BEAMS AND BENDING 105 
: =x. Hence the neutral axis XX must pass through the centre of gravity 
The moment of resistance is found as follows. The moment of the 
resultant stress on ss is Lay? or Lay, and the total moment for the 
l 2 
4 
“a 
‘whole section is LSay? or J23ay2,-which is equal to Ay or Z21, where Lis 
1 Yo 1 ¥ 
" the moment of inertia of the whole section about the axis XX. 
For equilibrium the bending moment must be equal to the moment 
of resistance, therefore M -4 I =f I. Putting Z, -~ and Zy= rr 
1 
~M=f,Z,=f,2,. Z, and Z, are called the moduli of the section. : 
Since E ak ge | —f2 therefore M = EI 
Ry wh % R 
111. Moments of Inertia, and Moduli of Various Sections.—The 
moments of inertia and the moduli of the more common sections required 
in connection with the moment of resistance to bending are tabulated 
on p. 106. The axis of moments is the neutral axis of the section, and 
_ passes through the centre of gravity of the section. y=distance of axis 
_ of moments from the top or bottom of the section. Where no value is 
given for y, it is equal to half the total depth of the section. Where the 
section is not symmetrical about the neutral axis, there are two values for 
the modulus, Z, =I+y,, and Z,=I+y,. 
112. Equivalent Beam Sections.—Since the moment of resistance of 
an element of a beam section is equal to its area multiplied by the stress 
on it and by its distance from the neutral axis, and since the stress on 
_ the element is proportional to its distance from 
the neutral axis, it follows that if the element 
_ be moved parallel to the neutral axis into 
another position its moment of resistance will 
not be altered. Hence if a beam section be 
divided into indefinitely narrow strips parallel 
_ to the neutral axis, these strips may be moved Fig. 144. 
_ parallel to the neutral axis so as to form another 
_ section, which will have the same moment of resistance as the original 
_ section. An example is shown in Fig. 144, where the original section, 
_ a hollow semicircle, shown in full lines, is converted into an equivalent 
solid section by collecting the area about a central axis. 
113. Section Modulus Figures.—Let MHNK (Fig. 145) be the 
cross section of a beam, and XX its neutral axis. The distribution of 
Stress on the section due to the bending is shown at (a), 7, being the 
stress at N, and /, the stress at M. Take an indefinitely narrow strip 
HK of the section parallel to XX. Draw the base line Y,Y, parallel. to 
XX and passing through N, the lowest point of the section. Draw Hh 
_ and Kk perpendicular to Y,Y,. Select a point O in XX. If the section is 
symmetrical about an axis perpendicular to XX, then O is preferably 
_ where this axis cuts XX. In Fig. 145, MON is an axis of symmetry per- 
_ pendicular to XX. Join’ and k to O by lines cutting HK at m and n. 
' The stress f at HK is such that //f,=OL/ON. By similar triangles 
 mnf/hik=OL/ON, but hk=HK, therefore mn/HK =OL/ON =//f,, and 
