Page 1 Hence it is required to make 12 assumptions to reduce the frame in to a statically determinate structure. From the deformed shape of the frame, it is observed that inflexion point (point of zero moment) occur at mid height of each column and mid point of each beam. This leads to 10 assumptions. Depending upon how the remaining two assumptions are made, we have two different methods of analysis: Portal method and cantilever method. They will be discussed in the subsequent sections. ) i ) ii 36.3.1 Portal method In this method following assumptions are made. 1) An inflexion point occurs at the mid height of each column. 2) An inflexion point occurs at the mid point of each girder. Page 2 Hence it is required to make 12 assumptions to reduce the frame in to a statically determinate structure. From the deformed shape of the frame, it is observed that inflexion point (point of zero moment) occur at mid height of each column and mid point of each beam. This leads to 10 assumptions. Depending upon how the remaining two assumptions are made, we have two different methods of analysis: Portal method and cantilever method. They will be discussed in the subsequent sections. ) i ) ii 36.3.1 Portal method In this method following assumptions are made. 1) An inflexion point occurs at the mid height of each column. 2) An inflexion point occurs at the mid point of each girder. 3) The total horizontal shear at each storey is divided between the columns of that storey such that the interior column carries twice the shear of exterior column. The last assumption is clear, if we assume that each bay is made up of a portal thus the interior column is composed of two columns (Fig. 36.6). Thus the interior column carries twice the shear of exterior column. This method is illustrated in example 36.2. Example 36.2 Analyse the frame shown in Fig. 36.7a and evaluate approximately the column end moments, beam end moments and reactions. Solution: The problem is solved by equations of statics with the help of assumptions made in the portal method. In this method we have hinges/inflexion points at mid height of columns and beams. Taking the section through column hinges we get, (ref. Fig. 36.7b). O N M , . 20 2 0 = + + ? = ? V V V F X or kN 5 = V Taking moment of all forces left of hinge R about R gives, 0 5 . 2 5 . 1 = × - × y M V ( ) ? = kN 3 y M Column and beam moments are calculates as, kN.m 5 . 7 kN.m 5 . 7 ; kN.m 5 . 7 5 . 1 5 - = + = = × = CF IH CB M M M Taking moment of all forces left of hinge about gives, S S () 0 kN 3 0 5 . 2 5 . 1 5 = ? = = × - × y y y N O O Taking a section through column hinges we get, (ref. Fig. 36.7c). L K J , , Page 3 Hence it is required to make 12 assumptions to reduce the frame in to a statically determinate structure. From the deformed shape of the frame, it is observed that inflexion point (point of zero moment) occur at mid height of each column and mid point of each beam. This leads to 10 assumptions. Depending upon how the remaining two assumptions are made, we have two different methods of analysis: Portal method and cantilever method. They will be discussed in the subsequent sections. ) i ) ii 36.3.1 Portal method In this method following assumptions are made. 1) An inflexion point occurs at the mid height of each column. 2) An inflexion point occurs at the mid point of each girder. 3) The total horizontal shear at each storey is divided between the columns of that storey such that the interior column carries twice the shear of exterior column. The last assumption is clear, if we assume that each bay is made up of a portal thus the interior column is composed of two columns (Fig. 36.6). Thus the interior column carries twice the shear of exterior column. This method is illustrated in example 36.2. Example 36.2 Analyse the frame shown in Fig. 36.7a and evaluate approximately the column end moments, beam end moments and reactions. Solution: The problem is solved by equations of statics with the help of assumptions made in the portal method. In this method we have hinges/inflexion points at mid height of columns and beams. Taking the section through column hinges we get, (ref. Fig. 36.7b). O N M , . 20 2 0 = + + ? = ? V V V F X or kN 5 = V Taking moment of all forces left of hinge R about R gives, 0 5 . 2 5 . 1 = × - × y M V ( ) ? = kN 3 y M Column and beam moments are calculates as, kN.m 5 . 7 kN.m 5 . 7 ; kN.m 5 . 7 5 . 1 5 - = + = = × = CF IH CB M M M Taking moment of all forces left of hinge about gives, S S () 0 kN 3 0 5 . 2 5 . 1 5 = ? = = × - × y y y N O O Taking a section through column hinges we get, (ref. Fig. 36.7c). L K J , , 60 ' ' 2 ' 0 = + + ? = ? V V V F X or '1 5 kN V = Page 4 Hence it is required to make 12 assumptions to reduce the frame in to a statically determinate structure. From the deformed shape of the frame, it is observed that inflexion point (point of zero moment) occur at mid height of each column and mid point of each beam. This leads to 10 assumptions. Depending upon how the remaining two assumptions are made, we have two different methods of analysis: Portal method and cantilever method. They will be discussed in the subsequent sections. ) i ) ii 36.3.1 Portal method In this method following assumptions are made. 1) An inflexion point occurs at the mid height of each column. 2) An inflexion point occurs at the mid point of each girder. 3) The total horizontal shear at each storey is divided between the columns of that storey such that the interior column carries twice the shear of exterior column. The last assumption is clear, if we assume that each bay is made up of a portal thus the interior column is composed of two columns (Fig. 36.6). Thus the interior column carries twice the shear of exterior column. This method is illustrated in example 36.2. Example 36.2 Analyse the frame shown in Fig. 36.7a and evaluate approximately the column end moments, beam end moments and reactions. Solution: The problem is solved by equations of statics with the help of assumptions made in the portal method. In this method we have hinges/inflexion points at mid height of columns and beams. Taking the section through column hinges we get, (ref. Fig. 36.7b). O N M , . 20 2 0 = + + ? = ? V V V F X or kN 5 = V Taking moment of all forces left of hinge R about R gives, 0 5 . 2 5 . 1 = × - × y M V ( ) ? = kN 3 y M Column and beam moments are calculates as, kN.m 5 . 7 kN.m 5 . 7 ; kN.m 5 . 7 5 . 1 5 - = + = = × = CF IH CB M M M Taking moment of all forces left of hinge about gives, S S () 0 kN 3 0 5 . 2 5 . 1 5 = ? = = × - × y y y N O O Taking a section through column hinges we get, (ref. Fig. 36.7c). L K J , , 60 ' ' 2 ' 0 = + + ? = ? V V V F X or '1 5 kN V = Taking moment of all forces about P gives (vide Fig. 36.7d) () () 015 1.5 5 1.5 3 2.5 2.5 0 15 kN 15 kN py y y MJ J L =× +× +× - × = =? =? ? Page 5 Hence it is required to make 12 assumptions to reduce the frame in to a statically determinate structure. From the deformed shape of the frame, it is observed that inflexion point (point of zero moment) occur at mid height of each column and mid point of each beam. This leads to 10 assumptions. Depending upon how the remaining two assumptions are made, we have two different methods of analysis: Portal method and cantilever method. They will be discussed in the subsequent sections. ) i ) ii 36.3.1 Portal method In this method following assumptions are made. 1) An inflexion point occurs at the mid height of each column. 2) An inflexion point occurs at the mid point of each girder. 3) The total horizontal shear at each storey is divided between the columns of that storey such that the interior column carries twice the shear of exterior column. The last assumption is clear, if we assume that each bay is made up of a portal thus the interior column is composed of two columns (Fig. 36.6). Thus the interior column carries twice the shear of exterior column. This method is illustrated in example 36.2. Example 36.2 Analyse the frame shown in Fig. 36.7a and evaluate approximately the column end moments, beam end moments and reactions. Solution: The problem is solved by equations of statics with the help of assumptions made in the portal method. In this method we have hinges/inflexion points at mid height of columns and beams. Taking the section through column hinges we get, (ref. Fig. 36.7b). O N M , . 20 2 0 = + + ? = ? V V V F X or kN 5 = V Taking moment of all forces left of hinge R about R gives, 0 5 . 2 5 . 1 = × - × y M V ( ) ? = kN 3 y M Column and beam moments are calculates as, kN.m 5 . 7 kN.m 5 . 7 ; kN.m 5 . 7 5 . 1 5 - = + = = × = CF IH CB M M M Taking moment of all forces left of hinge about gives, S S () 0 kN 3 0 5 . 2 5 . 1 5 = ? = = × - × y y y N O O Taking a section through column hinges we get, (ref. Fig. 36.7c). L K J , , 60 ' ' 2 ' 0 = + + ? = ? V V V F X or '1 5 kN V = Taking moment of all forces about P gives (vide Fig. 36.7d) () () 015 1.5 5 1.5 3 2.5 2.5 0 15 kN 15 kN py y y MJ J L =× +× +× - × = =? =? ? Column and beam moments are calculated as, (ref. Fig. 36.7f) kN.m 5 . 22 5 . 1 15 ; . 5 . 7 5 . 1 5 = × = = × = BA BC M m kN M kN.m 30 - = BE M kN.m 45 5 . 1 30 ; kN.m 15 5 . 1 10 = × = = × = ED EF M M kN.m 30 kN.m 30 - = - = EH EB M M kN.m 5 . 22 5 . 1 15 ; kN.m 5 . 7 5 . 1 5 = × = = × = HG HI M M kN.m 30 - = HE M Reactions at the base of the column are shown in Fig. 36.7g. 36.3.2 Cantilever method The cantilever method is suitable if the frame is tall and slender. In the cantilever method following assumptions are made. 1) An inflexion point occurs at the mid point of each girder. 2) An inflexion point occurs at mid height of each column. 3) In a storey, the intensity of axial stress in a column is proportional to its horizontal distance from the center of gravity of all the columns in that storey. Consider a cantilever beam acted by a horizontal load as shown in Fig. 36.8. In such a column the bending stress in the column cross section varies linearly from its neutral axis. The last assumption in the cantilever method is based on this fact. The method is illustrated in example 36.3. PRead More

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