Page 107 - 4749

dQ   x
 q   x . (6.1)
dx
From the second equation, neglecting the square of small size,
we find
dM   x
 Q   x . (6.2)
dx

Figure 6.4 Figure 6.5

After differentiation of (6.2) and taking into account the
condition (6.1), we obtain
2
d M   x
 q   x . (6.3)
dx 2
Analyzing the depending we can see the following:
- at the section of the beam, where q  , the function   x is
Q
0
decreasing and vice versa;
0
- at the section of the beam, where q  , shear force is a
constant;

q
- at the section of the beam, where   x   const , we obtain
q
Q   x  qx C ,
i.e. shear force varies linearly;

- If Q  , then M    0x  that means that in the area where the
0
shear force is positive, bending moment is increasing;

107