. Proceedings of the annual convention . ut Joint Confer Ja AlgebraicSummations Constant I Mt = ZM2JA M2eAB = KAg(-2e2KA~ ^ZKb) MZOAC = KAc(-2e2KA) ^ZeAD = *^AD ( 2QzKA~ ^ZKp) 9 = IM2JA-Z2KA(e2KA)-J:(KAie2i) Variable I Me = IMgJA MzSAB ~ ^AB (~CaB®ZKA~.^ ^ZKB J MeeAC = Kac (~ CacSzka) ^ZeAD - *^A0 (~ CApQkKA ~ C QgKD ) e= zmzJa-ikaCaOska)-i:(KAiCe2i) (C) Joint Equation for IM2eA~ ^ FIGURE 7 22 Iron and Steel Structures RIGID FRAME ANALYSIS Frame of No Side Sway STEP 2(b) Moments from Additional Rotations Continued _ IM2JA ) 1 2lKi 2IK/ ^^^S 2IKb ~ 2EKb RelativeValues A -A 2KB
. Proceedings of the annual convention . ut Joint Confer Ja AlgebraicSummations Constant I Mt = ZM2JA M2eAB = KAg(-2e2KA~ ^ZKb) MZOAC = KAc(-2e2KA) ^ZeAD = *^AD ( 2QzKA~ ^ZKp) 9 = IM2JA-Z2KA(e2KA)-J:(KAie2i) Variable I Me = IMgJA MzSAB ~ ^AB (~CaB®ZKA~.^ ^ZKB J MeeAC = Kac (~ CacSzka) ^ZeAD - *^A0 (~ CApQkKA ~ C QgKD ) e= zmzJa-ikaCaOska)-i:(KAiCe2i) (C) Joint Equation for IM2eA~ ^ FIGURE 7 22 Iron and Steel Structures RIGID FRAME ANALYSIS Frame of No Side Sway STEP 2(b) Moments from Additional Rotations Continued _ IM2JA ) 1 2lKi 2IK/ ^^^S 2IKb ~ 2EKb RelativeValues A -A 2KB - ZM2JB ECKgjC Ogi (d) ColcuJotion of 62K by Approximations ^^-^rrTA Mz^AD ..-.^rTTX \:/j^^^^^ M26AC \ M26AB VadAB = M2eAB+ MaGsA Constant I - ^AB [ 292KA~ ®2KB ^2e&A — •^abL~®2ka~ ^92kb] (•) Momont Incrcmtnts Mee VarlobI* I ^2eAB~ *^ab LCab02KA*C92kbJ*^2«BA~ *^AB L~C®2KA~Cba92KbJ FIGURE 7 Completed Equivalent Joint Method of Rigid Frame Analysis 23 SECONDARY STRESS ANALYSIS Frame of Figure 4 STEP I Conventional Solution. 3IK^f 2IK True Value Constant IM|KAB=Ka8 [3^^AB-2e,A-eiB} Mikac=Kac J3^^AC-2eiA-6ic) I (KAiCeii) True Volue M|KAD = KaD J3^^AD-2eiA-eiDJ 0= IKa3^ A- ZIKAOiAi-KKAiei) MiKAB = Kab HCab* C)-^ ab - Cab Gia- COib ]M|KAC= Kac [(Cac + C)^^ac -Cac 6,a- Ceic}MiKAD= Kao 1{Cad + C) —ao - Cad 9rA - Ceiol ,.>A o=iKA(CA+c)g^A-iKACA(eiA)-i(KAiCe,i) Joint Equotion for EMa= 0(o) Deflection Diogrom and 9| ComputationsM,da M, ^^^^^^ ^ ^^^^^ --^^^ M|ABM,AB = 2E^AB |3-^ AB-2 0IA- eie) \J M|AB=2E^ABJ(CAB+C)qAB-CAbeiA-CeiB} MlBA 2e{-AB {3AAB-Q,^-2e,B) \ M,BA=2EtABJ(C+CBA)f;j*«-CeiA-CBAe,B} (b) Moment Diagrom M„ Note: Conventional Solution (Step I) gives moments ot faces of supports. FIGURE 8 24 Iron and Steel Structures SECONDARY STRESS ANALYSIS Frome of Figure 4 STEP 2(a) Moments from Displocements at Feces of Supports Given: — for eoch member Required: MoA for eoch member6,, A, for eacti joint
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