(b) Displacement diagram
FIGURE 19(Fig. 196) line Ca parallel to member AC, with Ca = 0.0781 in (1.98 mm). Similarly,
draw Cb parallel to member BC, with Cb = 0.0845 in (2.146 mm).
- Determine the displacement
Erect perpendiculars to Ca and Cb at a and 6, respectively. Designate the intersection
point of these perpendiculars as C'.
Line CC represents, in both magnitude and direction, the approximate displacement
of joint C under the applied load. Scaling distance CC to obtain the displacement shows
that the displacement of C = 0.134 in (3.4036 mm).
AXIAL STRESS CAUSED BY IMPACT LOAD
A body weighing 18 Ib (80.1 N) falls 3 ft (0.9 m) before contacting the end of a vertical
steel rod. The rod is 5 ft (1.5 m) long and has a cross-sectional area of 1.2 in^2 (7.74 cm^2 ).
If the entire kinetic energy of the falling body is absorbed by the rod, determine the stress
induced in the rod.
Calculation Procedure:- State the equation for the induced stress
Equate the energy imparted to the rod to the potential energy lost by the falling body:
s = (P/A) { 1 + [1 + 2EhI(LPIA)]*-^5 ), where h = vertical displacement of body, ft (m). - Substitute the numerical values
Thus, PIA = 18/1.2 = 15 lb/in^2 (103 kPa); h = 3 ft (0.9 m); L = 5 ft (1.5 m); 2EhI(LPIA) =
2(30) x 106 )(3)/[5(15)] = 2,400,000. Then s = 23,250 lb/in^2 (160,285.5 kPa).
Related Calculations: Where the deformation of the supporting member is neg-
ligible in relation to the distance h, as it is in the present instance, the following approxi-
mation is used: s = [2PEh/(AL)]°-^5.
(a) Space diagram