NUMERICAL TO DETERMINE NORMAL THRUST AND RADIAL SHEAR FOR A THREE HINGED PARABOLIC ARCH

NUMERICAL TO DETERMINE NORMAL THRUST AND RADIAL SHEAR FOR A THREE HINGED PARABOLIC ARCH 

A three hinged parabolic arch of span 20m and rise 5m carries a uniformly distributed 

load of 20KN/m for entire left half of the span and a point load of 120KN at 5m from 

right support .Determine normal thrust and radial shear for the arch at section 4m from left span.

Step 1:

Applying vertical equilibrium condition for the arch

V _a + V _b = 20(10) +120 = 320KN….. (1)

Taking moment about support A

V _b (20) –120 (15) -20(10) (5) = 0….. (2)

Solving Eq 2

V _b= 140KN

Substitute the value of V _b in Eq 1

V _a= 180KN

Taking moment about Crown C

V _a (10) -20(10) (5)-H (5) = 0

180 (10) -20(10) (5)-H (5) = 0…….. (3)

Solving the above equation

H = 160KN

Step 2:

Determining the moment about the section 4m from left support

M _d = V _a (4) – 20(4) (2) – H (y _d) ……… (4)

Determining the value of vertical distance y

W k t

y = 4hx (L-x)/L² ………. (5)

y _d = 4(5)(4) (20-4)/20²  

y _d = 3.2m

Substituting the value of y _d in Eq (4)

M _d = 140 (4) – 20(4) (2) – 160 (3.2)

M _d = 48KN-m

Step 3:

Determining the normal thrust and radial shear

Differentiate the Eq (5) wrt x

dy/dx =tan Ꝋ= 4h (L-2x)/L²                                                                  

tan Ꝋ = 4(5) (20-2(4))/20²                                                                                                             

tan Ꝋ = 3/5                                                         

Sin Ꝋ = 3/√ {(3)² + (5)²} = 3/√ 34

Cos Ꝋ = 5/ √ 34

Normal thrust at D = P_n = H_d Cos Ꝋ +V_d Sin Ꝋ

Where,

H_d = Total horizontal force at section D

V_d = Total vertical force at section D

V_d = V _a – 20(4)

V_d = 180 – 80

V_d = 100KN

H_d = 160KN

P_n = H_d (5/ √ 34) +V_d Sin Ꝋ

P_n = 160(5/ √ 34) + 100 (3/ √ 34)

P_n = 188.65KN

Radial Shear at D = S_d = H_d Sin Ꝋ -V_d Cos Ꝋ

          

                                S_d = 160(3/ √ 34)- 100((5/ √ 34)

                     S_d = -3.43KN

Components of the arch

COMPONENTS OF THE ARCH STRUCTURE

The main structural components of the arches are

1.     Abutments: These are the structures are at the edges of the arch on which arch rests. They are also termed as supports, which are used to resist lateral loads due to arch action.

2.     Crown: Crown is the point on the Arch with highest elevation. It is also termed as voussoirs. The main function of crown is that it takes up the external load (in downward direction) on it and coverts the load into lateral force or thrust, which is transmitted to the supports and finally to the ground.

3.  Springing: The connecting junction between the Arch rib and the Abutment is called as springing.

4.   Span of arch: the clear horizontal distance between two supports of an arch is called as span of the arch. Generally, span of arch in case of tied arches carrying highway will be in the range 75m to 250m. In case of railways, the span range would be 50m to 200m.

5.     Rise: The perpendicular distance between the line of action of the support and the crown point of the arch. The height of rise in an arch as in case of parabolic arch subjected to entire udl is generally opted as 1/3 of span. If the arch is subjected to the unsymmetrical loading, then height of rise in the parabolic arch is generally taken as 1/5 to 1/6 of the span. (Ref: Lipson & Haque ,ASCE ,1980).

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