wsd WORKING STRESS DESIGN

DESIGN OF BEAMS FOR FLEXURE USING WSD (ALTERNATE DESIGN METHOD) Introduction In working-stress design a margin of safety is provided by permitting calculated flexure stresses to reach only a certain percentage of the ultimate strength of the concrete or of the yield strength of the reinforcing steel. These percentages are sufficiently small so that an approximately linear relation exists between the stress and strain in the concrete as well as in the reinforcing. The NSCP permits the design of reinforced concrete members using service loads and the working-stress design method, except that the method is now referred to as the Alternate Design Method. Section 424.4 of the NSCP provides expressions for the permissible service load stresses. The maximum permissible concrete compressive stress in the extreme fiber of a member is 0.45fc' . The maximum allowable tensile stress in the reinforcing is 140 MPa for grade 275 reinforcement and 170 MPa for grade 415 reinforcement or greater. Derivation of WSD Design Formulas Reference is made to Figure 4.1 for the beam cross section and the terms that will be used in the derivation. The steel is once again transformed into an equivalent area of fictitious concrete that can resist tension. This area is nA s . In WSD the most economical design possible is referred to as balanced design. A beam designed by this method will, under full service load, have its extreme fibers in compression stressed to their maximum permissible value fc and its reinforcing bars stressed to their maximum permissible value fs . Balanced design is the situation assumed in the derivation of the design formulas for WSD. b

fc kd/3 kd

C = (1/2)fcbkd

NA jd

d

h

d-kd As T = A s fs fs/n Original Beam Section

Stress Diagram

Figure 4.1 Derivation of WSD Design Formulas

The design formulas are derived on the basis of a consideration of the internal couple consisting of the two forces C and T. Once again, the total compression C equals the compression area bkd times the average compression fc / 2 and T equals A s fs . The sum of the horizontal forces in a beam in equilibrium is obviously zero, and thus C = T. The internal moment M can be written as Cjd or Tjd, and these are equated to the external moment M and the resulting expressions are solved for the beam dimensions and the steel area required. 25

As a straight-line variation of stress is assumed from fc to fs / n , the following ratio can be written and from it the design value of k obtained:

fc   fs / n  fc  kd d fc k fc   fs / n  In a similar manner j is determined: kd jd  d  3 k j 1 3 Now, using the internal couples, M  Cjd 1 M  fcbkd  jd 2 2M bd2  fckj M  Tjd M  A s fs jd

As 

M fs jd

(4.1)

(4.2)

(4.3)

(4.4)

Note: These equations (4.1 through 4.4) were derived for rectangular sections and they do not apply to sections where the compression area is not rectangular or to sections with compression reinforcing. Differences between Investigation and Design problems There are two general types of formulas that are presented for beams: those used for investigation or review and those used for design. In investigation problems the physical dimensions of the beam and the number, sizes, and placement of bars are given. It is desired to determine the flexural stresses for certain moments or the permissible moment a section can carry for certain allowable stresses. Design problems are those in which the member sizes are to be determined based on certain external moments and certain allowable steel and concrete fiber stresses. It is very important to realize that some of the formulas do not apply to both investigation and design. For this reason the formulas for the investigation and design of rectangular beams are repeated here.

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For investigation of rectangular beams: k  2n  (n)2  n

k 3 2M fc  2 bd kj M fs  A s jd j 1

For design of rectangular beams: k

fc fc   fs / n 

k 3 2M bd2  fckj j 1

As 

M fs jd

Miscellaneous topics Before the design of an actual beam is attempted, it is important to discuss the following topics: 1. Estimated Beam Weight. The weight of the beam to be selected must be included in the calculation of the bending moment to be resisted, as the beam must support itself as well as the external loads. 2. Beam proportions. Unless architectural or other requirements dictate the proportions of reinforced concrete beams, the most economical beam sections are usually obtained for the shorter beams (up to 6m or 7.5m in length) when the ratio of b to d is in the range of 1/2 to 2/3. For longer spans better economy is usually obtained if deep, narrow sections are used. The depth may be as large as 3 or 4 times the widths. 3. Selection of bars. After the required reinforcing area is calculated, the selection of bar sizes will follow. For the usual situations bars of sizes 36mm and smaller are practical. It is usually convenient to use bars of one size only in a beam, although occasionally two sizes will be used. 4. Cover. The reinforcing for concrete members must be protected from the surrounding environment; that is, fire and corrosion protection needs to be provided. To do this the reinforcing is located at certain minimum distances from the surface of the concrete so that a protective layer of concrete, called concrete cover, is provided. In addition the concrete cover improves the bond between the concrete and the steel. In Section 407.8.3.1 of the NSCP, minimum permissible concrete cover is given for reinforcing bars under different conditions. Values are 27

given for reinforced concrete beams, columns, and slabs, for members exposed to weather and earth, and so on. 5. Minimum spacing of bars. Section 407.7.1 of NSCP states that the minimum clear spacing between parallel bars in a layer be db but not less than 25 mm. Where parallel reinforcement is placed in two or more layers (Section 407.7.2), bars in the upper layer shall be placed directly above bars in the bottom layer with clear distance between layers not less than 25mm. Please refer to Section 407.7.3 until 407.7.5 for the spacing limitations of other conditions. Minimum Percentage of Steel Sometimes due to architectural or functional requirements beam dimensions are selected that are much larger than are required for bending alone. Such members theoretically require very small amounts of reinforcing. There is actually another possible mode of failure that can occur in very lightly reinforced beams. If the ultimate resisting moment of the section is less than its cracking moment, the section will fail immediately when a crack occurs. This type of failure would occur without warning, and the NSCP (Section 410.6) provides a minimum steel percentage equal to f ' 1.4 min  c  4fy fy so, Asmin = ρminbd And when As < Asmin, use smaller between 4/3As or Asmin. Example 4.1 Design the beam shown in Figure 4.1 for moment only. Compute stresses in the resulting section by using the review formulas. Use fc'  21 MPa, fy  276 MPa and n = 9.

w = 20 kN/m

5.50 m

Figure 4.1 Example 4.1

Solution: 1. Compute Design Moment, Assume beam weight, wbeam  5 kN/m

wL2 25(5.50)2   94.53 kN.m 8 8 2. Determine Allowable Stresses, fc  0.45fc'  0.45(21)  9.45 MPa M

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fs  140 MPa (for Grade 275) 3. Compute Design Constants, fc 9.45 k   0.378 fc   fs / n 9.45  140 / 9  k 0.378 1  0.874 3 3 4. Estimate beam size, 2M Assume d  1.5b bd2  fckj j 1

2(94.53x106 ) 9.45(0.378)(0.874) b  343.05 mm; d  514.58 mm (b)(1.5b)2 

Say use 350mm x 600mm (d = 540mm) 5. Compute required steel reinforcements, M 94.53x106 As    1431 mm2 fs jd 140(0.874)(540) Using 20mm bars, ( Ab  314 mm2 ) A s 1431   4.56 Ab 314 Say use 5 – 20mm bars in one layer 6. Check Spacing of bars b  2c  2ds  ndb s   db  25mm n 1 350  2(40)  2(10)  5(20) s  37.50 mm > 25mm Ok! 5 1 7. Check Weight of Beam, wbeam  23.50(0.35)(0.60)  4.935 kN/m < wassumed  5 kN/m Ok! 8. Draw Section Detail, n

350mm

540mm

600mm 5 - 20mm

60mm

Figure 4.2 Section at Midspan

Checking Stresses:

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1. Compute steel ratio  ,   A s  5  (20)2   1571 mm2 4  A 1571  s   0.008311 bd 350(540) 2. Compute k,

k  2n  (n)2  n k  2(0.008311x9)  (0.008311x9)2  (0.008311x9)

k  0.319 3. Compute j, k 0.319 j 1 1  0.894 3 3 4. Compute stresses, 24.935(5.50)2 M  94.29 kN.m 8 M 94.29x106 fs    124.32 MPa < 140 MPa A s jd 1571(0.894)(540) fc 

Ok!

2M 2(94.29x106 )   6.48 MPa < 9.45 MPa Ok! bd2kj 350(540)2 (0.319)(0.894)

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