Question 9.3: A simple beam AB supports a concentrated load P acting at di......

Bending moments in the beam. In this example the bending moments are expressed by two equations, one for each part of the beam. Using the free-body diagrams of Fig. 9-13. we arrive at the following equations:

M=\frac{Pbx}{L}     (0 ≤ x ≤ a)               (9-27a)

M=\frac{Pbx}{L}\ -\ P(x-a)     (a ≤ x ≤ L)               (9-27b)

Differential equations of the deflection curve. The differential equations for the two parts of the beam are obtained by substituting the bending-moment expressions (Eqs. 9-27a and b) into Eq. (9-12a). The results are

EIv″=M                               (9-12a)

EIv″=\frac{Pbx}{L}               (0 ≤ x ≤ a)             (9-28a)

EIv″=\frac{Pbx}{L}\ -\ P(x\ -\ a)               (a ≤ x ≤ L)             (9-28b)

Slopes and deflections of the beam. The first integrations of the two differential equations yield the following expressions for the slopes:

EIv^′=\frac{Pbx^2}{2L}+C_1               (0 ≤ x ≤ a)             (h)

EIv^′=\frac{Pbx^2}{2L}\ -\ \frac{P(x\ -\ a)^2}{2}+C_2               (a ≤ x ≤ L)             (i)

in which C_1 and C_2 are constants of integration. A second pair of integrations gives the deflections:

EIv=\frac{Pbx^3}{6L}+C_1x+C_3               (0 ≤ x ≤ a)             (j)

EIv=\frac{Pbx^3}{6L}\ -\ \frac{P(x\ -\ a)^3}{6}+C_2x+C_4               (a ≤ x ≤ L)             (k)

These equations contain two additional constants of integration, making a total of four constants to be evaluated.
Constants of integration. The four constants of integration can be found
from the following four conditions:

(1) At x = a, the slopes v′ for the two parts of the beam are the same
(2) At x = a, the deflections v for the two parts of the beam are the same
(3) At x = 0, the deflection v is zero
(4) At x = L, the deflection v is zero

The first two conditions are continuity conditions based upon the fact that the axis of the beam is a continuous curve. Conditions (3) and (4) are boundary conditions that must be satisfied at the supports.
Condition (1) means that the slopes determined from Eqs. (h) and (i) must be equal when x = a; therefore,

\frac{Pba^2}{2L}+C_1=\frac{Pba^2}{2L}+C_2\quad \quad or\quad \quad C_1=C_2

Condition (2) means that the deflections found from Eqs. (j) and (k) must be equal when x = a; therefore,

\frac{Pba^3}{6L}+C_1a+C_3=\frac{Pba^3}{6L}+C_2a+C_4

Inasmuch as C_1 = C_2, this equation gives C_3 = C_4.
Next, we apply condition (3) to Eq. (j) and obtain C_3 = 0; therefore,

C_3 = C_4=0                                       (l)

Finally, we apply condition (4) to Eq. (k) and obtain

\frac{PbL^2}{6}\ -\ \frac{Pb^3}{6}+C_2L=0

Therefore,

C_1 = C_2=-\frac{Pb(L^2\ -\ b^2)}{6L}                               (m)

Equations of the deflection curve. We now substitute the constants of integration (Eqs. l and m) into the equations for the deflections (Eqs. j and k) and obtain the deflection equations for the two parts of the beam. The resulting equations, after a slight rearrangement, are

v=-\frac{Pbx}{6LEI}(L^2\ -\ b^2\ -\ x^2)                    (0 ≤ x ≤ a)                      (9-29a)

v=-\frac{Pbx}{6LEI}(L^2\ -\ b^2\ -\ x^2)\ -\ \frac{P(x\ -\ a)^3}{6EI}            (a ≤ x ≤ L)                      (9-29b)

The first of these equations gives the deflection curve for the part of the beam to the left of the load P. and the second gives the deflection curve for the part of the beam to the right of the load.
The slopes for the two parts of the beam can be found either by substituting the values of C_1 and C_2 into Eqs. (h) and (i) or by taking the first derivatives of the deflection equations (Eqs. 9-29a and b). The resulting equations are

v^′=-\frac{Pb}{6LEI}(L^2\ -\ b^2\ -\ 3x^2)                    (0 ≤ x ≤ a)                      (9-30a)

v^′=-\frac{Pb}{6LEI}(L^2\ -\ b^2\ -\ 3x^2)\ -\ \frac{P(x\ -\ a)^2}{2EI}            (a ≤ x ≤ L)                      (9-30b)

The deflection and slope at any point along the axis of the beam can be calculated from Eqs. (9-29) and (9-30).
Angles of rotation at the supports. To obtain the angles of rotation θ_A and θ_B at the ends of the beam (Fig. 9-12b), we substitute x = L into Eq. (9-30a) and x = L into Eq. (9-30b):

θ_A=-v^′(0)=\frac{Pb(L^2\ -\ b^2)}{6LEI}=\frac{Pab(L+b)}{6LEI}                      (9-31a)

θ_B=-v^′(L)=\frac{Pb(2L^2\ -\ 3bL+b^2)}{6LEI}=\frac{Pab(L+a)}{6LEI}                      (9-31b)

Note that the angle θ_A is clockwise and the angle θ_B is counterclockwise, as shown in Fig. 9-12b.
The angles of rotation are functions of the position of the load and reach their largest values when the load is located near the midpoint of the beam. In the case of the angle of rotation θ_A, the maximum value of the angle is

(θ_A)_{max}=\frac{PL^2\sqrt{3}}{27EI}                           (9-32)

and occurs when b = L/\sqrt{3} = 0.577L (or a = 0.423L). This result is obtained by taking the derivative of θ_A (Eq. 9-31a) with respect to b and setting it equal to zero.
Maximum deflection of the beam. The maximum deflection δ_{max} occurs at point D (Fig. 9-12b) where the deflection curve has a horizontal tangent. If the load is to the right of the midpoint, that is, if a > b. point D is in the part of the beam to the left of the load. We can locate this point by equating the slope v′ from Eq. (9-30a) to zero and solving for the distance x which we now denote as x_1. In this manner we obtain the following formula for x_1:

x_1=\sqrt{\frac{L^2\ -\ b^2}{3}}                (a ≥ b)                      (9-33)

From this equation we see that as the load P moves from the middle of the beam (b = L/2) to the right-hand end (b = 0), the distance x_1 varies from L/2 to L/\sqrt{3} = 0.577L. Thus, the maximum deflection occurs at a point very close to the midpoint of the beam, and this point is always between the midpoint of the beam and the load.
The maximum deflection δ_{max} is found by substituting .v, (from Eq. 9-33) into the deflection equation (Eq. 9-29a) and then inserting a minus sign:

δ_{max}=-(v)_{x=x_1}=\frac{Pb(L\ -\ b^2)^{3/2}}{9\sqrt{3}LEI}               (a ≥ b)            (9-34)

The minus sign is needed because the maximum deflection is downward (Fig. 9-12b) whereas the deflection v is positive upward.
The maximum deflection of the beam depends on the position of the load P, that is, on the distance b. The maximum value of the maximum deflection (the “max-max” deflection) occurs when b = L/2 and the load is at the midpoint of the beam. This maximum deflection is equal to PL³/48EI.
Deflection at the midpoint of the beam. The deflection δ_C at the midpoint C when the load is acting to the right of the midpoint (Fig. 9-12b) is obtained by substituting x = L/2 into Eq. (9-29a), as follows:

δ_C=-v\left(\frac{L}{2}\right)=\frac{Pb(3L^2\ -\ 4b^2)}{48EI}                 (a ≥ b)               (9-35)

Because the maximum deflection always occurs near the midpoint of the beam, Eq. (9-35) yields a close approximation to the maximum deflection. In the most unfavorable case (when b approaches zero), the difference between the maximum deflection and the deflection at the midpoint is less than 3% of the maximum deflection, as demonstrated in Problem 9.3-10.
Special case (load at the midpoint of the beam). An important special case occurs when the load P acts at the midpoint of the beam (a = b = L/2). Then we obtain the following results from Eqs. (9-30a). (9-29a), (9-31), and (9-34), respectively:

v^′=-\frac{P}{16EI}(L^2\ -\ 4x^2)\quad \quad \left(0≤x≤\frac{L}{2}\right)           (9-36)

v=-\frac{Px}{48EI}(3L^2\ -\ 4x^2)\quad \quad \left(0≤x≤\frac{L}{2}\right)           (9-37)

θ_A=θ_B=\frac{PL^2}{16EI}           (9-38)

δ_{max}=δ_{C}=\frac{PL^3}{48EI}           (9-39)

Since the deflection curve is symmetric about the midpoint of the beam, the equations for v′ and v are given only for the left-hand half of the beam (Eqs. 9-36 and 9-37). If needed, the equations for the right-hand half can be obtained from Eqs. (9-30b) and (9-29b) by substituting a = b = L/2.