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For the frame shown below, q=1 kN/m and L=9 m. (1) Analyse the frame assuming beam = 21 and Icolumn = 1. (2) Determine the deformed shape and BMD of the frame. In BMD, determine (a) relevant values at the "strategic" points; (b) magnitude(s) and location(s) of the local maximum or minimum moment(s); and (c) location of contraflexural point(s). (Hint: make use of the principle of superposition.). 12 kN q q D 30 m 30 m 20 m mmm. B TIIM A m Part a) Task 1: B B TIIM L 6 kN, 6 kN ¡H 6 kN L H 2L 6 KN C 'H C ! L A F TTTM TTM E F 777M Symmetric frame: The boundary condition at H is: O Free end O Hinge O Sliding roller Determine the bending stiffness: Flexural stiffness: KBA = EI, KBE= End rotational stiffness: SBA= KBA= EI, KBH EI, SBE = D D Hmm EI (Symmetric frame) (Antisymmetric frame) EI, SBH = KBE= KBH = EI

Symmetric frame: Determine the distribution factor (DF) (absolute value; please present with four decimal places): DFBA= , DFBE = , DFBH = Determine the Fixed End Moment (FEM) (kNm) (assuming anti-clockwise positive; please present with two decimal places): FEMAB = , FEMBA= , FEMBE = , FEMBH = , FEMHB = , FEMEB = Part a) Task 3: Symmetric frame: Fill in the table below (actual values; please DO NOT present the values in fraction): Joint A B Member-end AB BA BE DF (fill in with negative values) Fixed End Moment (FEM) (kNm) Distribution (kNm) CO factor = Carry-over moment (kNm) Sum of member-end moment (kNm) (fill in with actual values, anti-clockwise positive) O (- O $ N/A → ↑ O N/A CO factor → -> N/A BH CO factor O ↓ 0 N/A OT H HB O N/A E EB 1-º -> N/A Submit part Unanswered

Part a) Task 4: Antisymmetric frame: The boundary condition at H is: O Free end O Hinge O Sliding roller Determine the bending stiffness: Flexural stiffness: KBA El, KBE= End rotational stiffness: SBA= EI, KBH- Part a) Task 5: El KBH = KBA= El, SBE = KBE= EI, SBH = Antisymmetric frame: Determine the distribution factor (DF) (absolute value; please present with four decimal places): DFBA= , DF BE = , DFBH = Determine the Fixed End Moment (FEM) (kNm) (assuming anti-clockwise positive; please present with two decimal places): FEMAB = , FEMBA = , FEMBE = , FEMBH = , FEMHB = , FEMEB = ΕΙ Submit part Unanswered Submit part Unanswered

Part a) Task 6: Antisymmetric frame: Fill in the table below (actual values; please DO NOT present the values in fraction): Joint A B Member-end AB BA BE DF (fill in with negative values) Fixed End Moment (FEM) (kNm) Distribution (kNm) CO factor Carry-over moment (kNm) Sum of member-end moment (kNm) (fill in with actual values, anti-clockwise positive) O J 10 O N/A T ↑ O N/A CO factor 1- -→-> O N/A BH 10- CO factor ↓ O N/A OT H HB N/A E EB -→-) O N/A

Part a) Task 7: Determine all the member end moments for the original frame combining symmetric and antisymmetric cases (fill in with absolute values): Member end moment at A: MAB kNm O N/A O O Member end moment at B: MBA kNm O O N/A O Member end moment at B: MBC= kNm O N/A O Member end moment at B: MBE= kNm Member end moment at E: MEB= kNm O O N/A Member end moment at C: MCB= kNm O O N/A O Member end moment at C: McD kNm O N/A O (- O Member end moment at D: Moc kNm O Member end moment at C: MCF kNm kNm Member end moment at F: MFC= O O N/A For column BE, the location of the contraflexure point (CP) from E: For column CF, the location of the contraflexure point (CP) from F: O N/A O N/A O N/A m m