Presented by: Civil Engineering Academy Beams Presented by: Civil - - PowerPoint PPT Presentation

Presented by: Civil Engineering Academy

Beams

Presented by: Civil Engineering Academy

 Concrete Beams (CERM Ch. 50)

• T-beams – Flexural Strength
• Reinforced Concrete Beams – Shear Design

 Steel Beams (CERM Ch. 59)

• Loads
• Flexure
• Shear
• Deflection (covered already)

 Due to the low tensile strength and brittle

nature, concrete beams must be reinforced with steel. The steel is placed at the tensions side of the beam (called singly reinforced). Double reinforced places steel in the compression zone too.

 For a composite material where steel bars

carry tension and the concrete carries compression the stress distribution differs. Use Whitney stress block assumptions to simplify computation of moment capacity.

The depth of a is proportional to the depth to the neutral axis and is given as: a= β1 c Where β = 0.85 for concrete strength not exceeding 4.0 ksi and reduced at 0.05 for each 1.0 ksi over 4.0 ksi, but not less than 0.65.

When asked to determine the moment capacity of a singly reinforced concrete beam, you don’t know where the neutral axis, c is. By equating the tensile force in the steel with the compressive force in the concrete, the assumed depth of the compression zone can be determined: a=

𝐵𝑡𝑔𝑧 0.85𝑔′

𝑑𝑐

Similar to Eq. 50.26

With a determined, the nominal moment capacity of the cross section can be determined by using the force-couple system that develops on either side of the neutral axis or: Mn=Asfy(d-𝑏

2)

Since in Ultimate Strength Design or LRFD, the nominal strength must be reduced by a reduction factor, ϕ, the design moment capacity is given as: ϕMn= ϕAsfy(d-𝑏

2)

• Eq. 50.28

Where ϕ =0.90 is used.

Continued… ϕ Mn= ϕAsfy(d-𝑏

2)

Note that β1 is not needed to find the moment capacity for a singly reinforced concrete beam. If the actual location of the neutral axis was needed, it can be found by: C= 𝑏 β1

 The analysis of a T-section is dependent on

the location of the neutral axis, a. If a is in the flange, the same analysis procedure can be used as with a singly reinforced beam.

a = 𝐵𝑡𝑔𝑧−0.85𝑔′

𝑑ℎ𝑔(𝑐𝑓−𝑐𝑥)

0.85𝑔′

𝑑𝑐𝑥

ϕMn= ϕ0.85f’c[hf(be-bw)(d-ℎ𝑔

2 )+bwa(d-𝑏 2)]

You won’t know the location of a. Therefore, assume a is located in the flange

• portion. Calculate and verify that it is

located in the flange. The moment capacity can then be calculated as before.

 Flexural shear stresses result in diagonal cracking

in the beam. Concrete resists shear up to a certain level (Vc). Beyond that, the shear strength must come from steel stirrups. Therefore the nominal shear capacity (Vn) is given as:

 Vn = Vc+Vs

• Eq. 50.51

 Vc = 2 𝑔′ 𝑑(bw)(d)

• Eq. 50.53a

 Vs = Asfy(𝑒

𝑡)

• Eq. 50.54

 Stirrups are required in the beam section any time

when the design shear force exceeds one-half the concrete design shear stress. The required stirrup spacing can be found by combining the previously shown equations and re-arranging as:

 Sreq= 𝐵𝑡𝑔𝑧𝑒

𝑊𝑣

ϕ−𝑊𝑑

• Eq. 50.64

 There are several limits placed on the max spacing of

stirrups and are a function of the percentage of strength of the cross section that comes from the stirrups (see Eqs. 50.59(a)-60.60(b)).

• If Vs,req≤2Vc

Smax = d/2 ≤24”

• If 2Vc<Vs,req ≤4Vc

Smax=d/4 ≤12”

 Also:  Sreq=

𝐵𝑡𝑔𝑧𝑒

𝑊𝑣

ϕ−𝑊𝑑

• Eq. 50.64

 Maximum shear occurs at supports. ACI 318 allows the

maximum shear to be reduced to take into account the compression stress that results from the reaction and helps to suppress the tensile cracks.

 As shear increases, the stirrups are spaced to meet the

• demand. However, max spacing limits are imposed to

insure the stirrups adequately intersect the shear cracks. The stirrups shear strength cannot exceed 4 times the concrete shear strength. The max shear capacity of a concrete beam with stirrups is given as 5 times the concrete shear strength. If this is less than the shear demand (Vu), a larger cross section must be used.

 Typical properties of steel are listed in CERM

Table 58.1. The important properties are E = 29,000 ksi, Fy and Fu. Shapes found in Table 58.2.

 Compactness

• A cross section that can reach its plastic moment

without buckling locally is called a Compact

• Section. Most hot-rolled shapes are compact.

Compactness depends on the steel strength; in fact, compact sections have thicker webs and

• flanges. Compact sections are permitted higher

allowable strengths. Most rolled wide-flange shapes are compact. To check whether a shape is compact or not, check the following equations.

 The plastic moment of a compact section is given

by:

Mp=FyZ

• Eq. 59.1

 Where Z is the plastic Modulus of the cross

• section. Provided that the beam is adequately

braced to prevent lateral torsional buckling, the nominal moment capacity is:

Mn=Mp

For design strength, the nominal strength mist be modified by: ϕ = 0.90 (LRFD) or 1.67 (ASD)

 Lateral Torsional Buckling (LTB)

• If the compression flange is not continuously

braced, the beam may buckle laterally prior to reaching its plastic moment. This buckling is called lateral torsional buckling. The beam can be continuously braced. The distance between bracing points is called the unbraced length, Lb. The greater the unbraced length, the more susceptible the member is to LTB.

 The maximum unbraced length that can be

used and still have a compact section reach its plastic moment is given as: If Lb>Lp, the nominal moment capacity of the cross section must be reduced to prevent LTB.

 Shear

• For strong axis bending, the shear capacity of a

steel I-shaped beam is provided by the web as this cross section element is in-line with the shear force. The design shear capacity of the cross section based on yielding is given as:

 Determine the design moment capacity

(ϕ=0.90) for the reinforced concrete cross section shown. Assume f’c = 4000 psi and Grade 60 reinforcing steel.

Solution: As = 2(0.79) = 1.58 in² (or use Table 50.3) a=

𝐵𝑡𝑔𝑧 0.85𝑔′

𝑑𝑐 = 1.58(60𝑙𝑡𝑗)

0.85(4𝑙𝑡𝑗)(9) = 3.09 in

Mn=Asfy(d-𝑏

2) = 1.58(60ksi)(17”-3.09" 2 ) = 1465 k-in

ϕMn = 0.90(1465) = 1319 k –in = 110 k-ft

 Beam practice problems!  Columns – Beam’s cousin.