][ STEEL STRUCTURES ][

						Beam Design
						   Deflection
					              P&A 278 - 285

Methods:					  
		1. Formula
			 D = 5/384 * W*L*L*L / E* I	

		2. Deflection of Steel Beams with
			bending stress of 24 ksi
			Graph figure 9.1 P&A 280

		3. Load-Span Values for Beams
			Safe Load table 9.2 P&A 282-284

The formula yields the most accurate value,
but the level of accuracy that is significant for real design work 
can be obtained by any of the methods illustrated.

D = 5/384  x  [W x L x L x L  /  (E x I)]
D = maximum deflection in inches
W = total uniformly distributed load in pounds or kips
	(for a beam on which the loading is not
	typical, we may find the W that would 
	produce the same bending moment) 
L = Length of the span in inches
E = modulus of elasticity of the beam in psi or ksi
       (for structural steel E= 29,000,000 psi)
I = moment of inertia of the cross section of the beam in inches to the fourth power
for simple beam with a uniform disributed
load only !

DEFLECTION Calculation :
====================
Example:
A W 12 x 22 is used for a simple span beam
to carry a total uniformly distributed load of 10 kips, 
including the beam weight.
The span is 16 ft.
Find the maximum deflection.

A. Given:
1. W 12 x 22
2. simple span beam
3. W =10 kips/ft
4. L = 16 ft
5. A 36 steel

B. Asked:
	deflection in inches

C. Graph:
		10kips/ft
	----------------------------
                V                               V
                                16 ft
                         simple span

D. Calculations:
1. I = 156 in 4
	around the axis x-x
	from Table 4.1
	P & A 172

2. E = 29,000,000 psi)
	for structural steel
	P&A 53

3. D = 5 /384  x    W L 3 / E I
    D = 5 / 384 x    W x L x L x L  / E I 
	   		     3 
    D = 5 /384 x   10 ( 16 x 12)    / 29,000 (156)
		           x 12 to convert into inches
    D = 0.204 inches

E. Answer: 
    Actual deflection 0.20 inches

For a uniformly loaded simple span beam,
with a fixed value for the maximum bending stress, a formula may be derived that expresses
the maximum deflection in terms of
		span
		beam depth

For a stress value of 24 ksi
the formula has the form:

	D = 0.02483 x L x L /d

D = maximum deflection in inches
L = span length in feet
d = beam depth in inches

Example:
L = 16 ft
d = 12 inches
D = 0.02483 x L x L /d
D = 0.02483 x 16 x 16 /12
D = 0.5297 inches maximum deflection in inches
D = 0.53 inches

Method 2. Graph:
=============
Deflection of steel beams with bending stress of 24 ksi

Figure 9.1 P&A 280
graph can be used for:
1. uniformly loaded simple span beam
2. with fixed value for maximum bending stress
3. A 36 steel
4. 24 ksi maximum bending stress

This graph is based on the following formula:
D = (0.02483 x L x L ) / d
D = maximum deflection in inches
L = span length in feet
d = beam depth in inches

pa280	 Fig 9.1
Deflection of steel beams with bending stress
of 24 ksi (165 Mpa)

Example:
L = 16 ft				(y-axis)
d = 12 inches			(12" curve)
intersection read 0.53"  max. deflection in inches
DEFLECTION GRAPH 

Example Graph Method:
A W 12 x 22 is used for a simple span beam
to carry a total uniformly distributed load of 10 kips, including the beam weight. The span is 16 ft.Find the maximum deflection

A. Given:
1. W 12 x 22
2. simple span beam
3. W =10 kips/ft
4. L = 16 ft

B. Asked:
	deflection in inches

C. Graph:

		10kips/ft
	----------------------------
               V                                V
                                16 ft
                         simple span

D. Calculations:
1. M = W x L / 8
2. M = 10 x 16 /8
3. M = 20 kip-ft
4. Section Modulus
	P&A 172
	Table 4.1
	W 12 x 22 	=	25.4 in 3
5. Bending Moment
    fb = M / S
6. fb = 20 x 12 / 25.4
7. fb = 9.449 ksi
8. P&A 280
	Figure 9.1
	read the chart as follows
	span 		16 feet 	(y-axis)
	depth		12 " 	(curve)
	deflection l/360		(diagonal)
	deflection	0.53	(x-axis) read
9. D = fb/ 24 x deflection
    D = 9.449/24 x 0.53
	D = 0.209 in = 0.21inches

E. Answer: Actual deflection 0.21 inches

Method 3.
Load-Span Values for Beams
Figure 9.2 P&A 281-284

To be used under the folowing conditions:
	simple span beam
	uniform distributed load
	A 36 steel

Maximum deflection = deflection factor / depth of 								beam
Deflection is the deflection for the full load table load!

You need to correct this number by the 
ratio of the load

D = actual load / full table load x deflection factor / actual beam depth
(nominal beam depth actual = more accurate actual beam depth)

Example:
A W 12 x 22 is used for a simple span beam
to carry a total uniformly distributed load of 10 kips, including the beam weight. The span is 16 ft.Find the maximum deflection

A. Given:
simple span 
A 36
W = 10 kips
L = 16 ft
d = 12 inches
W 12 x 22

B. Asked 
Actual Deflection

C. Graph

		10kips/ft
	----------------------------
               V                                V
                                16 ft
                         simple span

D. Calculations
1. D = actual load / full table load x deflection factor / actual beam depth
(nominal beam depth actual = more accurate)

2. P&A 282
	Table 9.2
	span = 16 ft
	shape = 12 x 22

	read deflection factor = 6.36

D = 10 / 25.4   x 6.36 /12
D = 0.209 inches
D = 0.21 inches

E. Answer
D = 0.21 inches deflection


Equivalent Tabular Loads
The tables for
simple span beam
total uniformly distributed load
A 36 steel
can be expanded for other load cases:
for 2 loads at the third point of the beam
W = 2.67 x P