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3hinged_archs_stranal

3hinged_archs_stranal

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3hinged_archs_stranal

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  1. THREE HINGED ARCHES - 2 By: Dr. N Suresh, NIE, Mysore e-Notes for the lecture on VTU EDUSAT Programme Structural Analysis, Dr. N Suresh, NIE, Mysore 1

  2. Bending moment diagram for a 3-hinged arch (Refer comparison between arch and beam). Hy is called H-Moment. It varies with respect to Y. Therefore the shape of BM due to Hy should be the shape of the arch. Therefore to draw the BMD for an arch, draw the BMD for the beam over that superimpose the H-moment diagram as shown in fig. W C We know that for an arch, bending moment at any point is equal to beam BM-Hy h A B a b L Wab L + Shape – same as arch BEAM B.M. Hh – H – Moment diagram Wab L + – Hh Structural Analysis, Dr. N Suresh, NIE, Mysore 2

  3. Normal thrust and radial shear in an arch Normal Radial A VERTICAL VA 90 - φ NORMAL M φ HA HORIZONT Normal HA Radial A RADIAL VA along the radial direction is called radial shear. For the case shown in fig normal thrust = + HA Cos φ + VA Cos (90 - φ) = HA Cos φ + VA Sin φ (Treat the force as +ve if it is acting towards the arch and -ve if it is away from the arch). Radial shear = + HA Sin φ -VA Sin (90 - φ) = HA Sin φ + VA Cos φ (Treat force up the radial direction +ve and down the radial direction as -ve). Note: 1) To determine normal trust and tangential shear at any point cut the arch into 2 parts. Consider any 1 part. Determine net horizontal and vertical force on to the section. Using these forces calculate normal thrust and tangential shear. 2. Parabolic arch: If the shape of the arch is parabolic then it is called parabolic arch. Total force acting along the normal is called normal thrust and total force acting Normal C φ y h Horizontal A B ? ?? 2 Structural Analysis, Dr. N Suresh, NIE, Mysore 3

  4. If A is the origin then the equation of the parabola is given by y = cx [L – x] where C is a constant. L y = h We have at X = 2 L L L L ? ? − = L C . . h = C ?? ?? 2 2 2 2 4 H C= 2L Equation of parabola is = 4 hx ( ) − y L x L 2 φ is given by the following equation. ( 2 Lx L dy tan = φ 4 h ) = − y x 2 4 h ( ) = − L 2 x dx 4 L 2 h ( ) φ = − tan L 2 x L 2 1. A UDL of 4kN/m covers left half span of 3-hinged parabolic arch of span 36m and central rise 8m. Determine the horizontal thrust also find (i) BM (ii) Shear force (iii) Normal thrust (iv) Radial shear at the loaded quarter point. Sketch BMD. 4 kN/m C h = 8 m B HB HA A VA 36 m VB ? =0 Fx Structural Analysis, Dr. N Suresh, NIE, Mysore 4

  5. 0 H = − H A B ------ (1) = H H A B ? =0 V Fy + = V 4 x 18 A B ------ (2) + = V V 72 A B ? =0 V − MA + = x 36 4 x 18 x 9 0 B = ∴ = V 18 kN V 54 kN B A Mc= 0 + − = V x 18 H x 8 0 B B = H 40 5 . kN B = H 40 5 . kN A VERTICAL 4 kN/m NORMAL 66.04 M φ = 23.96 HORIZONTAL 40.5 kN y = 6 18 kN A RADIAL 40.5 kN 54 kN 9 m BM at M = 4 hx ( ) = − y L x - 40.5 x 6 + 54 x 9 L 2 4 x 8 x 9 ( ) 9 − = y 36 - 4 x 9 x 4.5 36 m 2 y= 6 = 81 kN.m Structural Analysis, Dr. N Suresh, NIE, Mysore 5

  6. Shear force at M = + 54 – 4 x 9 = 18 kN (only vertical forces) ( ) x 2 L L ( ) 9 x 2 36 36 96 . 230 = φ Normal thrust = N = + 40.5 Cos 23.96 + 18 Cos 66.04 = 44.32 kN S = 40.5 Sin 23.96 – 18 Sin 66.04 4 h φ = − tan 2 4 x 8 − = 2 S = - 0.0019 ? 0 4 kN/m ?? ?? ????? ????? ??????? ???????? ???????? ??????? ????????? ??????? ?????????? ? ?????????????? ??????? ????????? ?? ???????? − x 18 x = 54 18 − = x 54 3 x = x 13 5 . m 13 5 . 54 x 13.5 – 4 x 13.5 x 2 = 364 5 . kNm Structural Analysis, Dr. N Suresh, NIE, Mysore 6

  7. 364.5 kNm 324 kNm + – A C B 13.5 m A symmetrical 3-hinged parabolic arch has a span of 20m. It carries UDL of intensity 10 kNm over the entire span and 2 point loads of 40 kN each at 2m and 5m from left support. Compute the reactions. Also find BM, radial shear and normal thrust at a section 4m from left end take central rise as 4m. ?????? ?????? ???????? ?? ?? ???? ????? ?????????? ? =0 H Fx − = H 0 A B ------ (1) = H H A B ? =0 V Fy + − − − = V 40 40 10 x 20 0 A B ------ (2) + = V V 280 A B ? =0 40 + MA + + − = x 2 40 x 5 10 ( x 20 ) 10 V x 20 0 B = V 114 kN B = V 166 kN A Mc= 0 Structural Analysis, Dr. N Suresh, NIE, Mysore 7

  8. 10 x 114 4 x H 5 ) 10 x 10 ( B + − − = 0 = H 160 kN B = H 160 kN A VERTICAL 10 kN/m 40 kN NORMAL φ = 25.64 64.35 M HORIZONT 160 kN y = 2.56 86 kN REDIAL 4 m 160 kN 2 m 166 kN BM at M = - 160 x 2.56 + 166 x 4 – 40 x 2 - (10 x 4)2 = + 94.4 kNm ( L x 4 x 4 = 4 hx ) = − y L x 2 4 ( ) 4 − 20 20 56 . 2 2 y= m h 4 ( ) φ = − L 2 x tan L ( 20 2 4 x 4 ) 4 x − 2 = 20 φ Normal thrust = N = + 160 Cos 25.64 + 86 Cos 64.36 = 181.46 kN S = 160 Sin 25.64 - 86 x Sin 64.36 S = - 8.29 kN 2 = 250 . 64 Structural Analysis, Dr. N Suresh, NIE, Mysore 8

  9. Segmental arch A segmental arch is a part of circular curve. For such arches y = − 4 hx ( L ) x is 2 L not applicable since the equation is applicable only for parabolic arches. Similarly equation for φ will be different. To develop necessary equations for 3-hinged segmental arch = = = OA OC OM= OB R R C φ M E 90 h (90-φ)x x y Origin A B L/2 D L/2 φ R R O Relationship between R, L and h: From OAD ∆ OA = ? + 2 2 2 AD L OD [ R + 2 ? ] = − 2 R h ?? ?? 2 2 L 2 = + − + R R 2 Rh h 2 2 2 4 L 2 + = + 2 Rh h 2 4 L h 2 = + R 8 = h 2 x φ Sin R OE Cos = φ OM R − + h y φ = Cos R Structural Analysis, Dr. N Suresh, NIE, Mysore 9

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