Page 1 Solution: Since, the arch is symmetrical and the loading is also symmetrical, kN 40 = = by ay R R (1) Now the strain energy of the arch is given by, Page 2 Solution: Since, the arch is symmetrical and the loading is also symmetrical, kN 40 = = by ay R R (1) Now the strain energy of the arch is given by, ? ? + = s s EA ds N EI ds M U 0 2 0 2 2 2 (2) Let us choose and as redundants. Then we have, a H a M 0 = ? ? a M U and 0 = ? ? a H U (3) The bending moment at any cross section is given by, D a a ay y H M x R M ? ? = = - - = 0 (4) 2 / ) 10 ( 40 p ? ? = = - - - - = D a a ay x y H M x R M ? ? cos ) 90 cos( a a R H N + - = D a a R H N ? ? ? ? = = + = 0 cos sin (5) sin ( 40)cos aa NH R ? ? =+- /2 ? ?p = = (6) ? sin R y = ) cos 1 ( ? - =R x And ? Rd ds = See Fig 34.5. 0 ) 0 ( ) 1 ( 0 0 = + - = ? ? ? ? s s a ds EA N ds EI M M U 0 0 = ? s ds EI M Since the arch is symmetrical, integration need to be carried out between limits 2 / 0 p to and the result is multiplied by two. 0 2 2 / 0 = ? p ds EI M 0 ] 10 ) cos 1 ( [ 40 sin ) cos 1 ( 40 2 / 552 . 2 / 2 / 0 2 / 0 2 / 0 = - - - - - - ? ? ? ? ? ? ? ? ? ? ? p p p p p Rd R Rd R H Rd M Rd R a a Page 3 Solution: Since, the arch is symmetrical and the loading is also symmetrical, kN 40 = = by ay R R (1) Now the strain energy of the arch is given by, ? ? + = s s EA ds N EI ds M U 0 2 0 2 2 2 (2) Let us choose and as redundants. Then we have, a H a M 0 = ? ? a M U and 0 = ? ? a H U (3) The bending moment at any cross section is given by, D a a ay y H M x R M ? ? = = - - = 0 (4) 2 / ) 10 ( 40 p ? ? = = - - - - = D a a ay x y H M x R M ? ? cos ) 90 cos( a a R H N + - = D a a R H N ? ? ? ? = = + = 0 cos sin (5) sin ( 40)cos aa NH R ? ? =+- /2 ? ?p = = (6) ? sin R y = ) cos 1 ( ? - =R x And ? Rd ds = See Fig 34.5. 0 ) 0 ( ) 1 ( 0 0 = + - = ? ? ? ? s s a ds EA N ds EI M M U 0 0 = ? s ds EI M Since the arch is symmetrical, integration need to be carried out between limits 2 / 0 p to and the result is multiplied by two. 0 2 2 / 0 = ? p ds EI M 0 ] 10 ) cos 1 ( [ 40 sin ) cos 1 ( 40 2 / 552 . 2 / 2 / 0 2 / 0 2 / 0 = - - - - - - ? ? ? ? ? ? ? ? ? ? ? p p p p p Rd R Rd R H Rd M Rd R a a 0 92 . 135 304 . 41 571 . 1 8310 . 22 2 2 2 = + - - - R R R H R M R a a 0 92 . 135 56 . 169 15 571 . 1 477 . 342 = + - - - a a H M 0 837 . 308 15 571 . 1 = - + a a H M 0 ) (sin ) ( 0 0 = + - = ? ? ? ? s s a ds EA N ds y EI M H U ? 0 cos 40 ) (sin 1 ) (sin ) cos sin ( ]]} 10 ) cos 1 ( [ 40 ){[ sin ( 1 )} sin ( )] cos 1 ( 40 ){[ sin ( 1 2 / 552 . 2 / 2 / 0 2 / 552 . 2 / 2 / 0 = - + + - - - - - - - - ? ? ? ? p p p p p p ? ? ? ? ? ? ? ? ? ? ? ? ? ? Rd EA Rd EA R H Rd R R EI Rd R H M R R EI a a a a ? ? = - - - + - + + + + - 2 / 552 . 2 / 2 3 3 2 2 / 0 2 3 2 3 3 0 } cos sin 40 sin 400 cos sin 40 sin 40 { } cos sin ) ( sin sin sin cos sin 40 sin 40 { p p p ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? d EA R EI R EI R EI R d EA R R EA R H EI R H EI R M EI R EI R ay a a a 0 ) 0555 . 0 ( 40 ) 333 . 0 ( 400 ) 0554 . 0 ( 40 ) 333 . 0 ( 40 ) 2 1 ( 40 ) 785 . 0 ( ) 785 . 0 ( ) 1 ( ) 2 1 ( 40 ) 1 ( 40 2 2 2 = - - - + - + + + + - A R RI I I A R A R H I H IR M I I a a a 0 2 58 . 23 266 = + + - a a M H (7) Solving equations (6) and (7), and are evaluated. Thus, a H a M kN 28 . 28 = a H kN 42 . 466 - = a M (8) Page 4 Solution: Since, the arch is symmetrical and the loading is also symmetrical, kN 40 = = by ay R R (1) Now the strain energy of the arch is given by, ? ? + = s s EA ds N EI ds M U 0 2 0 2 2 2 (2) Let us choose and as redundants. Then we have, a H a M 0 = ? ? a M U and 0 = ? ? a H U (3) The bending moment at any cross section is given by, D a a ay y H M x R M ? ? = = - - = 0 (4) 2 / ) 10 ( 40 p ? ? = = - - - - = D a a ay x y H M x R M ? ? cos ) 90 cos( a a R H N + - = D a a R H N ? ? ? ? = = + = 0 cos sin (5) sin ( 40)cos aa NH R ? ? =+- /2 ? ?p = = (6) ? sin R y = ) cos 1 ( ? - =R x And ? Rd ds = See Fig 34.5. 0 ) 0 ( ) 1 ( 0 0 = + - = ? ? ? ? s s a ds EA N ds EI M M U 0 0 = ? s ds EI M Since the arch is symmetrical, integration need to be carried out between limits 2 / 0 p to and the result is multiplied by two. 0 2 2 / 0 = ? p ds EI M 0 ] 10 ) cos 1 ( [ 40 sin ) cos 1 ( 40 2 / 552 . 2 / 2 / 0 2 / 0 2 / 0 = - - - - - - ? ? ? ? ? ? ? ? ? ? ? p p p p p Rd R Rd R H Rd M Rd R a a 0 92 . 135 304 . 41 571 . 1 8310 . 22 2 2 2 = + - - - R R R H R M R a a 0 92 . 135 56 . 169 15 571 . 1 477 . 342 = + - - - a a H M 0 837 . 308 15 571 . 1 = - + a a H M 0 ) (sin ) ( 0 0 = + - = ? ? ? ? s s a ds EA N ds y EI M H U ? 0 cos 40 ) (sin 1 ) (sin ) cos sin ( ]]} 10 ) cos 1 ( [ 40 ){[ sin ( 1 )} sin ( )] cos 1 ( 40 ){[ sin ( 1 2 / 552 . 2 / 2 / 0 2 / 552 . 2 / 2 / 0 = - + + - - - - - - - - ? ? ? ? p p p p p p ? ? ? ? ? ? ? ? ? ? ? ? ? ? Rd EA Rd EA R H Rd R R EI Rd R H M R R EI a a a a ? ? = - - - + - + + + + - 2 / 552 . 2 / 2 3 3 2 2 / 0 2 3 2 3 3 0 } cos sin 40 sin 400 cos sin 40 sin 40 { } cos sin ) ( sin sin sin cos sin 40 sin 40 { p p p ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? d EA R EI R EI R EI R d EA R R EA R H EI R H EI R M EI R EI R ay a a a 0 ) 0555 . 0 ( 40 ) 333 . 0 ( 400 ) 0554 . 0 ( 40 ) 333 . 0 ( 40 ) 2 1 ( 40 ) 785 . 0 ( ) 785 . 0 ( ) 1 ( ) 2 1 ( 40 ) 1 ( 40 2 2 2 = - - - + - + + + + - A R RI I I A R A R H I H IR M I I a a a 0 2 58 . 23 266 = + + - a a M H (7) Solving equations (6) and (7), and are evaluated. Thus, a H a M kN 28 . 28 = a H kN 42 . 466 - = a M (8) Equations (34.7a) and (34.7b) are quite difficult to solve. However, they can be further simplified if the origin of co-ordinates is moved from to in Fig. 34.3. The distance d is chosen such that C O 1 ( ) y yd = - satisfies the following condition. ( ) 0 0 0 1 = - = ? ? s s ds EI d y ds EI y (34.10a) Solving which, the distance d may be computed as ? ? = s s EI ds ds EI y d 0 0 34.10b) The point is known as the elastic centre of the arch. Now equation (34.7a) can be written with respect to new origin . Towards this, substitute O O d y y + = 1 in equation (34.7a). 34.4 Elastic centre method Page 5 Solution: Since, the arch is symmetrical and the loading is also symmetrical, kN 40 = = by ay R R (1) Now the strain energy of the arch is given by, ? ? + = s s EA ds N EI ds M U 0 2 0 2 2 2 (2) Let us choose and as redundants. Then we have, a H a M 0 = ? ? a M U and 0 = ? ? a H U (3) The bending moment at any cross section is given by, D a a ay y H M x R M ? ? = = - - = 0 (4) 2 / ) 10 ( 40 p ? ? = = - - - - = D a a ay x y H M x R M ? ? cos ) 90 cos( a a R H N + - = D a a R H N ? ? ? ? = = + = 0 cos sin (5) sin ( 40)cos aa NH R ? ? =+- /2 ? ?p = = (6) ? sin R y = ) cos 1 ( ? - =R x And ? Rd ds = See Fig 34.5. 0 ) 0 ( ) 1 ( 0 0 = + - = ? ? ? ? s s a ds EA N ds EI M M U 0 0 = ? s ds EI M Since the arch is symmetrical, integration need to be carried out between limits 2 / 0 p to and the result is multiplied by two. 0 2 2 / 0 = ? p ds EI M 0 ] 10 ) cos 1 ( [ 40 sin ) cos 1 ( 40 2 / 552 . 2 / 2 / 0 2 / 0 2 / 0 = - - - - - - ? ? ? ? ? ? ? ? ? ? ? p p p p p Rd R Rd R H Rd M Rd R a a 0 92 . 135 304 . 41 571 . 1 8310 . 22 2 2 2 = + - - - R R R H R M R a a 0 92 . 135 56 . 169 15 571 . 1 477 . 342 = + - - - a a H M 0 837 . 308 15 571 . 1 = - + a a H M 0 ) (sin ) ( 0 0 = + - = ? ? ? ? s s a ds EA N ds y EI M H U ? 0 cos 40 ) (sin 1 ) (sin ) cos sin ( ]]} 10 ) cos 1 ( [ 40 ){[ sin ( 1 )} sin ( )] cos 1 ( 40 ){[ sin ( 1 2 / 552 . 2 / 2 / 0 2 / 552 . 2 / 2 / 0 = - + + - - - - - - - - ? ? ? ? p p p p p p ? ? ? ? ? ? ? ? ? ? ? ? ? ? Rd EA Rd EA R H Rd R R EI Rd R H M R R EI a a a a ? ? = - - - + - + + + + - 2 / 552 . 2 / 2 3 3 2 2 / 0 2 3 2 3 3 0 } cos sin 40 sin 400 cos sin 40 sin 40 { } cos sin ) ( sin sin sin cos sin 40 sin 40 { p p p ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? d EA R EI R EI R EI R d EA R R EA R H EI R H EI R M EI R EI R ay a a a 0 ) 0555 . 0 ( 40 ) 333 . 0 ( 400 ) 0554 . 0 ( 40 ) 333 . 0 ( 40 ) 2 1 ( 40 ) 785 . 0 ( ) 785 . 0 ( ) 1 ( ) 2 1 ( 40 ) 1 ( 40 2 2 2 = - - - + - + + + + - A R RI I I A R A R H I H IR M I I a a a 0 2 58 . 23 266 = + + - a a M H (7) Solving equations (6) and (7), and are evaluated. Thus, a H a M kN 28 . 28 = a H kN 42 . 466 - = a M (8) Equations (34.7a) and (34.7b) are quite difficult to solve. However, they can be further simplified if the origin of co-ordinates is moved from to in Fig. 34.3. The distance d is chosen such that C O 1 ( ) y yd = - satisfies the following condition. ( ) 0 0 0 1 = - = ? ? s s ds EI d y ds EI y (34.10a) Solving which, the distance d may be computed as ? ? = s s EI ds ds EI y d 0 0 34.10b) The point is known as the elastic centre of the arch. Now equation (34.7a) can be written with respect to new origin . Towards this, substitute O O d y y + = 1 in equation (34.7a). 34.4 Elastic centre method ? ? ? - = + + s s c s c ds EI M ds EI d y N EI ds M 0 0 0 1 0 ) ( (34.11) In the above equation, ? s ds EI y 0 1 is zero. Hence the above equation is rewritten as ? ? - = + s s c c EI ds ds EI M d N M 0 0 0 (34.12) Now, is the moment ) ( d N M c c + 0 ~ M at (see Fig. 34.3). Similarly the equation (34.7b) is also simplified. Thus we obtain, O ? ? - = + = s s c c EI ds ds EI M d N M M 0 0 0 0 ~ (34.13) and, ?? ?? + + - = = ss ss c ds EA ds EI y ds EA N ds EI y M N H 00 2 1 2 00 0 1 0 0 cos cos ~ ? ? (34.14) 34.4.1Temperature stresses Consider a symmetrical hinge less arch of spanL, subjected to a temperature rise of . Let elastic centre Obe the origin of co-ordinates and be the redundants. The magnitude of horizontal force be such as to counteract the increase in the span C T ° 0 0 ~ , ~ M H 0 ~ H 2 LT a due to rise in temperature Also from Symmetry, there must not be any rotation at the crown. Hence, . T ? = ? ? = = ? ? s O O ds M M EI M M U 0 0 ~ 0 (34.15) 2 ~ ~ ~ 0 0 T L ds H N EA N ds H M EI M H U s O s O O a ? ? = ? ? + ? ? = ? ? (34.16)Read More

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