Page 1 Instructional Objectives After reading this chapter the student will be able to 1. Compute stresses developed in the truss members due to temperature changes. 2. Compute stresses developed in truss members due to fabrication members. 3. Compute reactions in plane truss due to temperature changes and fabrication errors. 26.1 Introduction In the last four lessons, the direct stiffness method as applied to the truss analysis was discussed. Assembly of member stiffness matrices, imposition of boundary conditions, and the problem of inclined supports were discussed. Due to the change in temperature the truss members either expand or shrink. However, in the case of statically indeterminate trusses, the length of the members is prevented from either expansion or contraction. Thus, the stresses are developed in the members due to changes in temperature. Similarly the error in fabricating truss members also produces additional stresses in the trusses. Both these effects can be easily accounted for in the stiffness analysis. 26.2 Temperature Effects and Fabrication Errors Page 2 Instructional Objectives After reading this chapter the student will be able to 1. Compute stresses developed in the truss members due to temperature changes. 2. Compute stresses developed in truss members due to fabrication members. 3. Compute reactions in plane truss due to temperature changes and fabrication errors. 26.1 Introduction In the last four lessons, the direct stiffness method as applied to the truss analysis was discussed. Assembly of member stiffness matrices, imposition of boundary conditions, and the problem of inclined supports were discussed. Due to the change in temperature the truss members either expand or shrink. However, in the case of statically indeterminate trusses, the length of the members is prevented from either expansion or contraction. Thus, the stresses are developed in the members due to changes in temperature. Similarly the error in fabricating truss members also produces additional stresses in the trusses. Both these effects can be easily accounted for in the stiffness analysis. 26.2 Temperature Effects and Fabrication Errors Consider truss member of length L, area of cross section A as shown in Fig.26.1.The change in length l ? is given by T L l ? = ? a (26.1) where a is the coefficient of thermal expansion of the material considered. If the member is not allowed to change its length (as in the case of statically indeterminate truss) the change in temperature will induce additional forces in the member. As the truss element is a one dimensional element in the local coordinate system, the thermal load can be easily calculated in global co- ordinate system by 1 () t p AE L ' =? (26.2a) 2 () t p AE L ' =- ? (26.2b) or () {} ? ? ? ? ? ? - + ? = 1 1 ' L AE p t (26.3) The equation (26.3) can also be used to calculate forces developed in the truss member in the local coordinate system due to fabrication error. L ? will be considered positive if the member is too long. The forces in the local coordinate system can be transformed to global coordinate system by using the equation, () () () () () () ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? = ? ? ? ? ? ? ? ? ? ? ? ? ? ? t t t t t t p p p p p p ' 2 ' 1 4 3 2 1 sin 0 cos 0 0 sin 0 cos ? ? ? ? (26.4a) where () ( ) t t p p 2 1 , and () ( ) t t p p 4 3 , are the forces in the global coordinate system at nodes 1 and 2 of the truss member respectively Using equation (26.3), the equation (26.4a) may be written as, 1 2 3 4 () cos () sin () cos () sin t t t t p p AE L p p ? ? ? ? ?? ?? ?? ?? ?? ? ? =? ?? ? ? - ?? ? ? ?? ? ? - ?? ?? (26.4b) Page 3 Instructional Objectives After reading this chapter the student will be able to 1. Compute stresses developed in the truss members due to temperature changes. 2. Compute stresses developed in truss members due to fabrication members. 3. Compute reactions in plane truss due to temperature changes and fabrication errors. 26.1 Introduction In the last four lessons, the direct stiffness method as applied to the truss analysis was discussed. Assembly of member stiffness matrices, imposition of boundary conditions, and the problem of inclined supports were discussed. Due to the change in temperature the truss members either expand or shrink. However, in the case of statically indeterminate trusses, the length of the members is prevented from either expansion or contraction. Thus, the stresses are developed in the members due to changes in temperature. Similarly the error in fabricating truss members also produces additional stresses in the trusses. Both these effects can be easily accounted for in the stiffness analysis. 26.2 Temperature Effects and Fabrication Errors Consider truss member of length L, area of cross section A as shown in Fig.26.1.The change in length l ? is given by T L l ? = ? a (26.1) where a is the coefficient of thermal expansion of the material considered. If the member is not allowed to change its length (as in the case of statically indeterminate truss) the change in temperature will induce additional forces in the member. As the truss element is a one dimensional element in the local coordinate system, the thermal load can be easily calculated in global co- ordinate system by 1 () t p AE L ' =? (26.2a) 2 () t p AE L ' =- ? (26.2b) or () {} ? ? ? ? ? ? - + ? = 1 1 ' L AE p t (26.3) The equation (26.3) can also be used to calculate forces developed in the truss member in the local coordinate system due to fabrication error. L ? will be considered positive if the member is too long. The forces in the local coordinate system can be transformed to global coordinate system by using the equation, () () () () () () ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? = ? ? ? ? ? ? ? ? ? ? ? ? ? ? t t t t t t p p p p p p ' 2 ' 1 4 3 2 1 sin 0 cos 0 0 sin 0 cos ? ? ? ? (26.4a) where () ( ) t t p p 2 1 , and () ( ) t t p p 4 3 , are the forces in the global coordinate system at nodes 1 and 2 of the truss member respectively Using equation (26.3), the equation (26.4a) may be written as, 1 2 3 4 () cos () sin () cos () sin t t t t p p AE L p p ? ? ? ? ?? ?? ?? ?? ?? ? ? =? ?? ? ? - ?? ? ? ?? ? ? - ?? ?? (26.4b) The force displacement equation for the entire truss may be written as, {} []{} { } t p u k p ) ( + = (26.5) where , {} p is the vector of external joint loads applied on the truss and () { } t p is the vector of joint loads developed in the truss due to change in temperature/fabrication error of one or more members. As pointed out earlier. in the truss analysis, some joint displacements are known due to boundary conditions and some joint loads are known as they are applied externally.Thus,one could partition the above equation as, [] [ ] [] [ ] { } {} ( ) () ? ? ? ? ? ? ? ? ? ? + ? ? ? ? ? ? ? ? ? ? ? ? = ? ? ? ? ? ? t u t k k u u k p p u u k k k k p p 22 21 12 11 (26.6) where subscript u is used to denote unknown quantities and subscript k is used to denote known quantities of forces and displacements. Expanding equation (26.6), {} []{} [ ] { } ( ) { } t k k u k p u k u k p + + = 12 11 (26.7a) {} []{} [ ] { } ( ) { } 21 22 uu k u t pk u k u p =+ + (26.7b) If the known displacement vector { } { } 0 = k u then using equation (26.2a) the unknown displacements can be calculated as {} [] {} ( ) { } () t k k u p p k u - = -1 11 (26.8a) If {} 0 ? k u then {} [] {} [ ] { } ( ) { } () t k k k u u p u k p k u - - = - 12 1 (26.8b) After evaluating unknown displacements, the unknown force vectors are calculated using equation (26.7b).After evaluating displacements, the member forces in the local coordinate system for each member are evaluated by, {} [][ ]{} { } t p u T k p ' + ' = ' (26.9a) or () () ? ? ? ? ? ? ? ? ? ? + ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? - - = ? ? ? ? ? ? ? ? ? ? t t p p v u v u L AE p p ' 2 ' 1 2 2 1 1 ' 2 ' 1 sin cos 0 0 0 0 sin cos 1 1 1 1 ? ? ? ? Page 4 Instructional Objectives After reading this chapter the student will be able to 1. Compute stresses developed in the truss members due to temperature changes. 2. Compute stresses developed in truss members due to fabrication members. 3. Compute reactions in plane truss due to temperature changes and fabrication errors. 26.1 Introduction In the last four lessons, the direct stiffness method as applied to the truss analysis was discussed. Assembly of member stiffness matrices, imposition of boundary conditions, and the problem of inclined supports were discussed. Due to the change in temperature the truss members either expand or shrink. However, in the case of statically indeterminate trusses, the length of the members is prevented from either expansion or contraction. Thus, the stresses are developed in the members due to changes in temperature. Similarly the error in fabricating truss members also produces additional stresses in the trusses. Both these effects can be easily accounted for in the stiffness analysis. 26.2 Temperature Effects and Fabrication Errors Consider truss member of length L, area of cross section A as shown in Fig.26.1.The change in length l ? is given by T L l ? = ? a (26.1) where a is the coefficient of thermal expansion of the material considered. If the member is not allowed to change its length (as in the case of statically indeterminate truss) the change in temperature will induce additional forces in the member. As the truss element is a one dimensional element in the local coordinate system, the thermal load can be easily calculated in global co- ordinate system by 1 () t p AE L ' =? (26.2a) 2 () t p AE L ' =- ? (26.2b) or () {} ? ? ? ? ? ? - + ? = 1 1 ' L AE p t (26.3) The equation (26.3) can also be used to calculate forces developed in the truss member in the local coordinate system due to fabrication error. L ? will be considered positive if the member is too long. The forces in the local coordinate system can be transformed to global coordinate system by using the equation, () () () () () () ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? = ? ? ? ? ? ? ? ? ? ? ? ? ? ? t t t t t t p p p p p p ' 2 ' 1 4 3 2 1 sin 0 cos 0 0 sin 0 cos ? ? ? ? (26.4a) where () ( ) t t p p 2 1 , and () ( ) t t p p 4 3 , are the forces in the global coordinate system at nodes 1 and 2 of the truss member respectively Using equation (26.3), the equation (26.4a) may be written as, 1 2 3 4 () cos () sin () cos () sin t t t t p p AE L p p ? ? ? ? ?? ?? ?? ?? ?? ? ? =? ?? ? ? - ?? ? ? ?? ? ? - ?? ?? (26.4b) The force displacement equation for the entire truss may be written as, {} []{} { } t p u k p ) ( + = (26.5) where , {} p is the vector of external joint loads applied on the truss and () { } t p is the vector of joint loads developed in the truss due to change in temperature/fabrication error of one or more members. As pointed out earlier. in the truss analysis, some joint displacements are known due to boundary conditions and some joint loads are known as they are applied externally.Thus,one could partition the above equation as, [] [ ] [] [ ] { } {} ( ) () ? ? ? ? ? ? ? ? ? ? + ? ? ? ? ? ? ? ? ? ? ? ? = ? ? ? ? ? ? t u t k k u u k p p u u k k k k p p 22 21 12 11 (26.6) where subscript u is used to denote unknown quantities and subscript k is used to denote known quantities of forces and displacements. Expanding equation (26.6), {} []{} [ ] { } ( ) { } t k k u k p u k u k p + + = 12 11 (26.7a) {} []{} [ ] { } ( ) { } 21 22 uu k u t pk u k u p =+ + (26.7b) If the known displacement vector { } { } 0 = k u then using equation (26.2a) the unknown displacements can be calculated as {} [] {} ( ) { } () t k k u p p k u - = -1 11 (26.8a) If {} 0 ? k u then {} [] {} [ ] { } ( ) { } () t k k k u u p u k p k u - - = - 12 1 (26.8b) After evaluating unknown displacements, the unknown force vectors are calculated using equation (26.7b).After evaluating displacements, the member forces in the local coordinate system for each member are evaluated by, {} [][ ]{} { } t p u T k p ' + ' = ' (26.9a) or () () ? ? ? ? ? ? ? ? ? ? + ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? - - = ? ? ? ? ? ? ? ? ? ? t t p p v u v u L AE p p ' 2 ' 1 2 2 1 1 ' 2 ' 1 sin cos 0 0 0 0 sin cos 1 1 1 1 ? ? ? ? Expanding the above equation, yields {}{} L AE v u v u L AE p ? + ? ? ? ? ? ? ? ? ? ? ? ? ? ? - - = ' 2 2 1 1 1 sin cos sin cos ? ? ? ? (26.10a) And, {}{} 1 1 2 2 2 cos sin cos sin u v AE pAEL u L v ?? ? ? ?? ?? ?? '=- - - ? ?? ?? ?? ?? (26.10b) Few problems are solved to illustrate the application of the above procedure to calculate thermal effects /fabrication errors in the truss analysis:- Example 26.1 Analyze the truss shown in Fig.26.2a, if the temperature of the member (2) is raised by C o 40 .The sectional areas of members in square centimeters are shown in the figure. Assume 2 5 / 10 2 mm N E × = and 1/ 75,000 a = per C o . Page 5 Instructional Objectives After reading this chapter the student will be able to 1. Compute stresses developed in the truss members due to temperature changes. 2. Compute stresses developed in truss members due to fabrication members. 3. Compute reactions in plane truss due to temperature changes and fabrication errors. 26.1 Introduction In the last four lessons, the direct stiffness method as applied to the truss analysis was discussed. Assembly of member stiffness matrices, imposition of boundary conditions, and the problem of inclined supports were discussed. Due to the change in temperature the truss members either expand or shrink. However, in the case of statically indeterminate trusses, the length of the members is prevented from either expansion or contraction. Thus, the stresses are developed in the members due to changes in temperature. Similarly the error in fabricating truss members also produces additional stresses in the trusses. Both these effects can be easily accounted for in the stiffness analysis. 26.2 Temperature Effects and Fabrication Errors Consider truss member of length L, area of cross section A as shown in Fig.26.1.The change in length l ? is given by T L l ? = ? a (26.1) where a is the coefficient of thermal expansion of the material considered. If the member is not allowed to change its length (as in the case of statically indeterminate truss) the change in temperature will induce additional forces in the member. As the truss element is a one dimensional element in the local coordinate system, the thermal load can be easily calculated in global co- ordinate system by 1 () t p AE L ' =? (26.2a) 2 () t p AE L ' =- ? (26.2b) or () {} ? ? ? ? ? ? - + ? = 1 1 ' L AE p t (26.3) The equation (26.3) can also be used to calculate forces developed in the truss member in the local coordinate system due to fabrication error. L ? will be considered positive if the member is too long. The forces in the local coordinate system can be transformed to global coordinate system by using the equation, () () () () () () ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? = ? ? ? ? ? ? ? ? ? ? ? ? ? ? t t t t t t p p p p p p ' 2 ' 1 4 3 2 1 sin 0 cos 0 0 sin 0 cos ? ? ? ? (26.4a) where () ( ) t t p p 2 1 , and () ( ) t t p p 4 3 , are the forces in the global coordinate system at nodes 1 and 2 of the truss member respectively Using equation (26.3), the equation (26.4a) may be written as, 1 2 3 4 () cos () sin () cos () sin t t t t p p AE L p p ? ? ? ? ?? ?? ?? ?? ?? ? ? =? ?? ? ? - ?? ? ? ?? ? ? - ?? ?? (26.4b) The force displacement equation for the entire truss may be written as, {} []{} { } t p u k p ) ( + = (26.5) where , {} p is the vector of external joint loads applied on the truss and () { } t p is the vector of joint loads developed in the truss due to change in temperature/fabrication error of one or more members. As pointed out earlier. in the truss analysis, some joint displacements are known due to boundary conditions and some joint loads are known as they are applied externally.Thus,one could partition the above equation as, [] [ ] [] [ ] { } {} ( ) () ? ? ? ? ? ? ? ? ? ? + ? ? ? ? ? ? ? ? ? ? ? ? = ? ? ? ? ? ? t u t k k u u k p p u u k k k k p p 22 21 12 11 (26.6) where subscript u is used to denote unknown quantities and subscript k is used to denote known quantities of forces and displacements. Expanding equation (26.6), {} []{} [ ] { } ( ) { } t k k u k p u k u k p + + = 12 11 (26.7a) {} []{} [ ] { } ( ) { } 21 22 uu k u t pk u k u p =+ + (26.7b) If the known displacement vector { } { } 0 = k u then using equation (26.2a) the unknown displacements can be calculated as {} [] {} ( ) { } () t k k u p p k u - = -1 11 (26.8a) If {} 0 ? k u then {} [] {} [ ] { } ( ) { } () t k k k u u p u k p k u - - = - 12 1 (26.8b) After evaluating unknown displacements, the unknown force vectors are calculated using equation (26.7b).After evaluating displacements, the member forces in the local coordinate system for each member are evaluated by, {} [][ ]{} { } t p u T k p ' + ' = ' (26.9a) or () () ? ? ? ? ? ? ? ? ? ? + ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? - - = ? ? ? ? ? ? ? ? ? ? t t p p v u v u L AE p p ' 2 ' 1 2 2 1 1 ' 2 ' 1 sin cos 0 0 0 0 sin cos 1 1 1 1 ? ? ? ? Expanding the above equation, yields {}{} L AE v u v u L AE p ? + ? ? ? ? ? ? ? ? ? ? ? ? ? ? - - = ' 2 2 1 1 1 sin cos sin cos ? ? ? ? (26.10a) And, {}{} 1 1 2 2 2 cos sin cos sin u v AE pAEL u L v ?? ? ? ?? ?? ?? '=- - - ? ?? ?? ?? ?? (26.10b) Few problems are solved to illustrate the application of the above procedure to calculate thermal effects /fabrication errors in the truss analysis:- Example 26.1 Analyze the truss shown in Fig.26.2a, if the temperature of the member (2) is raised by C o 40 .The sectional areas of members in square centimeters are shown in the figure. Assume 2 5 / 10 2 mm N E × = and 1/ 75,000 a = per C o . The numbering of joints and members are shown in Fig.26.2b. The possible global displacement degrees of freedom are also shown in the figure. Note that lower numbers are used to indicate unconstrained degrees of freedom. From the figure it is obvious that the displacements 0 8 7 6 5 4 3 = = = = = = u u u u u u due to boundary conditions. The temperature of the member (2) has been raised by C o 40 . Thus, T L L ? = ? a ( )() 3 10 2627 . 2 40 2 3 75000 1 - × = = ?L m (1) The forces in member (2) due to rise in temperature in global coordinate system can be calculated using equation (26.4b).Thus, () () () ? ? ? ? ? ? ? ? ? ? ? ? ? ? - - ? = ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? sin cos sin cos ) ( 2 1 6 5 L AE p p p p t t t t (2) For member (2), 2 4 2 10 20 20 m cm A - × = = and o 45 = ?Read More

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- The Direct Stiffness Method: Temperature Changes & Fabrication Errors in Truss Analysis - 2
- The Direct Stiffness Method: Temperature Changes & Fabrication Errors in Truss Analysis - 3
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- The Direct Stiffness Method: Beams - 2
- The Direct Stiffness Method: Beams - 3
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