FLOW CHART FOR DESIGN OF BEAMS

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1 FLOW CHART FOR DESIGN OF BEAMS Write Known Data Estimate self-weight of the member. a. The self-weight may be taken as 10 percent of the applied dead UDL or dead point load distributed over all the length. b. If only live load is applied, self-weight may be taken equal to 5 percent of its magnitude. c. In case only factored loads are given, selfwt. may be taken equal to 3 % of the given loads.

2 Calculate Factored Loads Draw B.M. and S.F. Diagrams Calculate C b For Each Unbraced Segment Find M u,max, V u,max, L b for each segment and guess which segment is the most critical. Design this segment first and then check for others.

3 Assume the section to be compact without LTB in the start and calculate Z x accordingly. 6 M u 10 Assumed Z x,req = φ b = 0.9 ; F F y = 250 MPa for A36 steel Z sel Z req φ b y Selection of Section Minimum weight F d d y l min d min = For max = L/ = L /22 for A36 steel and simply supported beams

4 For max required to be lesser than L /360, like L /500 or L /800, find (I x ) req from the deflection formula, with only the live load acting, and select section such that I x (I x ) req. Method 1 Use Of Selection Tables 1. Enter the column headed Z x and find a value equal to or just greater than the plastic section modulus required. 2. The beam corresponding to this value in the shape column and all beams above it have sufficient flexural strength based on these parameters.

5 3. The first beam appearing in boldface type (top of a group) adjacent to or above the required Z x is the lightest suitable section. 4. If the beam must have to satisfy a certain depth or minimum moment of inertia criterion, proceed up the column headed Shape until a beam fulfilling the requirements is reached. 5. If C b > 1.0, use L m in place of L p for the approximate selection. 6. If L b is larger than L m of the selected section, use the unbraced design charts.

6 7. Apply moment capacity, shear capacity, deflection and all other checks. 8. The column headed φ b M p may also be used in place of the Z x column in the above method. Method 2: Use Of Unbraced Design Charts This method is applicable in cases where the above method is not fully applicable and L b L p. The design charts are basically developed for uniform moment case with C b = 1.0.

7 Following notation is used to separate full plastic, inelastic LTB, and elastic LTB ranges: Solid Circle Hollow Circle represents L p represents L r 1. According to M u in kn-m units and L b in meters, enter into the charts. 2. Any section represented by a curve to the right and above ( ) the point selected in No. 1 will have a greater allowed unbraced length and a greater moment capacity than the required values of the two parameters.

8 3. A dashed line section is not an economical solution. If dashed section is encountered while moving in top-right direction, proceed further upwards and to the right till the first solid line section is obtained. Select the corresponding section as the trial section, and it will be the lightest available section for the requirements. 4. If C b > 1.0, use M u,req = M u / C b but check that the selected section has φ b M p > M u.

9 Check the three conditions of compact section for internal stability, namely, 1. web continuously connected with flange, 2. flange stability criterion, and 3. web stability criterion. If any one out of the above three is not satisfied, revise the section. Either calculate L p, L r, and L m or find their values from beam selection tables.

10 1.76 r L p = (mm) E y F yf = 0.05 r y (m) for A36 steel L r = 1.95r ts E 0.7F y J c S h x o F E y S x J h c o 2 M r = 0.7F y S x /10 6 kn-m 2 r ts c I y = = S C x w I y 2S for doubly symmetric I-sections h o x = 1.0 for a doubly symmetric I-shape h o = d t f

11 BF = M L p r M L p r M p Cb 1 L m = L p + L BF r Cb Calculate design flexural strength: 1- If L b L m M n = M p = Z x F y / 10 6 (kn m) 2- If L m < L b L r M n = C b [M p BF(L b L p )] M p (kn m) 3- If L b > L r M n = C b F cr S x M p

12 where F cr = C π b L r b ts 2 E S J c x h o L r b ts 2 Bending strength check: M u φ b M n If not satisfied revise the trial selection. OK Shear check:

13 h For 2.24 E / (= 63.4 for A36 steel) t w C v = 1.0 F yw φ v V n = F yw A w C v (kn) 1000 If not satisfied, revise the section. Deflection check: Find act due to service live loads. act L/360 or other specified limit OK

14 Check self-weight: Calculated self weight 1.2 assumed self weight OK Otherwise, revise the loads and repeat the calculations. Write final selection using standard designation.

15 1- For buildings, L/d ratio is usually limited to a maximum of 5500 / F y. L /d 5500 / F y d min = F y L /5500 (L /22 for A36 steel) 2- For bridge components and other beams subjected to impact or vibratory loads, L /d For roof purlins, DEFLECTIONS L /d 6900 / F y (27.5 for A36 steel, sometimes relaxed to a value equal to 30)

16 The actual expected deflections may be calculated using the mechanics principles. However, results given in Manuals and Handbooks may also be used directly. Some of the typical deflection formulas are reproduced here. 1- For uniformly loaded and simply supported beams, max = 5w l 384 l 4 EI

17 2- For uniformly loaded continuous beams, 5l midspan = 2 [M c 0.1(M a + M b )] 48 EI Where M c = magnitude of central moment, M a, M b = magnitude of end moments. 3- For simply supported beams subjected to point load (refer to Figure 4.16), P a ( midspan = 3 ) 2 2 l a 4 where a L/ 2 12EI

18 a P b 4- For overhanging part of beam subjected to UDL, l a No Load w per meter w a max ( 4l + 3a) 3 24EI

19 5. For the above case, with UDL also present within supports, wa max ( a l l + 3a ) 24EI 6- For overhanging part of beam subjected to point load, P No load l a max = P a 2 ( l + a) 3EI

20 Example 4.1: BEAMS WITH CONTINUOUS LATERAL SUPPORT Design a 7m long simply supported I-section beam subjected to service live load of 5 kn/m and imposed dead load of 6 kn/m, as shown in Figure The compression flange is continuously supported. Use (a) A36 steel and (b) steel with f y = 345 MPa and permissible live load deflection of span / 450.

21 Solution: In beams with continuous lateral support, unbraced length is not applicable or it may be assumed equal to zero in calculations. Assumed self weight = 10% of superimposed DL = 0.6 kn/m w u = 1.2D + 1.6L = = kn/m

22 15.92 kn/m kn 7 m kn kn kn S.F.D kn-m B.M.D.

23 If the beam is continuously braced, C b value is not applicable but may be considered equal to 1.0 in case it is required in the formulas. M u = M max = kn-m V u = V max = kn Compression flange continuously braced. (a) A36 Steel Assuming the section to be internally compact and knowing that there is no LTB,

24 6 6 M u (Z x ) req = = = mm 3 φ F b y 7000 d min = = 318 mm 22 Z sel Z req Selection of section Min. weight section d d min Consulting beam selection tables (Reference-1), following result is obtained: W provides sufficient strength but depth is little lesser.

25 Select W ; Z x = mm 3 Check internal compactness of section as under: 1. web is continuously connected OK bf 2. 2t = 7.5 < λ p = 10.8 OK h f 3. = 53.3< λ p =107 OK t w The section is internally compact. Continuously braced conditions imply: L b < L p and C b = not included in formulas

26 φ b M n = φ b M p = / kn-m M u = kn-m < φ b M p OK h/t w =53.3 < 63.4 shear yield formula is applicable φ v V n = F yw A w C v = = kn

27 V u = (kn) < φ v V n OK act due to service live load = 5 w l 4 l 384 EI ( ) = = 9.44 mm all = L/360 = 7000 / 360 = mm act < all OK Self-weight = = 0.32 kn/m

28 This is lesser than assumed self-weight of 0.6 kn/m in the start. OK Final Selection: W (b) Steel With F y = 345 MPa Assuming the section to be internally compact and knowing that there is no LTB, 6 6 M u (Z x ) req = = = mm 3 φb Fy d min = = 438 mm 16

29 This d min is much larger, so the direct deflection criteria will be used. I req = l 5 = wll 384 E w L 4 = l EI = mm 4 3 Z sel Z req Selection of section Min. weight section I I min

30 Consulting beam selection tables (Reference-1), following result is obtained: W provides sufficient strength and moment of inertia. I x = mm 4 ; Z x = mm 3 Check internal compactness of section as under: 1. web is continuously connected OK 2. b f / 2t f = 5.7 < λ p = 9.1 OK 3. h / t w = 46.2 < λ p =90.5 OK The section is internally compact.

31 Continuously braced conditions imply: and C b = N.A. L b < L p φ b M n = φ b M p = / kn-m M u = kn-m < φ b M p OK 2.24 E / F y h/t w =53.3 < = 53.9 shear yield formula is applicable

32 φ v V n = F yw A w C v = = kn V u = (kn) < φ v V n OK act < all Already satisfied OK Self-weight = = 0.28 kn/m

33 This is lesser than assumed self-weight of 0.6 kn/m in the start. OK Final Selection: W Example 4.2: BEAMS WITH COMPRESSION FLANGE LATERALY SUPPORTED (BRACED) AT REACTION POINTS ONLY Design the beam of Figure 4.19 with the lateral bracing only provided at the reaction points.

34 P D = 60 kn P L = 50 kn w D = 8 kn/m w L = 10 kn/m 3.5m 3.5m

35 Solution: Self load = = 1.7kN/m w u = = kn/m P u = = 152 kn The factored loading and the shear force and bending moment diagrams are shown in Figure 4.20.

36 P u = 152 kn w u = kn/m kn 3.5m 3.5m kn kn S.F.D kn kn-m B.M.D kn-m

37 Calculation of C b : M B = M max = kn-m M A = M C = /2 = kn-m C b = R m = (where R m = 1.0) M max = kn-m : V max = kn : L b = 7 m

38 Assuming the section to be internally compact with no LTB, (Z x ) req = d min. M u φ b 6 10 F y = = mm 3 = L/22 = 7000/22 = 318 mm Z sel Z req Selection of section Min. weight section d d min Using beam selection tables of Reference-1,

39 Trial section: W M p = kn-m, M r = kn-m, BF = 64.99, L r = 5.10 m L p = 1.67 m : L m = = 3.33 m L b >> L m revise the section using beam selection charts of Reference-1 For L b = 7m and M u,eq = / 1.24 = knm, from design charts of Reference-1,

40 Select W φ b M p = kn-m > M u d > d min OK OK Check internal compactness of section: 1- web is continuously connected OK b f 2- = 7.7< λ p = 10.8 OK 2t h f 3- = 35.9 < λ p = 107 OK t w The section is internally compact.

41 L p L r BF M p = 3.11 m = m = kn = kn-m M C 1 BF Cb p b L m = L p + L r L m = = 6.72 m L m < L b L r : φ b M n =C b φ b [M p BF(L b L p )] φ b M p = [ ( )] = kn-m

42 M u = kn-m φ b M n OK h/t w = 35.9 < φ v V n = F yw A w C v = = kn V u = kn < φ v V n OK

43 act = ( 7000) = = 8.44 mm L/360 = mm ( 0.75 ( 7000) ( 3500) ) act < L/360 OK Self weight = = kn/m 1000 < 1.7 kn/m assumed in the start OK Final Selection: W

44 Example 4.3: BEAMS WITH COMPRESSION FLANGE LATERALLY SUPPORTED AT INTERVALS Design the beam of Figure 4.21, ignoring self-weight and deflection check. Compression flange is laterally supported at points A, B and C. Solution: The factored loading and the shear force and bending moment diagrams are shown in Figure 4.21.

45 w D =3 kn/m w L =2 kn/m P D =60 kn P L =50 kn P D =35 kn P L =30 kn A C B D 7m 8m 6m

46 w u =6.8 kn/m P u =152 kn P u =90 kn kn kn S.F.D kn-m (+) ( ) B.M.D kn-m

47 M u = M max = 540 kn-m M u = kn-m for portion AC M u = kn-m for portions CB and BD V u = V max = kn L b for portion AC = 7 m L b for portion CB = 8 m L b for portion BD = 6 m Calculation of C b : i) Portion AC

48 M = x 6.8 x 2 /2 M max = kn-m M A = kn-m (x = 1.75 m) M B = kn-m (x = 3.50 m) M C = kn-m (x = 5.25 m) C b = R m (where R m = 1.0) = 1.49 ii) Portion CB + M = x (x starts from the end B)

49 M max = kn-m M A = kn-m (x = 2 m) M B = kn-m (x = 4 m) M C = kn-m (x = 6 m) C b = R m (where R m = 1.0) = 2.22 iii) Portion BD C b =

50 Portion (M u ) max (kn-m) (V u ) max (kn) L b (m) C b AC CB BD

51 Greater value of M u critical. makes a segment more Similarly, a longer unbraced length reduces the member capacity. However, smaller value of C b is more critical for a particular segment. The effect of these three factors together determines whether a selected segment is actually more critical than the other segments.

52 The decision about which part of the beam is more critical for design depends on experience but may not always be fully accurate. If a wrong choice is made for this critical section, the procedure will correct itself. Only the calculations will be lengthy in such cases with the end result being always the same. For this example, assume that portion BD is more critical and first design it. Later on, check for the other two portions.

53 l d min = = 682 mm Depth for portion AB may be relaxed a little due to the presence of cantilever portion. However, the portion BD may have its own larger depth requirements. Portion BD Assuming the section to be internally compact with no LTB, u Assumed (Z x ) req = = φb Fy 0.9 = mm 3 M

54 From beam selection tables, the trial section is: W L p = 1.75 m, L r = 5.33 m, L b = 6 m, L m = L p (as C b = 1.0) L b >> L m use beam selection charts For M u = 540 kn-m and L b = 6m From unbraced beam curves W is the first choice but select W to satisfy d min.

55 Trial Section: W Actual d = 678 (still a little lesser but within permissible limits) L p = 2.62 m ; L r = 7.67 m ; BF = 77.0 kn ; L m = L p ; M p = kn-m ; M r = kn-m Compactness Check: 1- web is continuously connected OK 2- b f /2t f = 7.8 λ p = 10.8 OK 3. h/t w = 52.7 λ p = 107 OK Section is internally compact.

56 L m < L b L r φ b M n =C b φ b [M p BF(L b L p )] φ b M p = [ (6 2.62)] = kn-m kn-m M u = 540 kn-m < φ b M n OK h/t w = φ v V n = = kn

57 V u = kn φ v V n OK Check for Portion AC L b = 7m ; C b = 1.49 L p < L b L r φ b M n = C b φ b [M p BF(L b L p )] φ b M p = [ (7 2.62)] = kn-m kn-m

58 φ b M n = kn-m M u = kn-m < φ b M n OK Check For Portion CB For W : S x = mm 3 M p = kn-m ; M r = kn-m

59 φ b M p = kn-m J = mm 4 ; r y = 52.6 mm L b = 8m ; C b = 2.22 I y = mm 4 ; r y = 52.6 mm d = 678 mm ; t f = 16.3 mm h o r ts = d t f = = mm = I y h 2S o x = = mm

60 L b >L r L r b ts = φ b M n = φ b C π b L r b ts 2 E 2 = J c S h x o L r b ts 2 S x / 10 6 φ b M p = 2.22 π /10 6 φ b M n = kn-m kn-m Hence φ b M n = φ b M p = kn-m and portion BD is most critical as expected earlier.

61 M u = 540 kn-m < φ b M n OK Deflection Check: Requires detailed calculations for actual that is not asked for in this example. Very approximate calculations are performed as under: 2 Pa l + a max for cantilever 3EI where a is the cantilever length and l is the span = 30,000 6,000 2 ( ) ( 15, ,000) 3 200, ,

62 max = mm (will be reduced by loads on span l ) Also max considering one end of the overhang as fixed = 3 Pa 3EI 30,000 6, , , = 9.08 mm l/360 = 6000/360 = 16.67mm Deflection may be critical, detailed calculations are recommended.

63 Self weight = = 1.23 kn/m (asked to be ignored in the problem statement) Final Selection: W Example 4.4: Select suitable trial section for a 8m long simply supported W-section beam subjected to service live load of 10 kn/m and imposed dead load of 4 kn/m. The compression flange is continuously supported in the lateral direction and the deflection is limited to span/1500.

64 Solution: Assumed self weight = 10% of superimposed dead load = 0.4 kn/m say 0.8 kn/m as live load is greater and deflection requirements are strict. w u = 1.2D + 1.6L = = kn/m M u = M max = = kn-m 8 2

65 Assuming the section to be internally compact and knowing that there is no LTB, M u (Z x ) req = = φ b 10 F y = mm d min = = 364 mm 22 act due to service live load = ( 8,000) ,000 I 6 4 min w 4 l 5 l 384 EI = 8,000 / 1500

66 I min = 50, mm 4 Z sel mm 3 Selection of section Minimum weight section d 364 mm From beam selection tables, Trial section no.1: W I min = 50, mm 4 d = 352 mm < d min I x = 12, mm 4

67 Moving upwards in the same tables, Trial section no.2: W d = 403 mm > d min I x = 15, mm 4 Trial section no.3: W d = 355 mm < d min I x = 14, mm 4 Trial section no.4: W Ix = 21, mm 4

68 Skipping W and W460 60, Trial section no.5: W I x = 35, mm 4 Trial section no.6: W I x = 41, mm 4 Trial section no.7: W I x = 56, mm 4 Trial Section: W After selection of the trial section, all the checks are to be performed. This part of the exercise is left for the reader to be completed.

69 End of Chapter 4

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