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ARCH5023/ARCH4333 - ARCHITECTURAL STRUCTURES I

][ STEEL STRUCTURES ][
THE UNIVERSITY OF OKLAHOMA - COLLEGE OF ARCHITECTURE
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SAMPLE HOMEWORK/EXAM FORMAT FOR CALCULATIONS

STUDENT NAME:

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

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