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