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ARCH5023/ARCH4333 - ARCHITECTURAL STRUCTURES I

][ STEEL STRUCTURES ][
THE UNIVERSITY OF OKLAHOMA - COLLEGE OF ARCHITECTURE
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STEEL BEAMS BENDING CALCULATIONS

STEEL BEAM DESIGN

Manual of Steel Construction
-----------------------------------

Weights of Building Materials
P 6-9  
 
Beam Diagrams & Formulas
P 2-293 - 313	* P 296
Parker & Ambrose P 109

Beam & Girder Design
Allowable Stress Design
P 2-1 - 13

Beam & Girder Design
Plastic Design 
based on plastic section modulus
P 2-15 - 29

Allowable Loads on Beams
P 2-30 - 140


Factors in Beam Design:
bending strength - primary concern
shear resistance
deflection
lateral support
web crippling
support details

ASTM A 36 steel with
Fy = 36 ksi
Fy = specified minimum yield stress of the type of steel being used 
        minimum yield point (P&A 268)
		(MSC 2-4 & 5-202)

Maximum allowable bending stress = 0.66 Fy

may be used when:
-----------------------
1. symmetric beam section
2. web and flanges with min. width-thickness ratios
3. braced compression flange

PROBLEM

BEAM DESIGN FOR BENDING

A girder has a span of 18 ft with floor beams framing into it from both sides at 6-ft intervals.
The reach of each floor beam is 4,000 lb,
so the girder receives two (2) concentrated 
loads of 8,000 lb each 
at the third points of the span.
The framing of the beams provides lateral support for the girder at 6'-0" intervals.
In addition, the girder has a uniformly distributed load of 400 lb/ft,
including its own weight,
extending over the full span.
Design the girder for bending,
assuming that A 36 steel is used.

A. GIVEN

	L 	= 18'-0"
	L 1  	= 6'-0"
	P 1	= 4,000 lb
	P 2	= 4,000 lb
	P 3	= 4,000 lb
	P 4	= 4,000 lb 
	w	= 400 lb/ft  including its own weight	
	Steel 	= A36

B. Diagram

	|                                  18'-0"                                |
               |       6'-0"          |        6'-0"        |     6'-0"             |
                                (2)8,000 LB      (2)8,000 LB
               |                         |                        |                          |   
                                             400 LB/FT including its own weight
                ===================================
               |                        H                      H                           |
                ===================================
               R                                                                             R

C. ASKED:

	Girder size for bending only

D. CALCULATIONS:

1.	Reaction R
	(P&A 109)
	(MSC 2-296)
	Because of the symmetric loading of the girder, 
	we know that the reaction will be
	half of the total load.
	R = 1/2 [P1 + P2 + P3 + P4 + (w x L)]	
	R = 1/2 [4,000+ 4,000+4,000+4,000lb+ (400lb x 18ft)]
	R = 0.5 (16,000 lb + 7,200lb)
	R = 0.5 (23,200 lb)
	R = 11,600 lb

2. Maximum bending moment M
	(P&A 109)
	(MSC 2-296)
	Because of the symmetric loading of the girder,
	we know that the maximum bending
	moment will occur at the center of the span.
	M(x=9)= (R x L/2) - [(P1+P2 x {L/2 - L1} ) + (w x L/2 x L/4)]
	M(x=9)= (11,600 x 18/2) - [(4,000 + 4,000 x {18/2ft - 6ft} + (400 x 18/2 x 18/4)
	M(x=9)= (11,660 x 9) - [(8,000 x {9-6} + (400 x 9x 4.5)]
	M(x=9)= (11,600lb x 9ft) - [(8,000lb x 3ft)	+ (400lb x 9ft x 4.5ft)]
	M(x=9)= 104,400 ft-lb -[24,000ft-lb + 16,200ft-lb]
	M(x=9)= 104,400 ft-lb - 40,200 ft-lb
	M(x=9)= 64,200 ft-lb

3. Allowable bending stress
	P&A 267
	Fb = Fy x 0.66
	Fb = 36 ksi x 0.66
	Fb = 24 ksi

4. Elastic Section Modulus (in3)
	P & A 269
	P & A 164
	S = section modulus
	S = I/c
		moment of inertia / distance of the most
		remote fiber from the neutral axis 	
	S = M / Fb

	M from step 2
	M is in ft-lb therefore has to be multiplied by 12 to get inches
	Fb from step 3

	S = 64,200 x 12
	      ----------------
	         24,000  
	
	S = 32.1 in 3

5. Section Modulus
	P&A 275, Table 9.1
	ASD Steel Construction Manual
	1-30		W 10 x 30		S = 32.4 in3
	1-28		W 12 x 26		S = 33.4 in3
	1-26		W 14 x 26		S = 35.3 in3