# H h p l b h ei ei ei kei a p i i b ki i b 1 2 r 11 z

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hhPl=bhEIEIEIkEIaPii/bkii/b12r11Z1=1r21M1b4ib4iZ2=1r22r12M24kij2(u2)2ij3(u)2ij3(u)4ij2(u)4ij2(u)abcdFig. 13.15Two-story frameSolution.The primary system is presented in Fig.13.15b. Let us assign member1–2 as basic one; its flexural stiffness per unit length equalsiDEI=h. The flexuralstiffness per unit length for each member are shown in Fig.13.15b. Parameters ofcritical force for member 1–2 and 2-Aare determined as follows:1DhrPEID;2DhrPC˛PkEIDr1C˛k:Bending moment diagrams caused byZ1D1andZ2D1are shown in Fig.13.15c, d,respectively.Unit reactions arer11D4i'2. /C4iˇIr12Dr21D2i'3. /Ir22D4i'2. /C4ki'2.2/C4iˇ:
47813 Stability of Elastic SystemsStability equation in general and expanded forms areˇˇˇˇr11r12r21r22ˇˇˇˇD04'2. /C1ˇ'2. /Ck'2.2/C1ˇ'23. /D0:Let˛D3; ˇD1; kD4. In this case,2Dand stability equation becomes4 Œ'2. /CŒ5'2. /C'23. /D0:The root of this equation isD4:5307. The critical load isPcrD2EIh2D4:53072EIh2:Example 13.5.The frame in Fig.13.16a is loaded by two forces at the joints. De-rive the stability equation and find the critical loadP.h=10ml=5m1.4PPEIEI2EIPrimary system12i=0.4EI34i=0.1EIabr12r22M2Z2=1Z2=1M31M42M13R2R1M13=4ij2(u) = 0.4EIj2(u)M13=6ij4(u) = 0.6EIj4(u)12il2R1=h2(u) = 0.12EIh2(u)3il2R2=h1(1.1832u) = 0.003EIh1(1.1832u)R=6ij4(u) = 0.6EIj4(u)r11Z1=13i=1.2EIr21M1M13M31RRM13M31RRcdFig. 13.16(a, b) Design diagram of the frame and primary system. (c, d) Unit bending momentdiagrams
13.4 Stability of Continuous Beams and Frames479Solution.The frame has two unknowns of the displacement method. They are theangle of rotationZ1of rigid joint and horizontal displacementZ2of the cross bar.The primary system is presented in Fig.13.16b.Bending moment diagrams caused by unit displacement of the introduced con-straints are presented in Fig.13.16c, d; elastic curves are shown by dotted line. Thediagrams within members 1–3 and 2–4 are curvilinear. Direction ofRforM1di-agram is explained for the left column (Fig.13.16c). Similarly may be explaineddirections forR1andR2.M2diagram).Parameters of a critical load are13D DhrPEII24Dhr1:4PEID1:1832.Canonical equations and unit reactions arer11Z1Cr12Z2D0;r21Z1Cr22Z2D0;(a)wherer11DŒ0:4'2. /C1:2ŁEIIr21Dr12D0:6EI'4. /Ir22DŒ0:0122. /C0:0031.1:1832 /ŁEIStability equation becomes:ˇˇˇˇ0:4'2. /C1:20:6'4. /0:6'4. /0:0122. /C0:0031.1:1832 /ˇˇˇˇD0Solution of this equation leads to parameter of critical loadD2:12. The criticalload isPcrD2EIh2D4:49EIh2:Thus the frame becomes unstable if it will be loaded by two forcesPand 1.4Psimultaneously.Example 13.6.Design diagram of the multispan frame is presented in Fig.13.17a.Derive the stability equation and calculate the critical force.Solution.The primary system is shown in Fig.13.17b. The introduced constraint 1prevents linear displacement. Canonical equation isr11Z1D0, so stability equationisr11D0. Bending moment diagram caused by unit displacement of the introducedsupport is presented in Fig.13.17b. Within the second and third columns, the bend-ing moment diagrams are curvilinear.

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