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Unformatted text preview: Arches Theory and Design of Structures I
• The threehinge arch is a type of statically threedeterminate structure. • Arches are used because they are: efficient economical aesthetic (curved structural form) Arches Arches x1 x2 P2 Reactions • 4 unknown reaction components: VA, HA, VB and HB • 3 equations for overall static equilibrium • an equation of condition at hinge C
x1 x2 P2 y1 yA Q1 P1
1 B C 2 y1 yA Q1 P1
1 C 2 Q2 y2 yB HB
B HA VB
A OK! VA xA L = xA + xB xB A threehinge arch under loading P1 Q1 C P2 Q2 HB Q2 y2 yB HB Reactions at A HA AV A B VB HA VB
A VA xA L = xA + xB xB Take moments of ACB about B, VA L  HA (yA  yB)  P1 (x1 + xB) + Q1 (yB  y1) – P2 (xB  x2)  Q2 (yB  y2) = 0 (1) A threehinge arch under loading M N
k C θ yk yi Qi yA Pi S P1 Q1 C VC HC i xk xi HA A HA
xA Substituting (3) into (1), Substituting VA [L  xA (1yB/yA)] + (1yB/yA) (P1 x1 + Q1y1) (1(1 P1 (x1 + xB) + Q1 (yB  y1)  P2 (xB  x2)  Q2 (yB  y2) = 0 A VA A free body of a threehinge arch VA Taking moments of AC about C, VA xA  HA yA  P1 x1  Q1 y1 = 0 ∴ HA = [VA xA  P1 x1  Q1 y1] / y1 (2) (3) Therefore VA is solved and hence HA can be obtained by substituting it back into (3). x1 x2 P2 y1 yA Q1 P1
1 C 2 Q2 y2 yB HB
B HA VB
A VA xA L = xA + xB xB Internal Forces A threehinge arch under loading Reactions at B Reactions Similarly, by taking moments of ACB about A and Similarly, CB about C, VB and HB can be solved. Alternatively, using vertical and horizontal equilibrium (i.e. ∑V = 0 and ∑H = 0 respectively), VB and HB can also be obtained after determining VA and HA. M N
k C θ yk Internal Forces
yA yi Qi Pi S R S M e=M/R R N i xk xi • If the bending moment M, axial force N (normally compressive) and shear force S are required at some point k on the arch, the arch segment to the left (or right) of k can be isolated as a free body for analysis. HA A xA VA R S M e=M/R R N Let θ be the slope of the tangent of the arch rib at k. Let Resolving parallel and normal to the tangent at k, and taking moments give (5) N = (VA  Pi) sin θ + (HA + Qi) cos θ S = (VA  Pi) cos θ  (HA + Qi) sin θ (6) M = VA (xA  xK)  HA (yA  yK)  Q1 (yi  yk)  P1 (xi  xk) (7) A free body of a threehinge arch Internal Forces Internal
• The effect of bending moment M, axial force N and shear force S at k can be replaced by a resultant force R • The curve of action of the resultant force at every section of the arch rib is called the resultant resultant pressure line or thrust line • The funicular polygon drawn through the 3 hinges of a threethreehinge arch is the resultant pressure line for the arch. • The corresponding ray of the force polygon gives the magnitude and direction of the resultant at a section. • In a well designed arch, the rib should be as close as possible to the pressure line. Internal Forces Internal
• From the force polygon and pressure line, the magnitude and line of action of the resultant force at any section are known. The axial force N, shear force S and bending moment M at the section can be determined. The eccentricity e can also be found as M / N. • The bending moments at the hinges must be zero. The pressure line must pass through the three hinges. Logical Arch Rib Profile Logical Arch Rib Profile
• When the pressure line coincides with the arch rib profile, there will be no bending moment at any section of the arch rib under that pattern of loading. The sections are subjected to axial stress only. • In the above case, the arch rib profile is called the logical arch rib profile. logical
P e Axial force P through centroid Uniform stress P σ= A Force P at eccentricity e Compression Tension Possible problems with eccentric internal force • It requires a larger cross section to resist the higher stresses – waste of materials • Certain materials (e.g. plain concrete or masonry) are not good at resisting tension. P P ⋅e⋅ y + =σ A I Logical Arch Rib Profile
The equation of ordinate of arch rib with supports at the same level can be shown to be M(x,y) = Mo  H y = 0 ∴ y = Mo / H • M(x,y) is the bending moment at point (x,y) in the arch rib. • Mo is the bending moment of the same loading in a simply supported beam with the same span.
w A case for verification: logical arch rib profile for uniformly distributed load uniformly Consider a symmetrical 3hinge arch with span L Consider 3and rise h subjected to a UDL w on the horizontal span
w h H H L V = wL/2 A threehinge arch under UDL V = wL/2 R R w
y H R V = wL/2 x L A threehinge arch under UDL V = wL/2 h H R L Mo w w
y H R V = wL/2 x L A threehinge arch under UDL V = wL/2 h H R wL/2 H
w w H = wL2 8h
h wL/2 H h L/2 H = wL 8h
2 h L/2
H R V = wL/2 L A threehinge arch under UDL V = wL/2 H R R = V 2 + H 2 V = wL/2 Half of threehinge arch as a free body M CROWN =
BASE R= V +H
2 2 V = wL/2 w Half of threehinge arch as a free body M CROWN =
V = wL/2 L
V = wL/2 VL wL2 − − Hh = 0 2 8 Mo
A simply support beam of the same span under the same UDL M BASE = wL2 − Hh = 0 8 The bending moment Mo of a simply supported beam with the same The span under the same loading is given by Mo = wLx/2 – wx2/2 = wx (Lx)/2 wLx wx At midspan, yc = h and Mo = wL²/8 (which is also the moment caused midwL only by vertical forces acting on the halfarch about the central hinge) half∴ H = Mo / yc = wL2 / 8h wL • In general, wL M = − Hh = 0 8 y = Mo / H = 4hx(Lx)/L2 which is a parabola having a span L and a rise h. • At any point k (u,v) on the arch rib, the bending moment is, Mk = Mo  H v = (wLu/2 – wu2/2)  (wL2/8h) 4hu (Lu)/L² = 0 ∴ It is indeed a logical arch rib profile.
2 VL wL2 − − Hh = 0 2 8 w
w wL/2 H H = wL2 8h
h h L/2 Other examples of logical arch rib profiles H R V = wL/2 L A threehinge arch under UDL V = wL/2 H R R = V 2 + H 2 V = wL/2 Half of threehinge arch as a free body M CROWN = M BASE = VL wL2 − − Hh = 0 2 8 wL2 − Hh = 0 8 The slope is dy/dx = 4h (L2u)/L2 = tan θ dy Nk = (wL/2  wu) sin θ + (wL2/8h) cos θ Sk = (wL/2  wu) cos θ  (wL2/8h) sin θ = … = 0 Example Example
(Threehinge Arch) (Three A threehinge circular arch with a radius of 10m is threeloaded as shown. • Calculate the reactions and the internal forces at points D and E. • Draw the pressure line diagram.
18kN/m
C D E HB B 90kN VB 4m 4m 45º
A HA 6m 10m
VA 8m A threehinge circular arch 18kN/m
C D E 90kN 18kN/m
C 90kN
D VB HB B 4m
E VB HB B 4m 4m 6m 45º
A HA
A HA 4m 6m 45º 10m 8m
VA Reactions 10m 8m VA A threehinge circular arch A threehinge circular arch Taking moment for free body ACB about B, Taking VA × 18  HA × 6  180 × 13  90 × 4 = 0 Taking moment for free body AC about C, VA × 10  HA × 10  180 × 5 = 0 Solving the above equations gives VA = 180 kN, HA = 90 kN kN, Taking moment for free body ACB about A, VB × 18  HB × 6 + 180 × 5 + 90 × 14 = 0 Taking moment for free body BC about C, VB × 8 + HB × 4 + 90 × 4 = 0 Solving the above equations gives VB = 90 kN, HB = 90 kN kN, 18kN/m 2.93 E 7.07 45º A HA = 90 10m VA = 180 θ 4m C D 90kN 2.93 4m HB = 90 B VB = 90 6m A 8m HA = 90 E 7.07 18kN/m C D 90kN 4m HB = 90 B VB = 90 6m 4m 45º θ 9.165 9.165 Internal Forces √(102 42) At D, y = = √84 = 9.165 m MD = 90 × 4  90 × 3.165 = 75.2 kNm kNSDB = 90 × 0.9165 + 90 × 0.4 =  46.5 kN NDB = 90 × 0.4 + 90 × 0.9165 = 118.5 kN SDC = 90 × 0.4 = 36 kN NDC = 90 × 0.9165 = 82.5 kN 10m 8m Internal Forces V = 180 At E, At ME = 180 × 2.93  90 × 7.07  0.5 × 18 × 2.932 = 186.2 kNm kNSE = 180 × 0.707  90 × 0.707  18 × 2.93 × 0.707 = 26.3 kN NE = 180 × 0.707 + 90 × 0.707  52.74 × 0.707 = 153.6 kN
A 18kN/m
C D E 90kN VB HB B 4m 4m 6m 45º
A HA 100 Bending Moment Diagram
10m 8m 75.2 VA A threehinge circular arch C D E B A Resultant pressure line Plotted on Tension Side The End The ...
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This note was uploaded on 12/01/2010 for the course CIVL 2007 taught by Professor Profchai during the Spring '10 term at HKU.
 Spring '10
 profchai
 Arches

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