Three Hinged Arch (Part - 2) Civil Engineering (CE) Notes | EduRev

Structural Analysis

Civil Engineering (CE) : Three Hinged Arch (Part - 2) Civil Engineering (CE) Notes | EduRev

The document Three Hinged Arch (Part - 2) Civil Engineering (CE) Notes | EduRev is a part of the Civil Engineering (CE) Course Structural Analysis.
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Example 32.2 
A three-hinged semicircular arch of uniform cross section is loaded as shown in Fig 32.7. Calculate the location and magnitude of maximum bending moment in the arch.

Solution:

Three Hinged Arch (Part - 2) Civil Engineering (CE) Notes | EduRev

Reactions:
Taking moment of all the forces about hinge B leads to

Three Hinged Arch (Part - 2) Civil Engineering (CE) Notes | EduRev                                (1)

Bending moment
Now making use of the condition that the moment at hinge C of all the forces left of hinge C is zero gives

Mc = Ray × 15 - Ha × 15 - 40 × 7 = 0 

Three Hinged Arch (Part - 2) Civil Engineering (CE) Notes | EduRev

Considering the horizontal equilibrium of the arch gives,

Three Hinged Arch (Part - 2) Civil Engineering (CE) Notes | EduRev

The maximum positive bending moment occurs below D and it can be calculated by taking moment of all forces left of D about D.

MD = Ray × 8 - Ha × 13.267                            (3)

= 29.33×8 - 10.66×13.267 = 93.213 kN

Example 32.3
A three-hinged parabolic arch is loaded as shown in Fig 32.8a. Calculate the location and magnitude of maximum bending moment in the arch. Draw bending moment diagram.

Solution:

Three Hinged Arch (Part - 2) Civil Engineering (CE) Notes | EduRev

Reactions:
Taking A as the origin, the equation of the three-hinged parabolic arch is given by

Three Hinged Arch (Part - 2) Civil Engineering (CE) Notes | EduRev                               (1)

Taking moment of all the forces about hinge B leads to,

Three Hinged Arch (Part - 2) Civil Engineering (CE) Notes | EduRev

Three Hinged Arch (Part - 2) Civil Engineering (CE) Notes | EduRev

Now making use of the condition that, the moment at hinge C of all the forces left of hinge C is zero gives,

Three Hinged Arch (Part - 2) Civil Engineering (CE) Notes | EduRev
Three Hinged Arch (Part - 2) Civil Engineering (CE) Notes | EduRev                               (3)

Considering the horizontal equilibrium of the arch gives,

Three Hinged Arch (Part - 2) Civil Engineering (CE) Notes | EduRev                                  (4)

Location of maximum bending moment
Consider a section x from end B . Moment at section x in part CB of the arch is given by (please note that B has been taken as the origin for this calculation),

Three Hinged Arch (Part - 2) Civil Engineering (CE) Notes | EduRev                                  (5)

According to calculus, the necessary condition for extremum (maximum or minimum) is that Three Hinged Arch (Part - 2) Civil Engineering (CE) Notes | EduRev

Three Hinged Arch (Part - 2) Civil Engineering (CE) Notes | EduRev                       (6)

= 40-4x = 0

x = 10 m.

Substituting the value of x in equation (5), the maximum bending moment is obtained. Thus,

Three Hinged Arch (Part - 2) Civil Engineering (CE) Notes | EduRev

Mmax = 200 kN.m                                                 (7)

Shear force at D just left of 40 kN load

Three Hinged Arch (Part - 2) Civil Engineering (CE) Notes | EduRev

The slope of the arch at D is evaluated by,

Three Hinged Arch (Part - 2) Civil Engineering (CE) Notes | EduRev                              (8)

Substituting x =10 m. in the above equation, θ= 21.80 

Shear force Sd at left of D is

Sd = Ha sin θ - Ray cos θ                                  (9)

Sd = 150sin(21.80) - 80cos(21.80)

= -18.57 kN.

Example 32.4 
A three-hinged parabolic arch of constant cross section is subjected to a uniformly distributed load over a part of its span and a concentrated load of 50 kN, as shown in Fig. 32.9. The dimensions of the arch are shown in the figure. Evaluate the horizontal thrust and the maximum bending moment in the arch.

Solution:

Three Hinged Arch (Part - 2) Civil Engineering (CE) Notes | EduRev

Reactions:

Taking A as the origin, the equation of the parabolic arch may be written as,

y = -0.03x2 + 0.6x                                              (1)

Taking moment of all the loads about B leads to,

Three Hinged Arch (Part - 2) Civil Engineering (CE) Notes | EduRev                         (2)

Three Hinged Arch (Part - 2) Civil Engineering (CE) Notes | EduRev

Taking moment of all the forces right of hinge C about the hinge C and setting Mc = 0 leads to,

Three Hinged Arch (Part - 2) Civil Engineering (CE) Notes | EduRev
Three Hinged Arch (Part - 2) Civil Engineering (CE) Notes | EduRev                                      (3)

Since there are no horizontal loads acting on the arch,

Ha = Hb = H (say)

Applying ∑Fy = 0 for the whole arch,

Three Hinged Arch (Part - 2) Civil Engineering (CE) Notes | EduRev

85 - 0.15 H + 75 + 0.45 H = 200 

H = 40/0.3 = 133.33 kN                         (4)

From equation (2),

Ray = 65.0 kN

Rby =135.0 kN                                   (5)

Bending moment
From inspection, the maximum negative bending moment occurs in the region AD and the maximum positive bending moment occurs in the region CB .

Span AD
Bending moment at any cross section in the span AD is

M = Ray x - Ha (-0.03x2 + 0.6 x)          0 < x < 5                               (6)

For, the maximum negative bending moment in this region,

Three Hinged Arch (Part - 2) Civil Engineering (CE) Notes | EduRev

x = 1.8748 m

M = −14.06 kN.m

For the maximum positive bending moment in this region occurs at D ,

MD = Ray 5 - Ha (-0.03 X 25 + 0.6 x 5)
= +25.0 kN.m

Span CB
Bending moment at any cross section, in this span is calculated by,

Three Hinged Arch (Part - 2) Civil Engineering (CE) Notes | EduRevThree Hinged Arch (Part - 2) Civil Engineering (CE) Notes | EduRev

For locating the position of maximum bending moment,

Three Hinged Arch (Part - 2) Civil Engineering (CE) Notes | EduRevThree Hinged Arch (Part - 2) Civil Engineering (CE) Notes | EduRev

x =17.5 m

Three Hinged Arch (Part - 2) Civil Engineering (CE) Notes | EduRevThree Hinged Arch (Part - 2) Civil Engineering (CE) Notes | EduRev

M = 56.25 kN.m

Hence, the maximum positive bending moment occurs in span CB.

Summary

In this lesson, the arch definition is given. The advantages of arch construction are given in the introduction. Arches are classified as three-hinged, two-hinged and hingeless arches. The analysis of three-hinged arch is considered here. Numerical examples are solved in detail to show the general procedure of threehinged arch analysis.

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