1. STRUCTURAL ANALYSIS - I
THREE HINGED ARCHES
SRINIVASA .M
IV Sem M.Tech
(Industrial Structures)
SJCE, Mysore

2. Arches:-
An arch is a structure whose general outline is curved in nature.
An arch is a structure which under vertical loads produces inclined reactions at both supports.

3. We have 3-hinged, 2-hinged and fixed arches.
1-Three-hinged arches are statically determinate; hence, horizontal displacement of the abutments does not produce any additional stresses on the structural system.
2- Two-hinged arches and the fixed arches are statically indeterminate; hence, displacement of the abutments produces additional stresses in the structural system. Furthermore, foundations of such arches should be on rock or on very solid gravel.

4. CLASSIFICATION OF ARCHES:-
Based on Materials Used for Construction :-
Steel Arches
R.C.C. Arches
Masonry Arches etc…
Based on shape:-
Parabolic Arches
Circular or Segmental Arches
Elliptical Arches etc…
Based on structural behaviour:-
Three hinged arches
Two hinged arches
Fixed or Hinge less arches
Fixed arch with a hinge

5. CLASSIFICATION OF ARCHES:-
Based on structural behaviour:-
Open Arches
Solid Arches
FORCES AT ANY SECTION OF AN ARCH
Bending moment (M)
Normal Thrust (N)
Radial Shear (S)

6. EXAMPLES FOR STATICALLY DETERMINATE & INDETERMINATE STRUCTURES Sl. No.
INDETERMINATE (HYPERSTATIC) 1
Simply supported beam
Continuous beam 2
Cantilever beam
Propped cantilever beam 3
Three-hinged arch
Fixed end arch 4
Three-hinged frame
Rigid frame

7. DIFFERENCES B/W BEAMS & ARCHES
SL.
NO.
BEAMS
ARCHES 1
Transfers the applied load to end supports by Bending & Shear action.
Transfers the applied load to end supports partly by axial comp. & partly by Flexure. 2
In the process of transferring a particular section subjected to maximum stress.
In the process of transferring loads, any part of the section subjected to equal stresses. 3
The material in most of the portion is under stress & hence the section is under- utilised.
Due to the equal distribution of stress, the section is fully utilised. 4
Very uneconomical for larger spans, Self weight of beam itself contributes to larger stresses.
Very economical for larger spans, Self weight of arches are comparatively low. 5
Bending moment at any section is very high for a given span.
Bending moment at any section is considerably less compare to beams of same span. 6
B.M. = W x x
B.M. = W x x – H x y

9. Forces Compression – a pushing or squeezing force
Tension – a pulling or stretching force

10. Abutments – the structures that support the ends of the bridge
CONSTRUCTION OF AN ARCH
Keystone – the wedge-shaped stone of an arch that locks its parts together
Abutments – the structures that support the ends of the bridge

11. Works by
Compression
Cold Spring Arch Bridge, Santa Barbara, CA

12. Cable-Stayed Bridges
Piers – the vertical supporting structures
Cables – thick steel ropes from which the decking is suspended
Decking – the supported roadway on a bridge

13. Cable-Stayed Bridges
Works by Tension AND Compression

14. Suspension Bridges Similar to Cable-Stayed
Different construction method

15. Suspension Bridges
Works by Tension and Compression

16. ANALYSIS OF THREE HINGED ARCHES
STEPS:-
Draw the free body diagram including all the forces, dimensions & reactions.
Calculate the support reactions using equilibrium conditions (i,e, ∑M=0, ∑V=0, ∑H=0).
∑MA=0 ; Find the support reaction VB
∑V=0 ; Calculate the value of VA
∑MC=0 ; Calculate the value of HA
∑H=0 ; Calculate the value of HB
Find max. positive B.M ( i,e, Mx-x=VA x x – HA x y – w x x x x/2

17. Find value of “x’ by using condition
; x = m
Substitute the value of ‘x’ in moment equation; calculate the value of max. B.M.
Then use the same procedure to find – ve B.M
Normal thrust & Radial shear at any point of the arch is given by
Normal Thrust, N = V sinq + H cosq
Radial Shear, S = V cosq + H sinq
where ‘q’ can be calculated as ;
Draw B.M.Diagram for the given arch.

18. PARABOLIC ARCHES
EX (1) A three hinged parabolic arch of span 60 m and a rise of 8m carries an UDL of intensity 20 KN/m over the left half span. Calculate i) The support reactions ii) 25m from left support iii) Max,+ve BM iv) Max,- ve BM v) Normal Thrust & Radial shear at left quarter span. Also sketch the BMD of the arch.
Soln:

19. To find support reactions.
a) ∑MA = 0 ( ve )
VB x 60 – 20 x 30 x 30/2 = 0
VB = 150 KN ( )
b) ∑V = 0 ( +ve)
VA + VB – 20 x 30 = 0
VA = 600 – 150 = 450 KN ( )
c) ∑MC = 0 ( ve ) ∑H = 0 ( +ve)
RB x 30 – HB x 8 = HA – HB = 0
150 x 30 – HB x 8 = HA = HB = KN ( )
HB = KN ( )

20. To find B.M @ 25 m from left support,
Mx-x= VA x 25 – HA x yD – 20 X 25 x 25/2
(Rise) yD can be calculated as follows
yD =
To find Max. + ve B.M.
To find Max. - ve B.M.

21. CIRCULAR or SEGMENTAL ARCHES
To find Normal Thrust & Radial Shear:-
Normal Thrust, N = V sinq + H cosq
Radial Shear, S = V cosq + H sinq
Draw B.M. Diagram.
Radius of circular arch is given by,
Arch profile at any point is given by,
CIRCULAR or SEGMENTAL ARCHES

22. Cold Spring Arch Bridge, Santa Barbara, CA

23. Marsh Rainbow Arch, Riverton, KS

24. Pont du Gard, Nimes, France

25. Iron bridge, River Severn, England, built by Abraham Darby, 1779
Iron bridge, River Severn, England, built by Abraham Darby, Members in compression; connections using dowels etc.

26. St Louis Rail Bridge, St Louis USA, Mississippi River
St Louis Rail Bridge, St Louis USA, Mississippi River. James Eades, First true steel bridge. Three spans, each 152 m.

27. THANK YOU