(a) Load on frame (b) Elastic curve (c) Moment diagramFIGURE 40Calculation Procedure:- Apply a unit horizontal load
Apply the unit horizontal load at A 9 directed to the right. - Evaluate the bending moments in each member
Let M and m denote the bending moment at a given section caused by the load P and by
the unit load, respectively. Evaluate these moments in each member, considering a mo-
ment positive if it induces tension in the outer fibers of the frame. Thus:
Member AB: Let x denote the vertical distance from A to a given section. Then M= O;
m = x.
Member BC: Let x denote the horizontal distance from B to a given section. Then M=
Px; m= a.
Member CD: Let x denote the vertical distance from C to a given section. Then M =
Pb; m = a - x. - Evaluate the required deflection
Calling the required deflection A, we apply A = / [MmI(EI)] dx; EIA = Jg Poxdx + /g Pb(a
~x)dx = Pax^2 /2]b 0 + Pb(ax - X^2 /2)]C 0 = Pab^2 /2 + Pabc - Pbc^2 /2; A = [Pb/(2EI)](ab + 2ac -
c^2 ).
If this value is positive, A is displaced in the direction of the unit load, i.e., to the right.
Draw the elastic curve in hyperbolic fashion (Fig. 4Ob). The above three steps constitute
the unit-load method of solving this problem. - Construct the bending-moment diagram
Draw the diagram as shown in Fig. 4Oc. - Compute the rotation and horizontal displacement by the
moment-area method
Determine the rotation and horizontal displacement of C. (Consider only absolute values.)
Since there is no rotation at D, EISc = Pbc\ EIA 1 = Pbc^2 /2. - Compute the rotation of one point relative to another and the
total rotation
Thus EIOj 8 C = Pb^2 /2; EIB 8 = Pbc + Pb^2 /2 = Pb(c + b/2). The horizontal displacement of B
relative to C is infinitesimal.