Deflections
8
In this chapter we will show how to determine the elastic deflections of a beam using the method of double integration and two important geometrical methods, namely, the moment-area theorems and the conjugate-beam method. Double integration is used to obtain equations which define the slope and the elastic curve. The geometric methods provide a way to obtain the slope and deflection at specific points on the beam. Each of these methods has particular advantages or disadvantages, which will be discussed when each method is presented.
8.1 Deflection Diagrams and the Elastic Curve Deflections of structures can occur from various sources, such as loads, temperature, fabrication errors, or settlement. In design, deflections must be limited in order to provide integrity and stability of roofs, and prevent cracking of attached brittle materials such as concrete, plaster or glass. Furthermore, a structure must not vibrate or deflect severely in order to “appear” safe for its occupants. More important, though, deflections at specified points in a structure must be determined if one is to analyze statically indeterminate structures. The deflections to be considered throughout this text apply only to structures having linear elastic material response. Under this condition, a structure subjected to a load will return to its original undeformed position after the load is removed. The deflection of a structure is caused
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TABLE 8–1 (1) u 0 roller or rocker
(2)
u
0 pin
(3)
by its internal loadings such as normal force, shear force, or bending moment. For beams and frames, however, the greatest deflections are most often caused by internal bending, whereas internal axial forces cause the deflections of a truss. Before the slope or displacement of a point on a beam or frame is determined, it is often helpful to sketch the deflected shape of the structure when it is loaded in order to partially check the results. This deflection diagram represents the elastic curve or locus of points which defines the displaced position of the centroid of the cross section along the members. For most problems the elastic curve can be sketched without much difficulty. When doing so, however, it is necessary to know the restrictions as to slope or displacement that often occur at a support or a connection. With reference to Table 8–1, supports that resist a force, such as a pin, restrict displacement; and those that resist moment, such as a fixed wall, restrict rotation. Note also that deflection of frame members that are fixed connected (4) causes the joint to rotate the connected members by the same amount u. On the other hand, if a pin connection is used at the joint, the members will each have a different slope or rotation at the pin, since the pin cannot support a moment (5).
0 u0 fixed support
(4)
u u
fixed-connected joint
8
(5) u2
u1
pin-connected joint
The two-member frames support both the dead load of the roof and a live snow loading. The frame can be considered pinned at the wall, fixed at the ground, and having a fixed-connected joint.
8.1
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DEFLECTION DIAGRAMS AND THE ELASTIC CURVE
If the elastic curve seems difficult to establish, it is suggested that the moment diagram for the beam or frame be drawn first. By our sign convention for moments established in Chapter 4, a positive moment tends to bend a beam or horizontal member concave upward, Fig. 8–1. Likewise, a negative moment tends to bend the beam or member concave downward, Fig. 8–2. Therefore, if the shape of the moment diagram is known, it will be easy to construct the elastic curve and vice versa. For example, consider the beam in Fig. 8–3 with its associated moment diagram. Due to the pin-and-roller support, the displacement at A and D must be zero. Within the region of negative moment, the elastic curve is concave downward; and within the region of positive moment, the elastic curve is concave upward. In particular, there must be an inflection point at the point where the curve changes from concave down to concave up, since this is a point of zero moment. Using these same principles, note how the elastic curve for the beam in Fig. 8–4 was drawn based on its moment diagram. In particular, realize that the positive moment reaction from the wall keeps the initial slope of the beam horizontal.
M
M
positive moment, concave upward
Fig. 8–1
M
M
negative moment, concave downward
Fig. 8–2
P1 A
B
C
P1
D
P2
P2 beam
beam M
M
x
x
moment diagram
moment diagram
inflection point
M
inflection point M
M deflection curve
Fig. 8–3
M
deflection curve
Fig. 8–4
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EXAMPLE 8.1 Draw the deflected shape of each of the beams shown in Fig. 8–5. SOLUTION In Fig. 8–5a the roller at A allows free rotation with no deflection while the fixed wall at B prevents both rotation and deflection. The deflected shape is shown by the bold line. In Fig. 8–5b, no rotation or deflection can occur at A and B. In Fig. 8–5c, the couple moment will rotate end A. This will cause deflections at both ends of the beam since no deflection is possible at B and C. Notice that segment CD remains undeformed (a straight line) since no internal load acts within it. In Fig. 8–5d, the pin (internal hinge) at B allows free rotation, and so the slope of the deflection curve will suddenly change at this point while the beam is constrained by its supports. In Fig. 8–5e, the compound beam deflects as shown. The slope abruptly changes on each side of the pin at B. Finally, in Fig. 8–5f, span BC will deflect concave upwards due to the load. Since the beam is continuous, the end spans will deflect concave downwards.
P
w B
A
B
A
(b)
(a)
P
8
A
D
M B
B
D
A
C
C
(d)
(c)
P w B
C
A
A
D C
B
(f)
(e)
Fig. 8–5
8.1
DEFLECTION DIAGRAMS AND THE ELASTIC CURVE
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EXAMPLE 8.2 Draw the deflected shapes of each of the frames shown in Fig. 8–6. B
C
A
D
P
(a)
P
B
C
D
A
F
E w
(b) B
C
D
E
A
H
G
F
SOLUTION In Fig. 8–6a, when the load P pushes joints B and C to the right, it will cause clockwise rotation of each column as shown. As a result, joints B and C must rotate clockwise. Since the 90° angle between the connected members must be maintained at these joints, the beam BC will deform so that its curvature is reversed from concave up on the left to concave down on the right. Note that this produces a point of inflection within the beam. In Fig. 8–6b, P displaces joints B, C, and D to the right, causing each column to bend as shown. The fixed joints must maintain their 90° angles, and so BC and CD must have a reversed curvature with an inflection point near their midpoint. In Fig. 8–6c, the vertical loading on this symmetric frame will bend beam CD concave upwards, causing clockwise rotation of joint C and counterclockwise rotation of joint D. Since the 90° angle at the joints must be maintained, the columns bend as shown. This causes spans BC and DE to be concave downwards, resulting in counterclockwise rotation at B and clockwise rotation at E. The columns therefore bend as shown. Finally, in Fig. 8–6d, the loads push joints B and C to the right, which bends the columns as shown. The fixed joint B maintains its 90° angle; however, no restriction on the relative rotation between the members at C is possible since the joint is a pin. Consequently, only beam CD does not have a reverse curvature.
(c)
P
C
P B
D
E
G
F
A
I (d)
Fig. 8–6
H
8
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DEFLECTIONS
FUNDAMENTAL PROBLEMS F8–1. Draw the deflected shape of each beam. Indicate the inflection points.
(a) (b)
(c)
F8–2
(b)
F8–3. Draw the deflected shape of each frame. Indicate the inflection points.
(c)
F8–1
8
(a)
F8–2. Draw the deflected shape of each frame. Indicate the inflection points.
(b) (a)
F8–3
