Lecture 9 - Chapter 13(1)

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Oct 30, 2023

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In Chapter 12 the second-order and fourth-order ordinary differential equations that govern the deflection of linearly elastic beams were derived. They were used to solve for the deflection and slop of statically determinate beams. In this chapter, we will extend the solution procedure to treat statically indeterminate beam deflection problems. Since the beam is statically indeterminate, the bending moment can not be obtained based on the equilibrium equations only. Therefore, we cannot use the second-ordinary equation method directly. To study the statically indeterminate beam, we can either use the fourth-order differential equation method, or the method of supervision. 1 Chapter 13 Deflection of statically indeterminate beams 13.1 Introduction

F L /2 L /2 v x Example 1 : Establish the deflection curve and find the maximum deflection and bending moment for a fixed-fixed beam with a concentrated load applied at the midpoint. Solution Because of symmetry, we only consider the left half of the beam (0  x  L ). Note that the distribution force q is zero, we can directly use Eq. (12.7) of Section 12.6. We need four Boundary Conditions (BCs) to obtain the solution. For the problem under consideration, the deflection and slope at the fixed end ( x = 0) are zero. The slope at the midpoint of the beam (i.e., at x = L / 2) is also zero. Each part of the beam (the left and right) carries half of the transverse shear force F . Therefore, the shear force at the midpoint of the beam is equal to negative F / 2 . (12.7) D Cx Bx Ax v     2 3 2 13.2 Use the fourth-order method

Mathematically, the boundary conditions (for the left half region) of the beam can be stated as: At x = L /2: d v  d x = 0 and V =  F / 2 (b) At x = 0: v = 0 and d v  d x = 0 (a) 3 EI F A 12   EI FL B 6 1  C = 0, D = 0 Substituting these constants into Eq. (12.7), we get the deflection curve of the beam as (d)   2 , 3 4 48 2 3 L x Lx x EI F v     EI M x v  2 2 d d (12.1) V x v EI  3 3 d d (12.2) Note that the shear force V is related to the deflection via Eq. (12.2). Therefore, Eq. (b) is equivalent to 2 d d 3 3 F x v EI   At x = L /2: d v  d x = 0 and (c) Applying Eqs. (a) and (c) to Eq. (12.7) yields: F L /2 L /2 v x

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By direct observation, the maximum deflection is at the midpoint of the beam. It is EI FL v L x 192 3 5 . 0 0     In order to obtain the internal bending moment, we use the moment-curvature equation , Eq. (12.1). With the substitution of Eq. (d), the moment is obtained as (e)   2 , 4 8 L x L x F M     With Eq. (e), we can draw a bending moment diagram. However we do not need to know the moments at everywhere of the beam since (for this problem) the moment is maximum either at the fixed ends or at the midpoint of the beam. At the fixed end ( x = 0), M = FL /8. At the midpoint ( x = L /2), M =  FL /2. 4 EI M x v  2 2 d d (12.1) V x v EI  3 3 d d (12.2) F L /2 L /2 v x (f) 2 , 2 L x F M    In order to obtain the internal transverse shear force, we can use Eq. (12.2) or the relationship between the shear force and the moment, Eq. (8.2). This yields

Example 2 : Establish the deflection curve, the internal bending moment and the transverse shear force for a fixed- fixed beam with a couple applied at the midpoint. What is the maximum deflection ? Solution The beam is symmetric about the midpoint. Again we only consider the left half of the beam (0  x  L ). Note that the distribution force q is zero, we can directly use Eq. (12.7) of Section 12.6. For the problem under consideration, the boundary conditions are different from Example 1. • The deflection and slope at the fixed end ( x = 0) are zero. These are same as example 1. • If we cut the beam at the midpoint, we see each part of the beam carries half of the applied couple M 0 . Therefore, the internal moment at the midpoint of the beam is equal to M 0 / 2. This gives another boundary condition. • Due to the action of the couple, the left part of the beam deflects upward, the right part of the beam deflects downward. Due to compatibility, the deflection at the midpoint of the beam must be zero. M 0 v x L /2 L /2 M 0 /2 L /2 M 0 /2 L /2 (b) (a) 5

In summary, the boundary conditions are: M 0 v x L /2 L /2 (a) At x = L / 2: v = 0 and M = M 0 / 2 (b) At x = 0: v = 0 and d v  d x = 0 (a) We can now apply the BCs (a) and (c) to Eq. (12.7) to obtain the constants A , B , C and D: (12.7) D Cx Bx Ax v     2 3 The procedure is quite straightforward, as we did in previous examples. We get 0 , 0 , 8 , 4 0 0      D C M B L M A Accordingly, the deflection curve is obtained as (d) 2 , 2 8 2 3 0 L x x L x EI M v            6 EI M x v  2 2 d d (12.1) V x v EI  3 3 d d (12.2) Note that the bending moment M is related to the deflection via Eq. (12.1). Therefore, Eq. (b) is equivalent to 2 d d 0 2 2 M x v EI  At x = L /2: v = 0 and (c)

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M 0 v x L /2 L /2 (a) In order to obtain the internal transverse shear force, we make use Eq. (12.2) or the relationship between the shear force and the moment, Eq. (8.2) With the substitution of Eq. (e), the shear force is obtained as (f) 2 , 1 6 4 0 L x L x M M          V x M  d d (8.2) In order to obtain the internal bending moment, we use the moment-curvature equation , Eq. (12.1). With the substitution of the deflection curve, Eq. (d), the moment is obtained as (e) 2 , 2 3 0 L x L M V   We see that the transverse force in the beam is constant but NOT zero. 7 EI M x v  2 2 d d (12.1) V x v EI  3 3 d d (12.2)

M 0 v x L /2 L /2 (a) In order to find the maximum deflection, we first use the following condition to find the location of it. From Eq. (d) and above condition, we get: 0 d d  x v The maximum deflection is obtained from Eq. (d) as: EI L M v L x 216 2 0 3 / max      3 / L x            x L x EI M x v 2 6 8 d d 2 0 0 2 6 2   x L x We see the left part of the beam deflects downwards. By anti- symmetry, the right part of the beam deflect upwards. 8

Example 3 : Determine the deflection at the midpoint of the beam shown. Solution : This is the same problem of example 1. Due to symmetry, we only consider the left half of the beam. We cut the beam at the midpoint as shown. The redundant couple M 0 is due to the action of the right part of the beam to the left part. It is unknown. We therefore need a condition to determine it (boundary condition or continuity condition). For this problem, we know that the slope (the rotation) of the beam at the midpoint (i.e., at x = L / 2) is zero. F L /2 L /2 v x F /2 L /2 v x = L /2 v x M 0 + (a) (b) EI FL a 16 2   The deflection of the original beam is the combination of beam (a) and beam (b). From the table of textbook or example 7 of Section 12.6, the rotation of the free end of beam (a) is EI L M b 2 0   From the table of the textbook or example 9 of Section 12.6, the rotation of the free end of beam (b) is (a) (b) The resultant rotation of the midpoint of the beam is (a) + (b). To give: EI L M EI FL 2 16 0 2    9 13.3 Use the method of supervision

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As described above, the rotation (the slope) at the midpoint of the beam must be zero. Therefore, 0 2 16 0 2   EI L M EI FL This yields the moment: 8 0 FL M   Finally, the deflection at the midpoint of the beam can be obtained, again, by adding the results of the table of the textbook or examples 7 and 9 of Section 12.6: EI FL EI L M EI L F 192 2 ) 2 / )( ( 3 ) 2 / )( 2 / ( 3 2 0 3     The negative sign means that the bending moment is opposite to the direction shown in Fig. (b) . The results by superposition are same as those obtained by solving the differential equation directly. 10 F L /2 L /2 v x F /2 L /2 v x = L /2 v x M 0 + (a) (b)

The superposition procedure allows us to get solutions for even more complex structures. Example 4 : Horizontal beam ABC and vertical rod BD are joined at B through a pin connection. A vertical force P (downward) is applied at B . Find the deflection of point B. Assume the flexural rigidity of the beam is EI and the axial rigidity of the rod is EA. A B C Beam Rod D 2 L 2 L H P P 2  P 1  Solution : The vertical force P is hold by the beam and the rod. Assume the beam holds P 1 and rod holds P 2 . 11

P 2  P 1  The equilibrium equation is:      3 1 3 1 48 48 L EI P EI L P P P P   2 1 We need an additional equation to determine two unknowns. We know that the deflection of the beam at B and the shortening of the rod are equal (the compatibility condition). For the beam, the deflection has been obtained in Example 3 of Section 12.6. Alternatively, it can be found from the table. At the midpoint of the beam, the deflection is: For the rod: (b)      H EA P EA H P 2 2 (c) (a) Putting (b) and (c) into (a) and solving H EA L EI P    3 48 (d) Once the deflection is obtained, the force P 1 in the beam and the axial force P 2 in the rod can be obtained from (b) and (c). 12

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