Mechanics of Structures and Their Analysis 75

FIGURE 5.28 Simply supported beam subject to uniform load

[UDL] throughout the entire length.

FIGURE 5.31 Simply supported beam carrying a load, the inten-

sity of which varies from zero at each end to w per unit runs in the
middle of the span.

FIGURE 5.32 Simply supported beam with a load of uniform

intensity from zero at one end to w per unit run at the other end.

FIGURE 5.33 Simply supported beam with a couple at midspan.

FIGURE 5.29 Simply supported beam with an UDL for a certain

distance from one end. (a) Beam. (b) M/El diagram.

FIGURE 5.34 Beam with overhang at one end and carrying an

UDL over the whole length.

FIGURE 5.30 Simply supported beam carrying an UDL on an

intermediate part of the span.

It must be noted that Z is the actual section width at the

position where s is computed and I is the total moment of
inertia at the neutral axis. As shown in Figure 5.39, there will
be a parabolic variation of shear stress with y. The maximum
shear force occurs at the neutral axis and is given by:

3F FIGURE 5.35 Simply supported beam with equivalent overhangs

s= (5.51)
2bd and UDL of w per unit run over the whole length.
76 Practical Civil Engineering

FIGURE 5.36 Relation between load, shear, and bending moment.

FIGURE 5.37 Theory of simple bending.

FIGURE 5.38 Transverse section subjected to shear stress.

Mechanics of Structures and Their Analysis 77

FIGURE 5.39 Parabolic variation of shear stress.

F
If is represented as the mean stress, then For the rectangular section displayed in Figure 5.40:
bd
s = 1.5 × smean (5.52) V  h2 
=  − y12  (5.55)
2I  4 
5.11.1 SHEAR FLOW
Shear stresses vary quadratically from the neutral axis to the
The shear ow in each section is the product of the shear distance y1. The maximum shear stress on the neutral axis is
stress and the correlating width. Shear ow, q, refers to the nil on the top and underside surfaces of the beam. The max
longitudinal force per unit length transmitted throughout the shear stress for a rectangular cross section is as follows:
section.
3V
V = (5.56)
q = × z = ×Q (5.53) 2Amax
I
For a circular cross section:
where q is the shear ow, V is the shear force at a given sec-
tion, I is the moment of inertia at the neutral axis, b is the 4V
width of section, and Q be the rst moment of area about the = (5.57)
3Amax
∫
neutral axis = y dA. Figure 5.40 shows a rectangular beam in
pure bending. 5.11.2 BUILT-UP BEAMS

VQ The built-up beams are made of two or more pieces of mate-

Shear stress, = (5.54) rials combined to form a single solid beam. Shear ow, f, in
Ib
such beams at critical locations is offered by

VQ
f= (5.58)
I

5.12 DEFLECTIONS
A beam or a frame de ects when it is loaded. If the de ection
exceeds the permissible value, the structure will not look aes-
thetic and may result in psychological upsetting of the occu-
pants. This also may cause cracking in the materials of the
structure.
The extent of the de ection of a member under a speci ed
load is related directly to the slope of the de ected member’s
shape under that load. De ection is measured by integrating
the function that characterizes the member’s slope under this
load arithmetically.

5.12.1 CORRELATION BETWEEN SLOPE, DEFLECTION,

AND R ADIUS OF CURVATURE

FIGURE 5.40 Rectangular beam in pure bending. (a) Both end As shown in Figure 5.37b, a small portion PQ of the beam
xed. (b) One end xed other end free. (c) One end xed other that is bent into an arc is considered. In Cartesian coordinates,
pinned. (d) Both ends pinned. the radius of curvature is given by the relation.
78 Practical Civil Engineering

3 points. While the other theorem is used to compute the verti-

  dy  2  2 cal distance between a point on the elastic curve and a line
1+   
dx  tangent to the elastic curve at a second point (called tangential
R=  (5.59) deviation). For this purpose, the bending moment chart for the
d2y
dx 2 beam is rst drawn and then divided by the exural rigidity
(EI) to obtain the “M/EI” chart.
Substituting for R in Equation 5.48c, we get

d2y Theorem 1:
M = EI (5.60)
dx 2
dy The change of slope between two points on the elastic
Since (represents the slope at any point in the center line curve corresponds to the area of the M/EI diagram between
dx 2
dy
of the bent beam) is an extremely small quantity,   is
these two points as shown in Figure 5.42.
 dx 
B
M
much smaller still and hence can be neglected. The expres-
3
AB =
∫ A EI
dx (5.62)
 dy 
2
sion 1+    is, therefore, equal to unity. To be con-
2

 dx   where M is the moment, EI is the exural rigidity, and θAB

 is the change of slope between points A and B of the elastic
sistent with the sign conventions adopted, a negative sign is curve.
introduced and the equation of the bending of the beam can
be represented as:
Theorem 2:
d2y
EI = −M (5.61)
dx 2 The vertical divergence of a point A on an elastic tangent
curve extending from another point B equates to the moment
This equation is regarded as the elastic curve differential
of the area under the M/EI diagram between these two points
equation of the beam. The product EI is termed the beam’s
(A and B). This moment is calculated on point A to evaluate
exural rigidity. Table 5.2 shows the beam de ection for-
the deviation from B to A.
mula for different types of beam with different loading
conditions. B
M
tA B =
∫ A EI
xdx (5.63)
5.12.2 MOMENT AREA METHOD
where t A B is the deviation of tangent at point B as regards
The rotation, inclination, and de ection of beams and frames tangent at point A.
are derived from the moment area theorem. The theorem
was formulated by Otto Mohr and disclosed later in 1873 by 5.12.2.1 Sign Rules
Charles Greene. In this technique, area of the BMD is con-
sidered for the measurement of the slope and/or the de ec- 1. The deviation is said to be positive at any point if the
tions along the axis of the beam or frame at any point. For point is above the tangent.
the computation of the de ection, two theorems known as 2. When computed from the left tangent if θ is in the
the moment area theorems are used. One theorem is used to counterclockwise direction, the slope change is
compute the change of slope in the elastic curve between two positive.

TABLE 5.2
Beam De ection Formulae
De ection at Any Section
Type of Beam Slope at Free End in Terms of x Maximum De ection
1. Cantilever beam—concentrated load P at the free end
Pl 2 Px 2 Pl 3
= y= 3l − x =
2EI 6EI 3EI max

(Continued )
Mechanics of Structures and Their Analysis 79

TABLE 5.2 (Continued)

Beam De ection Formulae
De ection at Any Section
Type of Beam Slope at Free End in Terms of x Maximum De ection
2. Cantilever beam—concentrated load P at any point
Pa2 Px 2 Pa 2
= y= 3a − x for 0 < x < a = 3l − amax
2EI 6EI 6EI
Px 2
y= 3x − a for a < x < 1
6EI

3. Cantilever beam—uniformly distributed load ω (N/m)

wl 3 wx 2 2 wl 4
= y= x + 6l 2 − 4lx =
6EI 24EI 8EI max

4. Cantilever beam—uniformly varying load: maximum intensity ω0 (N/m)

w0 l 3 w0 x 2 w0 l 4
= y= 10l 3 − 10l 2 x + 5lx − x 3 =
24EI 120lEI 30EI max

5. Cantilever beam—couple moment M at the free end

ml Mx 2 Ml 2
= y= =
EI 2EI 2EI max

6. Beam simply supported at ends—concentrated load P at the center

Pl 2 Px  3l 2  l Pl 3
= = y= − x2  for 0 < x < =
12EI  4
1 2
16EI  2 48EI max

7. Beam simply supported at ends—uniformly distributed load ω (N/m)

wl 3 wx 3 5wl 4
1 = 2 = y= l − 2lx 2 + x 3 =
24EI 24EI 384EI max

8. Beam simply supported at ends—couple moment M at the right end

Ml Mlx  x2  Ml 2
1 = y= 1− 2  =
6EI 
6EI  l  9 3EI max
Ml l
2 = at x =
3EI 3

9. Beam simply supported at ends—uniformly varying load: maximum intensity ω0 (N/m)

7w0 l 3 w0 x w0 l 4 at
1 = y= 7l 4 − 10l 2 x 2 + 3x 4 max = 0.00652
360EI 360lEI EI
w0 l 3 x = 0.519l
2 =
45EI