Module Anlysis of Stticlly Indeterminte Structures by the Mtrix Force Method Version CE IIT, Khrgpur
esson 9 The Force Method of Anlysis: Bems (Continued) Version CE IIT, Khrgpur
Instructionl Objectives After reding this chpter the student will be ble to. Clculte dditionl stresses developed in stticlly indeterminte structures due to support settlements.. Anlyse continuous bems which re supported on yielding supports. 3. Sketch the deflected shpe of the member. 4. Drw bnding moment nd sher force digrms for indeterminte bems undergoing support settlements. 9. Introduction In the lst lesson, the force method of nlysis of stticlly indeterminte bems subjected to externl lods ws discussed. It is however, ssumed in the nlysis tht the supports re unyielding nd the temperture remins constnt. In the design of indeterminte structure, it is required to mke necessry provision for future unequl verticl settlement of supports or probble rottion of supports. It my be observed here tht, in cse of determinte structures no stresses re developed due to settlement of supports. The whole structure displces s rigid body (see Fig. 9.). Hence, construction of determinte structures is esier thn indeterminte structures. Version CE IIT, Khrgpur
Version CE IIT, Khrgpur
The stticlly determinte structure chnges their shpe due to support settlement nd this would in turn induce rections nd stresses in the system. Since, there is no externl force system cting on the structures, these forces form blnced force system by themselves nd the structure would be in equilibrium. The effect of temperture chnges, support settlement cn lso be esily included in the force method of nlysis. In this lesson few problems, concerning the effect of support settlement re solved to illustrte the procedure. 9. Support Displcements Consider two spn continuous bem, which is stticlly indeterminte to second degree, s shown in Fig. 9.. Assume the flexurl rigidity of this bem to be constnt throughout. In this exmple, the support B is ssumed to hve settled by n mount s shown in the figure. Δ b This problem ws solved in the lst lesson, when there ws no support settlement (vide section 8.). In section 8., choosing rection t B nd C s the redundnt, the totl deflection of the primry structure due to pplied externl loding nd redundnt nd R is written s, R ( Δ ) + R + Δ (9.) R ( Δ ) + R + Δ (9.b) R ( ) wherein, R nd R re the redundnts t B nd C respectively, nd Δ, nd ( Δ ) re the deflections of the primry structure t B nd C due to pplied loding. In the present cse, the support B settles by n mount Δ b in the direction of the redundnt R. This support movement cn be redily incorported in the force method of nlysis. From the physics of the problem the totl deflection t the support my be equl to the given mount of support movement. Hence, the comptibility condition my be written s, Δ Δ b (9.) Δ 0 (9.b) It must be noted tht, the support settlement Δ b must be negtive s it is displces downwrds. It is ssumed tht deflections nd rections re positive in the upwrd direction. The eqution (9.) nd (9.b) my be written in compct form s, Version CE IIT, Khrgpur
R R Δ Δ ( Δ ) ( Δ ) (9.3) [ A]{ R} { Δ} { ( Δ )} (9.3b) Solving the bove lgebric equtions, one could evlute redundnts due to externl loding nd support settlement. R nd R 9.3 Temperture Stresses Internl stresses re lso developed in the stticlly indeterminte structure if the free movement of the joint is prevented. For exmple, consider cntilever bem AB s shown in Fig. 9.3. Now, if the temperture of the member is incresed uniformly throughout its length, then the length of the member is incresed by n mount Δ T Δ α T (9.4) T In which, is the chnge in the length of the member due to temperture chnge, α is the coefficient of therml expnsion of the mteril nd T is the chnge in temperture. The elongtion (the chnge in the length of the member) nd increse in temperture re tken s positive. However if the end B is restrined to move s shown in Fig 9.4, then the bem deformtion is prevented. This would develop n internl xil force nd rections in the indeterminte structure. Next consider cntilever bem AB, subjected to different temperture, t the top nd T t the bottom s shown in Fig. 9.5() nd (b). If the top temperture T is higher thn the bottom bem surfce temperture T, then the bem will deform s shown by dotted lines. Consider smll element dx t distnce x from A. The deformtion of this smll element is shown in Fig. 9.5c. Due to rise in temperture T C on the top surfce, the top surfce elongtes by Δ T dx (9.5) T α Similrly due to rise in temperture T, the bottom fibers elongte by Δ T dx (9.5b) T α T Version CE IIT, Khrgpur
As the cross section of the member remins plne, the reltive ngle of rottion dθ between two cross sections t distnce dx is given by ( T T ) dx dθ α (9.6) d where, d is the depth of bem. If the end B is fixed s in Fig. 9.4, then the differentil chnge in temperture would develop support bending moment nd rections. The effect of temperture cn lso be included in the force method of nlysis quite esily. This is done s follows. Clculte the deflection corresponding to redundnt ctions seprtely due to pplied loding, due to rise in temperture (either uniform or differentil chnge in temperture) nd redundnt forces. The deflection in the primry structure due to temperture chnges is denoted by th ( Δ T ) i which denotes the deflection corresponding to i redundnt due to temperture chnge in the determinte structure. Now the comptibility eqution for stticlly indeterminte structure of order two cn be written s R R Δ Δ ( Δ ) ( Δ ) [ A]{ R} { Δ} { ( Δ )} { ( Δ )} T ( Δ ) T ( ΔT ) (9.7) wherein,{ Δ } is the vector of displcements in the primry structure corresponding to redundnt rections due to externl lods; { Δ T } is the displcements in the primry structure corresponding to redundnt rections nd Δ is the mtrix of support displcements due to temperture chnges nd { } corresponding to redundnt ctions. Eqution (9.7) cn be solved to obtin the unknown redundnts. Exmple 9. Clculte the support rections in the continuous bem ABC (see Fig. 9.6) hving constnt flexurl rigidity EI throughout, due to verticl settlement of the 4 4 support B by 5 m m s shown in the figure. E 00 GP nd I 4 0 m. Version CE IIT, Khrgpur
Version CE IIT, Khrgpur
As the given bem is stticlly indeterminte to second degree, choose rection t B ( R ) nd C ( R ) s the redundnts. In this cse the cntilever bem AC is the bsic determinte bem (primry structure). On the determinte bem only redundnt rections re cting. The first column of flexibility mtrix is evluted by first pplying unit lod long the redundnt R nd determining deflections nd respectively s shown in Fig. 9.6b. 5 3 3EI 5 3EI 5 5 65 + 5 () 3EI EI 6EI Simply by pplying the unit lod in the direction of redundnt evlute flexibility coefficients nd (see Fig. 9.6c). R, one could 65 6EI nd 000 3 EI () The comptibility condition for the problem my be written s, 3 R + R 5 0 R + R 0 (3) The redundnt rections re, R R [ A] 5 0 0 3 (4) R 3EI 000 R 7343.75 3.5 3.5 5 0 5 0 3 (5) Substituting the vlues of E nd I in the bove eqution, the redundnt rections re evluted. R 43.885 kn nd R 3.7 kn R cts downwrds nd R cts in the positive direction of the rection i.e. upwrds. The remining two rections R3 nd R4 re evluted by the equtions of equilibrium. Version CE IIT, Khrgpur
F 0 R + R + R 0 y 3 Hence R 3 30.75 kn M 0 R + 5 R + 0 R 0 A 4 Solving for R 4, R 4 8.35 kn.m (counter clockwise) The sher force nd bending moment digrms re shown in Figs. 9.6d nd 9.6e respectively. Exmple 9. Compute rections nd drw bending moment digrm for the continuous bem ABCD loded s shown in Fig. 9.7, due to following support movements. Support B, 0.005 m verticlly downwrds. Support C, 0.0m verticlly downwrds. 3 4 Assume, E 00GP; I.35 0 m. Version CE IIT, Khrgpur
The given bem is stticlly indeterminte to second degree. Select verticl rections t B ( R ) nd C ( R ) s redundnts. The primry structure in this cse is simply supported bem AD s shown in Fig. 9.7b. The deflection ( Δ ) nd ( Δ ) of the relesed structure re evluted from unit lod method. Thus, Version CE IIT, Khrgpur
3 3 45833.33 0 45833.33 0 69 9 3 EI 00 0.35 0 ( ) 0. m Δ 3 45833.33 0 Δ 69 () EI ( ) 0. m The flexibility mtrix is evluted s explined in the previous exmple, i.e. by first pplying unit lod corresponding to the redundnt R nd determining deflections nd respectively s shown in Fig. 9.7c. Thus, 444.44 EI 388.89 () EI 444.44 EI 388.89 EI In this cse the comptibility equtions my be written s, 0.69 + R + R 0.69 + R + R 0.005 0.0 (3) Solving for redundnt rections, R EI 444.44 R 469.48 388.89 388.89 0.64 444.44 0.59 (4) Substituting the vlue of E nd I in the bove eqution, R 64. 48kN nd R 40. 74 kn Version CE IIT, Khrgpur
R R 3 Both nd cts in the upwrd direction. The remining two rections R nd R 4 re evluted by the equtions of sttic equilibrium. M A Hence R 4 6.74 kn 0 4 0 R + 0 R + 30 R 5 30 5 0 F y 0 R 3+ R R R + + 4 5 30 0 Hence R 3 8.6 kn (5) The sher force nd bending moment digrms re now constructed nd re shown in Figs. 9.7e nd 9.7f respectively. Version CE IIT, Khrgpur
Summry In this lesson, the effect of support settlements on the rections nd stresses in the cse of indeterminte structures is discussed. The procedure to clculte dditionl stresses cused due to yielding of supports is explined with the help of n exmple. A formul is derived for clculting stresses due to temperture chnges in the cse of stticlly indeterminte bems. Version CE IIT, Khrgpur