chapter 6 - non-composite beams
DESCRIPTION
For reference onlyTRANSCRIPT
CE134-
STRUCTURAL STEEL
DESIGN
J. Berlin P. Juanzon PhD
Beams
Beams are the most common
members found in a typical steel
structure. Beams are structural members
that support transverse loads and are
therefore subjected primarily to flexure
and bending
Beams
When a beam is subjected to bending loads,the bending stress in the extreme fiber is defined as:
𝑓𝑏 =𝑀𝑐
𝐼=
𝑀
𝑆Eqn 6-1
and the yield moment is:
My
= FyS Eqn 6-2
Where:
fb = Max bending stress,
My = Yield moment,
Fy = Yield stress,
M = Bending moment,
c = Distance from the neutral axis to the extreme fiber,
I = Moment of inertia, and
S = Section modulus
Where c is the perpendicular distance from the
neutral axis to the extreme fiber, and Sx is the
elastic section modulus of the cross section
A plastic hinge occurs when the entire cross
section of the beam is at its yield point, not just the
extreme fiber. The moment at which a plastic hinge is
developed in a beam is called the plastic moment and is
defined as:
MP
= FyZ Eqn 6-2
where
Mp = plastic moment, and
Z = plastic section modulus
The plastic moment is the maximum moment, or nominalbending strength of a beam with full lateral stability. For standardwide flange shapes, the ratio of the plastic moment, Mp, to theyield moment, My, usually varies from 1.10 to 1.25 for strong axisbending (Zx, Sx), and 1.50 to 1.60 for weak axis bending
bf
h
tf
tw
tf
cg
cg
C = AcFy
T = AtFy
a
Mp = Fy(Ac)a = Fy(At)a = Fy(A/2)a
Since Z = (A/2)a – Plastic section modulus
Therefore Mp= FyZ
For the built-up section shown in the figure, determine: (Bending is along x-axis and use grade 50steel)
a) the elastic section modulus S
b) yield moment My
c) the plastic modulus Z
d) the plastic moment Mp
8in
12in
1in
1in
½ in
All flexural members are classified as eithercompact, non-compact, or slender, depending on thewidth-to-thickness ratios of the individual elementsthat form the beam section.
When the width-to-thickness ratio is less than λ p,then the section is compact. When the ratio is greaterthan λp but less than λr, then the shape is non-compact.When the ratio is greater than λr, the section isclassified as slender.
There are also two type of elements that aredefined in the AISC specification stiffened andunstiffened elements.
The basic design strength equation for beams inbending is:
Mu ≤ ϕb Mn Eqn 6-4
where
Mu = Factored moment,
ϕb = 0.9,
Mn = Nominal bending strength, and
ϕMn = Design bending strength
The nominal bending strength, Mn, is a function of the following:
1. Lateral–torsional buckling (LTB),
2. Flange local buckling (FLB), and
3. Web local buckling (WLB).
When full lateral stability is provided for a beam, the nominal moment strength is the plastic moment capacity of the beam:
Mn = Mp = FyZx
Lateral–torsional buckling occurs when the distancebetween lateral brace points is large enough that the beamfails by lateral, outward movement in combination with atwisting action.
The beam shown in the figure is a W16 x 31 ofA992 steel. It supports a reinforced concrete floorslab that provides continuous lateral support of thecompression flange. The service dead load is450lb/ft. This load is superimposed on the beam; itdoes not include the weight of the beam itself. Theservice live load is 550lb/ft. Does this beam haveadequate moment strength?
30ft
In the design process for steel beams, shearrarely controls the design; therefore, most beamsneed to be designed only for bending anddeflection. Special loading conditions, such asheavy concentrated loads or heavy loads on a shortspan beam, might cause shear to control the designof beams.
From mechanics of materials, the general formula forshear stress in a beam is
𝑓𝑣 =𝑉𝑄
𝐼𝑏 Where:
fv = shear stress at the point under consideration,
V = vertical shear at a point along the beam under consideration
I = moment of inertia about the neutral axis, and
b = thickness of the section at the point under consideration
Q = first moment, about the neutral axis
In the AISC specification, the shear yield stress is taken as 60% of the yield stress, Fy.
The design shear strength is defined as:
ϕvVn = ϕv 0.6FyAwCv
where
Fy = Yield stress,
Aw =Area of the web (d)(tw)
Cv = Web shear coefficient, and
ϕv = 0.9 or 1.0
Since the shear stress is concentrated in thebeam web, localized buckling of the web needs tobe checked. A web slenderness limit for local webbuckling if there are I-shaped members is definedas
When this limit is satisfied, local web bucklingdoes not occur and Cv = 1.0 and ϕv = 1.0.
Other values of Cv :
ϕv = 0.90
where kv = 5 for unstiffened webs with h/tw < 260, except that kv = 1.2 for the stem of T-shapes.
>
Other values of Cv :
where kv = 5 for unstiffened webs with h/tw < 260, except that kv = 1.2 for the stem of T-shapes.
Allowable stress :
When
Required shear strength
Fv = 0.40 Fy
fa = Va / Aw
fa < Fa
A simply supported beam W14 x 90 with a span of 45 feet is laterally supported at the its ends and is subjected to the following service loads:
Dead load = 400lb/ft (Including the weight of beam)
Live load = 1,000lb/ft
Use steel Fy = 50ksi.
d = 14in, tw = 7/16in , h=11.33in
Check for shear strength of the beam.
In addition to being safe, a structure must beserviceable. A serviceable structure is one thatperforms satisfactorily, not causing any discomfortor perceptions of un-safety for occupants or usersof the structure. There are two main serviceabilityrequirements: deflection and floor vibrations.
The typical design procedure for beams involvesselecting a member that has adequate strength in bendingand adequate stiffness for serviceability. Shear typicallydoes not control, but it should be checked as well. Thedesign process is as follows:
1. Determine the service and factored loads on thebeam. Service loads are used for deflection calculations andfactored loads are used for strength design. The weight ofthe beam would be unknown at this stage, but the self-weight can be initially estimated and is usuallycomparatively small enough not to affect the design.
2. Determine the factored shear and moments on thebeam.
3. Select a shape that satisfies strength and deflectioncriteria. One of the following methods can be used:
a. For shapes listed in the AISC beam design tables,select the most economical beam to support the factoredmoment. Then check deflection and shear for the selectedshape.
b. Determine the required moment of inertia usingAISC. Select the most economical shape based on themoment of inertia calculated, and check this shape forbending and shear.
c. For shapes not listed in the AISC beam designtables, an initial size must be assumed. An estimate of theavailable bending strength can be made for an initial beamselection; then check shear and deflection. A more accuratemethod might be to follow the procedure in step b above.
For the floor plan shown , design members B1 and G1for bending, shear, and deflection. Compare deflectionswith L/240 for total loads and L/360 for live loads. Thesteel is ASTM A992, grade 50; assume that Cb = 1.0 forbending. The dead load (including the beam weight) isassumed to be 85 psf and the live load is 150 psf.
Assume that the floor deck provides full lateralstability to the top flange of B1. Ignore live loadreduction. Use the design tables in the AISCM whereappropriate.
8ft 8ft 8ft
20ft
G-1
G-1
B-1
B-1
A simply supported beam 28ft long is a W16 x 31 ofA992 steel. It supports a reinforced concrete floor slab thatprovides continuous lateral support of the compressionflange. The service dead load is 500lb/ft. This load issuperimposed on the beam; it does not include the weight ofthe beam itself. The service live load is 600lb/ft.
a) Is the beam adequate for bending moment?
b) Is the beam adequate for beam shear?
c) Is the beam adequate for allowable deflection?
28ft