Resolving by using calculatorVB 110kNH B 75kN Apply static equation Fy 0VA 110 80 100 0VA 70kN Fx 0H A H B 75kNby Saffuan Wan Ahmad

Consider LHS, taking moment under load 80kN4hxy80 2 ( L x)L4(16)(15) (40 15)240 15m'y80 15 12 3m M80 0 M 80 70(5) 75(3) 0M 80 125kNmby Saffuan Wan Ahmad

Consider RHS, taking moment under load 100kN4hxy100 2 ( L x)L4(16)(10) (40 10)240 12m M100 0M 100 75(12) 110(10) 0M 100 200kNmby Saffuan Wan Ahmad

THANKSby Saffuan Wan Ahmad

Author InformationMohd Arif Bin SulaimanMohd Faizal Bin Md. JaafarMohammad Amirulkhairi Bin ZubirRokiah Binti OthmanNorhaiza Binti GhazaliShariza Binti Mat Arisby Saffuan Wan Ahmad

– Mechanics of Materials, R.C. Hibbeler, 7th Edition, Prentice Hall – Structural Analysis, Hibbeler, 7th Edition, Prentice Hall – Structural Analysis, SI Edition by Aslam Kassimali,Cengage Learning – Structural Analysis, Coates, Coatie and Kong – Structural Analysis - A Cla

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