NUMERICAL ON 3 HINGED PARABOLIC ARCH WITH SUPPORTS AT DIFFERENT LEVELS

NUMERICAL ON 3 HINGED PARABOLIC ARCH WITH SUPPORTS AT DIFFERENT LEVELS

A three hinged parabolic arch ACB is hinged at supports A and B with rise of 3m and 6.75 m respectively from crown. Span of the arch is 22.5 m .The arch carries a uniformly distributed load of 30KN/m from A to C. Determine the maximum positive and negative bending moments.

Step 1:

Determining the values of L₁ and L₂

By the property of parabola

(L₁)²/ (h₁) = (L₂)²/ (h₂)

L₁ = (L₁ + L₂) (√ (h₁) / (√ (h₁) + √ (h₂))

wkt

L₁ + L₂ = L

Therefore,

L₁ = (L) (√ (h₁) / (√ (h₁) + √ (h₂))

L₁ = (22.5) (√ (3) / (√ (3) + √ (6.75))

L₁ = 9m

L₂ = L- L₁

L₂ = 13.5m

Step 2:

Applying vertical equilibrium condition for the arch

V _a + V _b = 30(9) = 270KN….. (1)

Taking moment from left support about C

V _a (L₁) –H (h₁) -30(9) (4.5) = 0

V _a (9) –H (3) -30(9) (4.5) = 0….. (2)

Solving Eq 2

V _a = 0.33H + 135

Taking moment from right support about C

V _b(13.5) - H (h₂) = 0

V _b(13.5) - H (6.75) = 0….. (3)

Solving Eq 3

V _b = 0.5H

Substitute values of V _a and V _b in Eq 1

V _a + V _b = 270KN

0.33H + 135 +0.5H =270….. (4)

By solving the above equation (4) , H = 162 KN

Therefore,

V _a = 189KN

V _b = 81KN

Step 3: Determination of Maximum positive and Negative BM

Consider the section X-X at the Horizontal distance x from support A and vertical distance y from the arch rib

M _x-x = V _a (x) - 30(x) ²/2 –H (y)

y = 4hx (2 L₁-x)/ 2L₁²     (In case of arch with supports at different levele L is twice the span of individual side)

y = 4(3) x (18 - x)/ 18²   

M _x-x = 189 (x) - 30(x) ²/2 –162 (4(3) x (18 - x)/ 18²))

Simplifying the above equation

M _x-x = 81x – 9x²

Therefore differentiating the M _x-x wrt x and equating to zero to determine the value of x

(d M _x-x) / (d x) = 0

81 – 18x = 0

x = 4.5m

M _x-x = 81(4.5) – 9(4.5)²

M _x-x = 182.25 KN-m

Consider the section X-X at the Horizontal distance x from support B and vertical distance y from the arch rib

M _x-x = V _b (x) –H (y)

y = 4hx (2 L₂-x)/ 2L₂²    

y = 4(3) x (27 - x)/ 27²   

M _x-x = 81 (x) –162 (4(3) x (27 - x)/ 27² )

M _x-x = -81x + 6x²

Therefore differentiating the M _x-x wrt x and equating to zero to determine the value of x

(d M _x-x) / (d x) = 0

-81 + 12x = 0

x = 6.75m

M _x-x = -81(6.75) + 6(6.75)²

M _x-x = 273.38KN-m

Therefore, Max positive moment = 182.25 KN-m

Max negative moment = 273.38KN-m

FUNICULAR NATURE OF 3 HINGED PARABOLIC ARCH

FUNICULAR NATURE OF 3 HINGED PARABOLIC ARCH

whenever a three hinged parabolic arch is subjected to the uniformly distributed load throughout the entire span with same support level , then in such case the bending moment at any section of the arch is equal to zero. This is because the shape of bending moment diagram with respect to the geometry of the arch is same.This can be experienced only in case of flexible structure.In case of steel and Rcc structures this cannot be adapted due to its rigid nature.Apart from this ,if the rigid structure is designed by assuming the zero BM then the whole structure will experience moment under different load conditions and lead to failure of structure.

NUMERICAL

A three hinged parabolic arch of length L and rise h carries a uniformly distributed load of w/m span .Show that the there is zero bending moment at any section of the arch.

Step 1:

Applying vertical equilibrium condition for the arch

V _a + V _b = w (L)….. (1)

Since the arch is symmetric, the support reactions V _a = V _b = w (L) / 2

Taking moments about C

V _a (L/2) - H x h - w ((L) / 2) (L) / 4)) = 0

Solving the above equation

H = wL² / 8h

Step 2:

Consider the section X-X at the Horizontal distance x from support A and vertical distance y from the arch rib

M _x-x = V _a (x) - w(x) ²/2 –H (y)

W k t

M _x-x = V _a (x) - w(x) ²/2 –H (y)

M _x-x = [(w (L)/2(x)) - w(x) ²/2 – {(wL² / 8h) (4hx (L-x)/L²)}]

M _x-x = 0

NUMERICAL ON 3 HINGED PARABOLIC ARCH TO DETERMINE MAXIMUM BM

A three hinged parabolic arch ACB of span 20m and rise 4m carries a uniformly distributed load of 20KN/m run on the left half of the span. Find the Maximum bending moment off the arch.

Step 1:

Applying vertical equilibrium condition for the arch

V _a + V _b = 20(10)

V _a + V _b = 200KN……… (1)

Taking moments about support A

V _b (20) = 20 (10) (5)

V _b = 50KN

Substitute the value of V _b in eq (1)

Therefore, V _{a =} 150KN

Taking moments about crown C

H x 4 - V _b x 10 = 0

H = 125 KN

Step 2:

Consider the section X-X at the Horizontal distance x from support A and vertical distance y from the arch rib

Taking moment about the section x-x

M _x-x = 150x- 20(x²/2) - H(y)

W k t

y = 4hx (L-x)/L²     (Since it is a parabolic Arch)

y = 4(4) (x) (20-x)/20²

y = 0.8x – 0.04x² ……. (2)

Therefore, Substitute the value of y in M _x-x Equation

M _x-x = 150x - 20(x²/2) – 125(0.8x – 0.04x²)

M _x-x = 50x – 5x² ………. (3)

In order to obtain Maximum moment, the value of x is essential. Therefore differentiating the M _x-x wrt x and equating to zero to determine the value of x

(d M _x-x) / (d x) = 0

50 – 10x = 0

x = 5m

Substitute the value of x in the eq 3

M _x-x = 50(5) – 5(5)²

M _x-x = 125 KN -m

Therefore Maximum Bending moment is 125 KN –m

Arch Analysis

Assumptions and limitations adopted while analyzing an arch

1.     The cross section of the arch is assumed to be very small compared to its length.

2.    Action of torsion or twist is neglected, since the load act in the transverse direction of longitudinal axis.

3.     Self-weight of the arch is neglected.

4. The material of arch is isotropic and homogeneous with a constant Modulus of elasticity throughout.

5.     The resultant moment of bending stress is equal to the external moment along entire length of beam.

6.     The neutral axis neither undergo stresses nor change in length.

7.     Deflections are considered as very small compared to the length of the arch.

8.     In case of circular arch, the deflected shape follows a circular arc whose radius of curvature is large compared to its other dimensions.

Expression for radius of curvature of a circular arch

Consider the above figure Let R = Radius of the arch, L = Span of the Arch, r = Rise of the Arch, x and y are the co- ordinates of the point P from Origin O.

From Triangle OEP,

OP² = OE² + PE²

R² = (OC – EC)² + x²

R² = (R – (r-y))² + x²

R² = (R – r + y)² + x²

From the figure, x = OP sin θ = R sin θ

Similarly, y = OE – OD

                 

y = R Cos θ – R Cos α

We Know that in a segment of a circle, (2R – r) r = L²/4

Therefore, 2R = (L²/4r) + r

Hence, R= (L²/8r) + (r/2)

Expression for rise of an arch in a parabolic arch

Consider the above figure Let AB = L = Span of the Arch, CD = r = Rise of the Arch, 

x and y are the co- ordinates of the point P from Origin O

The general Equation of Parabola is given by

y=K x(L-x)

Where K is a Constant

At x = L/2, y = r

Substitute the above values in the general equation

r = K(L/2) (L – (L/2))

Therefore, K = 4r/L²

Substitute the value of K in the general equation

y = 4rx (L- x)/L²

Slope of the arch is obtained by differentiating the above equation wrt x

Tan θ = dy/dx = 4r(L-2x) / L²

ARCHES - Introduction and types of arches

ARCHES

Arches are the upward convex shaped curved structure comparatively stronger than beams, supported at ends to resist both horizontal and vertical displacements.

Arches are economic for long span compared to beam

1.     Bending moment of the beam always varies with square of the span.

2.      In case of arches the total moment is always obtained by moment below the load detected by the horizontal thrust action at the same span. (M-Hy)

By the above discussion we can conclude that for the same amount of the load Arch structure is efficient than compared to that of long beams.

Arch Action

1.     Arch is basically a compressive member (zero tensile stress member), whenever the external load is applied on the arch structure it is generally resolved into two components

·        Axial compressive stress

·        Thrust at the base instead of bending moments

2.   The important feature of the arch action is one in which the horizontal reaction at the supports is the governing force to resist the externally applied load by preventing arch from collapsing.

3.  As the height of the arch decreases the horizontal reaction at the supports increases in order sustain the serviceability of the arch for the applied load.

Types of Arches

Basically, there are three types of arches used in practice they are

1.     Two hinged arches

Ø The arch supported with only two hinges.

Ø It is statically indeterminate of degree 1.

Ø It is structurally easier to construct.

Ø The normal thrust along the rib which is compressive in nature causes the rib to shorten.

Ø It is likely developing stresses due to sinking of support.

2.     Three hinged Arch:

Ø This is an arch which consists of three hinges.

Ø It is statically determinate structure.

Ø This arch can be analyzed easily, but difficult to construct.

Ø Since it is determinate in nature, there will be no stresses due to sinking of support.

3.     Fixed Arch:

Ø It is the arch which is supported by fixed supports at both the ends.

Ø The degree of redundancy is 3.

Ø Since the fixed ends are restraint for all reactions at both the ends, hence it creates additional stresses in the arch.

Ø This arch can be analyzed by using strain energy method, least energy method, column energy method etc.…