The point of inflection of the the moment diagram (slope = 0 = flat) is where the shear force = 0.it increases linearly from 0 to L/2 with a maximum value equivalent to the area of the shear diagram. Just like the evenly distributed load affected the right side of the shear/moment diagram, we can use the same concept on the moment diagram. The first half of the moment diagram is a slam dunk.We find the slope of the shear by subtracting Ra from Rb and dividing by the acting length (L/2) At L/2 we know the shear force must change linearly as the evenly distributed load acts form L/2 to L. Now that we have Ra, we know the shear force from support A to L/2.To find the reactions, we place a statically equivalent load in the middle and sum the forces and moments about on of the support locations and calculate Ra and Rb.Both ends are pinned so there can be no moment at either end. The simply supported beam above is subjected to a evenly distributed load from L/2 to L.The other two reactions are calculated by static equilibrium equations (videFig. Hence, the compatibility equation may be written as In the actual structure, the deflection at B is zero. Now, deflection at B in the primary structure due to redundant R B is, In thenextstep, apply a unit load at B in the direction ofĬalculate the deflection at B of the following structure. This is accomplished by unit load method. Now, compute the deflection at B, in the released structure due to uniformly distributed load and concentrated load.
The primary structure is a simply supported beam as shown in Fig.1.11. Choose the reaction at B, R By as the redundant. It is observed that the continuous beam is statically indeterminate to first degree. Assume EI to be constant for all members. The negative sign indicates that ( L )is downwards and rotation( is 1 L2) clockwise.Ī continuous beam ABC is carrying a uniformly distributed load of 1 kN/m in addition to a concentrated load of 10kN as shown in Fig.7.5a, Draw bending moment and shear force diagram.
The deflection(? L 1 )and(? L 2 )of the released structure can be evaluated from unit load method. The positive directions of the selected redundant are shown in Fig.8.3b. Let ( L ) be the transverse deflection at1 B and( L 2 bethe slope at B due to external loading. Now, calculate deflection at B duetoonly applied loading. The R1 is assumed to positive in the upward direction and R2 is assumed to be positive in the counterclockwise direction. The primary structure in this case is acantilever beam which could be obtained by releasing the redundant R1 andR2. Select vertical reaction (R1)and the support moment(R2) at B as the redundant. Draw Shear force and bending moment diagrams by force method.įig 1.3 Fixed Beam with R1 and R2 as Redundant Using equations of static equilibrium, R3 = 0.771 KN m and R4 = ?0.755 KN mĪ Fixed beam AB of constant flexural rigidity is shown in Fig.1.3 The beam is subjected to auniform distributed load of w moment M= wL 2 kN.m. Substituting the value of E and I in the above equation, Thus theĬompatibility conditions for the problem may be written as, a11 R1+ a12 R2 + (? L) 1 = 0 In the actual problem the displacements at B and Care zero. Thusįor the present problem the flexibility matrix is, In the pre sent case, the deflections (? L)1 and (? L) 2 of the released structure at B and C can be readily calculated by moment-area method. The primary structure with a given loading is shown in Fig. In this case the primary structure is a cantilever beam AC. Select two reactions vise, at B(R1 ) and C(R2 ) as redundant, since the given beam is statically indeterminate to second degree. Calculate the support reactions in the continuous beam ABC due to loading as shown in Fig.1.1 Assume EI to be constant throughout.
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