odue 2 naysis of Staticay ndeterminate Structures by the atri Force ethod Version 2 E T, Kharagpur
esson 12 The Three-oment Equations- Version 2 E T, Kharagpur
nstructiona Objectives fter reading this chapter the student wi be abe to 1. Derive three-moment equations for a continuous beam with unyieding supports. 2. Write compatibiity equations of a continuous beam in terms of three moments. 3. ompute reactions in staticay indeterminate beams using three-moment equations. 4. nayse continuous beams having different moments of inertia in different spans using three-moment equations. 12.1 ntroduction Beams that have more than one span are defined as continuous beams. ontinuous beams are very common in bridge and buiding structures. Hence, one needs to anayze continuous beams subjected to transverse oads and support settements quite often in design. When beam is continuous over many supports and moment of inertia of different spans is different, the force method of anaysis becomes quite cumbersome if vertica components of reactions are taken as redundant reactions. However, the force method of anaysis coud be further simpified for this particuar case (continuous beam) by choosing the unknown bending moments at the supports as unknowns. One compatibiity equation is written at each intermediate support of a continuous beam in terms of the oads on the adjacent span and bending moment at eft, center (the support where the compatibiity equation is written) and rigid supports. Two consecutive spans of the continuous beam are considered at one time. Since the compatibiity equation is written in terms of three moments, it is known as the equation of three moments. n this manner, each span is treated individuay as a simpy supported beam with eterna oads and two end support moments. For each intermediate support, one compatibiity equation is written in terms of three moments. Thus, we get as many equations as there are unknowns. Each equation wi have ony three unknowns. t may be noted that, apeyron first proposed this method in 1857. n this esson, three moment equations are derived for unyieding supports and in the net esson the three moment equations are modified to consider support moments. 12.2 Three-moment equation continuous beam is shown in Fig.12.1a. Since, three moment equation reates moments at three successive supports to appied oading on adjacent spans, consider two adjacent spans of a continuous beam as shown in Fig.12.1b., and respectivey denote support moments at eft, center and right supports. The moments are taken to be positive when they cause tension at Version 2 E T, Kharagpur
bottom fibers. The moment of inertia is taken to be different for different spans. n the present case and denote respectivey moment of inertia of; eft and right support and and are the eft and right span respectivey. t is assumed that supports are unyieding. The yieding of supports coud be easiy incorporated in three-moment equation, which wi be discussed in the net esson. Now it is required to derive a reation between, and. This reationship is derived from the fact that the tangent to the eastic curve at is horizonta. n other words the joint may be considered rigid. Thus, the compatibiity equation is written as, θ θ 0 (12.1) The rotation eft of the support, θ and rotation right of the support, θ may be cacuated from moment area method. Now, θ ' Defection of from tangent drawn at () oment of diagram between and about 1 1 1 1 2 2 3 2 3 θ (12.2) 3 Note that the actua moment diagram on span is broken into two parts (1) due to oads appied on span when it is considered as a simpy supported beam and, (2) due to support moments. n the above equation and denote respectivey area of the bending moment diagrams due to appied oads on eft and right supports. and denote their respective.g.(center of gravity) distances from the eft and right support respectivey. Simiary, defection of from tangent drawn at (') θ oment of diagram between and about Version 2 E T, Kharagpur
3 θ (12.3) Substituting the vaues of θ and θ in the compatibiity equation (12.1), 0 3 3 (12.4) which coud be simpified to, 2 (12.5) The above equation (12.5) is known as the three-moment equation. t reates three support moments, and with the appied oading on two adjacent spans. f in a span there are more than one type of oading (for eampe, uniformy distributed oad and a concentrated oad) then it is simper to cacuate moment diagram separatey for each of oading and then to obtain moment diagram. Version 2 E T, Kharagpur
Version 2 E T, Kharagpur
12.3 ternate derivation The above three moment equations may aso be derived by direct appication of force method as foows. Now choose, and the, the three support moments at eft, centre and right supports respectivey as the redundant moments. The primary determinate structure is obtained by reeasing the constraint corresponding to redundant moments. n this particuar case, inserting hinges at, and, the primary structure is obtained as beow (see Fig. 12.2) et dispacement (in the primary case rotations) corresponding to rotation Δ, which is the sum of rotations θ and θ. Thus, be Δ θ θ (12.) t is observed that the rotations θ and θ are caused due to ony appied oading as shown in Fig.12.2.This can be easiy evauated by moment area method as shown previousy. Δ (12.7) n the net step, appy unit vaue of redundant moments at, and cacuate rotation at (i.e. feibiity coefficients). and a a a 21 22 (12.8) 3 3 23 Version 2 E T, Kharagpur
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n the actua structure the reative rotation of both sides is zero. n other words the compatibiity equation is written as, 0 23 22 21 Δ a a a (12.9) Substituting the vaues of feibiity coefficients and Δ in the above equation, 0 3 3 Or, 2 (12.10) when moment of inertia remains constant i.e.,the above equation simpifies to, ( ) { } ( ) 2 (12.11) Eampe 12.1 continuous beam BD is carrying a uniformy distributed oad of 1 kn/m over span in addition to concentrated oads as shown in Fig.12.4a. acuate support reactions. so, draw bending moment and shear force diagram. ssume B to be constant for a members. Version 2 E T, Kharagpur
From inspection, it is assumed that the support moments at is zero and support moment at, 15 kn.m (negative because it causes compression at bottom at ) needs to be evauated. ppying three- Hence, ony one redundant moment moment equation to span B, { 10 10} ( 10) B 2 (1) The bending moment diagrams for each span due to appied uniformy distributed and concentrated oad are shown in Fig.12.4b. Version 2 E T, Kharagpur
Equation (1) may be written as, 40 B 83.33 5 125 5 83.33 5 150 10 10 10 Thus, 18.125 kn.m B fter determining the redundant moment, the reactions are evauated by equations of static equiibrium. The reactions are shown in Fig.12.4c aong with the eterna oad and support bending moment. Version 2 E T, Kharagpur
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n span B, can be cacuated by the condition that 0. Thus, 10 10 5 10 5 18.125 0 B 8.1875 kn ( ) ( ) 11.8125 kn B Simiary from span B, 4.7125 kn ( ) ( ) 5.3125 kn B The shear force and bending moment diagrams are shown in Fig.12.4d. Eampe 12.2 continuous beam B is carrying uniformy distributed oad of 2 kn/m as shown in Fig.12.5a.The moment of inertia of span B is twice that of span B. Evauate reactions and draw bending moment and shear force diagrams. Version 2 E T, Kharagpur
By inspection it is seen that the moment at support is zero. The support moment at and B needs to be evauated.for moment at B, the compatibiity Version 2 E T, Kharagpur
equation is written by noting that the tangent to the eastic curve at B is horizonta.the compatibiity condition corresponding to redundant moment at is written as foows. onsider span B as shown in Fig.12.5b. The sope at, θ may be cacuated from moment-area method. Thus, ( ) B θ (1) 3 Now, compatibiity equation is, θ 0 (2) t is observed that the tangent to eastic curve at remains horizonta. This can aso be achieved as foows. ssume an imaginary span of ength eft of support having a very high moment of inertia (see Fig. 12.5c). s the imaginary span has very high moment of inertia, it does not yied any imaginary span has very high moment of inertia it does not yied any diagram and hence no eastic curve. Hence, the tangent at to eastic curve remains horizonta. Now, consider the span B, appying three-moment equation to support, 2 ' 10 2 B 10 2 2(10) (3) The above equation is the same as the equation (2). The simpy supported bending moment diagram is shown in Fig.12.5d. Thus, equation (3) may be written as, 20 B (1.7) 5 (10) 10 Version 2 E T, Kharagpur
20 10 500 (4) B Now, consider span B, writing three moment equation for support B, 10 2 2 B 10 2 5 1.7 5 20.837 2.5 2 (10) (5) 5 20 Soving equation (4) and (5), B 250 2.5 312.5 (5) B.25 kn.m 37.5 kn.m The remaining reactions are cacuated by equiibrium equations (see Fig.12.5e) Version 2 E T, Kharagpur
n span B, B 0 10 37.5 2 10 5.25 0 Version 2 E T, Kharagpur
13.125 kn ( ) Simiary from span B, ( ).875 kn B 3.75 kn ( ) ( ).25 kn B The shear force and bending moment diagrams are shown in Fig. 12.5f. Summary n this esson the continuous beam with unyieding supports is anaysed by threemoment equations. The three-moment equations are derived for the case of a continuous beam having different moment of inertia in different spans. The threemoment equations aso beong to force method of anaysis and in this case, redundants are aways taken as support moments. Hence, compatibiity equations are derived in terms of three support moments. Few probems are soved to iustrate the procedure. Version 2 E T, Kharagpur