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An arch is defined as a curved girder, having convexity upwards and supported at its ends.

**ARCHS**

**ARCH**

An arch is defined as a
curved girder, having convexity upwards and supported at its ends.

The supports must
effectively arrest displacements in the vertical and horizontal directions.
Only then there will be arch action.

**2.Linear arch**

If an arch is to take loads, say W_{1},
W_{2}, and W_{3} and a vector diagram and funicular polygon are
plotted as shown; the funicular polygon is known as the linear arch or
theoretical arch.

The polar distance ‘ot’represents the horizontal thrust.

The links AC, CD, DE
and EB will br under compression and there will be no bending moment. If an
arch of this shape ACDEB is provided, there will be no bending moment.

**3.Eddy’stheorem.**

Eddy’stheorem states
that “The bending moment at any section of an arch is proportional to the
vertical intercept between the linear arch (or theoretical arch) and the center
line of the actual arch”.

BM_{x} = ordinate O_{2}
O_{3} * scale factor

**Degree of static indeterminacy of a
three hinged parabolic arch**

For a three-hinged
parabolic arch, the degree of static indeterminacy is zero. It is statically
determinate.

**1.A three hinged parabolic arch hinged
at the crown and springing has a horizontal span of 12m and a central rise of
2.5m. it carries a udl of 30 kN/m run over the left hand half of the span.
Calculate the resultant at the end hinges.**

Let us take a section X
of an arch. Let ?be the inclination of the tangent at X. if H is the horizontal
thrust and V the net vertical shear at X, from theb free body of the RHS of the
arch, it is clear that V and H will have normal and radial components given by,

N = H cos ?+ V sin ? R
= V cos?–H sin ?

**The normal thrust and radial shear in an
arch rib.**

Parabolic arches are preferable to carry
distributed loads. Because, both, the shape of the arch and the shape of the
bending moment diagram are parabolic. Hence the intercept between the
theoretical arch and actual arch is zero everywhere. Hence, the bending moment
at every section of the arch will be zero. The arch will be under pure
compression that will be economical.

**Difference between
the basic action
of an arch
and a suspension cable**

An arch is essentially
a compression member, which can also take bending moments and shears. Bending
moment and shears will be absent if the arch is parabolic and the loading
uniformly distributed.

A cable can take only
tension. A suspension bridge will therefore have a cable and a stiffening
girder. The girder will take the bending moment and shears in the bridge and
the cable, only tension.

Because of the thrust
in cables and arches, the bending moments are considerably reduced.

If the load on the
girder in uniform. The bridge will have only cable tension and no bending
moment on the girder.

**Under what conditions will the bending
moment in an arch be zero throughout**

The bending moment in an arch throughout the span
will be zero, if

(i)
The arch is parabolic and

(ii)
The arch carries udl throughout the span

**2.A three-hinged
semicircular arch carries a point load of 100 kN at the crown. The radius of
the arch is 4m. Find the horizontal reactions at the supports.**

V_{A} = V_{B} = 50 kN

Equating the moment about C to Zero, V_{A} *
4 –H*4 = 0

H = V_{A}

Horizontal reaction, H = 50 kN

**3.A three-hinged semicircular arch of
radius 10m carries a udl of 2 kN/m over the span. Determine the horizontal and
vertical reactions at the supports.**

**Determine H, V _{A} and V_{B}
in the semicircular arch shown in fig**

Equating moments about A to Zero,

V_{B} * 12 –12 * 9 = 0;

V_{B} = 9 kN and V_{A}
= 3 kN

Equating moments to the left of C to zero,

H = V_{A} = 3 kN; H= 3 kN

**Distinguish between two hinged and three
hinged arches.**

**Two hinged arches**

Statically indeterminate
to first degree

Might develop
temperature stresses.

Structurally
more efficient.

Will develop
stresses due to sinking of supports

**Three hinged arches **

Statically
determinate

Increase in
temperature causes increases in central rise. No stresses

Easy to
analyse. But, in construction, the central hinge may involve additional
expenditure.

Since this is
determinate, no stresses due to support sinking

**Rib –shorting in the case of arches.**

In a 2-hinged arch, the
normal thrust, which is a compressive force along the axis of the arch, will
shorten then rib of the arch. This is turn will release part of the horizontal
thrust.

Normally, this effect
is not considered in the analysis (in the case of two hinged arches). Depending
upon the important of the work we can either take into account or omit the
effect of rib shortening. This will be done by considering (or omitting) strain
energy due to axial compression along with the strain energy due to bending in
evaluating H.

**Effect of yielding of support in the
case of an arch.**

Yielding of supports has no effect in
the case of a 3 hinged arch which is determinate. These displacements must be
taken into account when we analyse 2 hinged or fixed arches as under

Here U is the strain energy of the arch
?H and ?V_{A} are the displacements due to yielding of supports.

**5.A three-hinged parabolic arch has a
horizontal span of 36m with a central rise of 6m. A point load of 40 kN moves
across the span from the left to the right. What is the absolute maximum
positive bending moment that wills occur in the arch**

For a single concentrated load moving
from one end to the other, Absolute maximum positive bending moment

= 0.096wl = 0.096*40 *
36=138.24 kNm This occurs at 0.211 l = 0.211 * 36 = 7.596 m from the ends.

Absolute maximum
positive bending moment = 138.24 kNm at 7.596 m from the ends.

**6.A 3 hinged arch of span 40m and rise
8m carries concentrated loads of 200 kN and 150 kN at a distance of 8m and 16m
from the left end and an udl of 50 kN/m on the right half of the span. Find the
horizontal thrust.**

**Solution:**

**(a) Vertical reactions V _{A} and
V_{B} :**

Taking moments about A,

200(8) + 150(16) + 50 * 20 * (20 + 20/2)
–V_{B} (40) = 0

1600 + 2400 + 30000 –40 V_{B} = 0

V_{B}= = 850 kN

V_{A} = Total load –V_{B} = 200 +
150 + 50 * 20 –850 = 500 kN

**(b) Horizontal thrust (H)**

Taking moments about C,

-H x 8 + V_{A}
(20) –200 (20 –8) –150 (20 –16) = 0 -8H + 500 * 20 –200 (12) –150 (4) = 0

H = 875 kN

**7.A parabolic 3-hinged arch carries a
udl of 30kN/m on the left half of the span. It has a span of 16m and central
rise of 3m. Determine the resultant reaction at supports. Find the bending
moment, normal thrust and radial shear at xx, and 2m from left support.**

** Solution:**

**(1)
****Reaction at A nd B; **

(i)
Vertical components of reactions;

Taking moments about A,

-V_{B} (16) + 30 x 8^{2} /2 = 0

- V_{B} (16) + 30 * 32 = 0 V_{B}
= 60 kN

V_{A} = Total
load –V_{B} = 30 * 8 –60 kN V_{A} = 180 kN

(ii)
Horizontal components of reactions at A
and

**(2)
****Bending moment at x = 2m from A:**

Bending moment = V_{A}
(2) –30
* 2 *1 –H_{A}(y) ---- (1)

Substitute in (1)y = 1.3125 m at x = 2m fromA’‘.

Bending moment at x = 2m from A = 180
(2) –30 * 2 * 1 –160 * 1.3125

Bending moment at xx = 90 kNm

**(3) Radial shear force at x = 2m from A**

Shear force, R_{X}
= V_{x} cos ?–H sin ? Where, V = Net vertical shear force at x = 2m
from A

= V_{A} - w (2)
= 180 –30 * 2 V = 120 kN

H = Horizontal shear force = 160 kN

**(4) Normal thrust at x = 2m from A:**

Normal thrust P_{N} = V_{x}
sin?+ H cos ?= 120 sin 29º21’+160cos 29º21’ P_{N} = 198.28 kN.

**8.A parabolic 3-hinged arch carries
loads as shown in fig. Determine the resultant reactions at supports. Find the
bending moment, normal thrust and radial shear at D, 5m from A. What is the
maximum bending moment**

**Solution:**

**(1)Reaction at supports: (R _{A}
and R_{B})**

(i) Vertical components of R_{A}
and R_{B} : (V_{A} and V_{B}) Taking moments about A,

20 * 3 + 30 (7) + 25 * 10 * (10 +10/2)
–V_{B} *(20) = 0 V_{B} = 201 kN

V_{A} = Total
load –V_{B} = 20 + 30 + 25 * 10 -201 V_{A} = 99 kN

(ii) Horizontal thrust (H):

Taking moments about
the crown point C, considering the right

side of ‘C’,

-V_{B} (20/2) +
H (5) + 25 * 10 *5 = 0 -201 * (20/2) + 5 H + 1250 + 0

H = 125 kN (iii)
Resultant reactions R_{A} and R_{B} ;

**2. Bending moment, normal thrust and
radial shear force (at D):**

(iii) Normal thrust

P = V sin ?+ H cos ?

V = Net beam shear
force = V_{A} –20 V = 99 –20 = 79 kN

Substitute in (iii) P =
79 sin ?+ 152 cos ?= 179.28 kN (iv) Radial shear force

F = V cos ?–H sin ?

F = 79 cos ?–152 sin ?= 2.683 kN

**3. Maximum Bending Moment in CB:**

Considering a section xx at a distance
of ‘x’ fromm B‘’ BM_{xx} = 254KNM

**9.A 3-hinged
arch is circular, 25 m in span with a central rise of 5m. It is loaded with a
concentrated load of 10 kN at 7.5m from the left hand hinge. Find the**

**(a)
****Horizontal thrust **

**(b)
****Reaction at each end hinge **

**(c)
****Bending moment under the load **

**Solution:**

**ertical reactions V _{A} and V_{B}:**

Taking moments about A, 10(7.5) –V_{B}
(25) = 0

V_{B} = 3 kN

V_{A} = Total load –V_{B} = 10 -3 =
7 kN

**1.Horizontal thrust (H):**

Taking moments about C

**10.A three hinged circular arch of span
16m and rise 4m is subjected to two point loads of 100 kN and 80 kN at the left
and right quarter span points respectively. Find the reactions at supports.
Find also the bending moment, radial shear and normal thrust at 6m from left
support.**

**Solution:**

**(a)
****Reaction at A and B: **

(i)
Vertical components of reactions at A
and B:

Taking moment about A,

100 (4) + 80(12) –V_{B}(16)
= 0 V_{B} = 85 kN.

V_{A} = Total
load - V_{B} = (100+80)-85 V_{A} = 95 kN.

b. Horizontal
components of reactions at A and B; Taking moments about the crone points C

V_{A}(8) -H(Y_{C})
- 100(4) = 0 95 (8) –H (y_{C} ) –100 (4) = 0

H = 90 kN

(iii)Resultant
reactions at A and B:

**b)
Bending moment at 6m from the left support:**

**(c)
Radial shear force ‘F’ :**

R = V cos ?- H sin ?

V = net shear force at x = 6m from A

= V_{A} -100 =
95 –100 = -5 kN H = 90 kN

R = -5 cos (11º32’)–90
sin (11º32’)= - 22.895 R = -22.89 kN

**11.A symmetrical three hinged parabolic
arch of span 40m and rise 8m carries an udl of 30 kN/m over left of the span.
The hinges are provided at these supports and at the center of the arch.
Calculate the reactions at the supports. Also calculate the bending moment,
radial shear, normal thrust at distance of 10 m from the left support.**

**Solution:**

**(1)
****Reactions at the supports: **

(i)Vertical components;

Taking moments
about A,

Vertical component of R_{B}, V_{B} = 150 kN

V_{A} = Total load –V_{B} = 30 * 20
–150 = 450 kN

(iii)Horizontal
components

Taking moments
about the crown, ‘C’,

**(22) ****Bending
moment at 10 m from A:**

**(3)Radial shear force at x = 10m:**

R = Radial shear force
= V cos ?–H sin ? Where, V = Net vertical shear force at x = 10m from A

H = Horizontal thrust.

Radial shear force, R = V cos ?–H sin ?

R = 150 cos 21º–37548’sin 21º48’

R = 0

**(4) Normal thrust at x = 10m from ‘A’:**

Normal thrust, N = V sin ?+ H cos ?= 150
sin 21º48’375 cos+ 21?48’ N = 403.89 kN

**12.A parabolic
3-hinged arch of span ‘l’issubjected to an u.d.l of w/m run over the entire
span. Find the horizontal thrust and bending moment at any section XX.**

**Solution:**

**(a) Reactions (Vertical) at the
supports:**

As the loading is symmetrical, vertical
reactions at A and B are equal V_{A} = V_{B} = Total load/2 =
wl/2

**(b)
****Horizontal thrust:**

Taking moments about
the crown point C,

**(c)
****Bending moment at xx;**

Tags : Civil - Structural Analysis - Archs

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