154 APPLIED MECHANICS 
and parallel to P be applied as shown at points F and H in YY produced, 
these added forces will not affect in any way the stresses at XX, because 
P, and P, balance one another. Pand P, being equal and parallel forces 
acting in opposite directions form a couple which produces a pure turning 
or bending action at XX. The bending moment is Pr, and the distribu- 
tion of stress at XX due to Pr is shown at (a), where /, is the tensile stress 
along AC, and f, is the compressive stress along BE. Pr=/,Z) = fos 
1 
also Pr=/,Z,= Tee where I is the moment of inertia of the section XX 
Yo 
about its neutral axis 0. From these equations /, = “rH and at: 5 
The force P, will produce a direct tension, and the tensile stress at 
XX due to P, is a — =f,, where a is the area of the cross section XX. 
The distribution of this stress is shown at (0). 
Combining the stresses due to the pure bending and the direct tension 
as shown at (c), it is seen that there is a tensile stress f along AC equal 
to f, +3, and a compressive stress along BE equal to f,—jf,. If f, is less 
than f,, then f, —/; is a tensile stress. 
Exampcre.—Referring to Fig. 215, the section at XX is a rectangle, 
depth XX=6 inches, breadth=3 inches, y=5 inches. Total tensile 
stress along AC =5 tons per square inch. It is required to find P and 
the stress along BE. ? 
Pr = }0d%f,, that is, 5P=1 x 3 x 6%, hence f, = = 
\ pubes 5P 
F255 faB=f, thw + 7.= 4) therefore P=15 tons, 
Ya ae =4i fp= re: Hence f, —/,= 3} tons per square 
inch, and is a compressive stress. 
150. Strength of a Ring.—Fig. 216 shows a ring of uniform cross 
Fia. 216. , Fia. 217. Fig. 218. Fra. 219. 
section carrying a load W. The mean radius of the ring is r. There is 
at the horizontal section at B a bending action tending to diminish the 
curvature of the ring at B, and at the vertical section at A there is a 
