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.
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.
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/360 5500 = L /22 for A36 steel and simply supported beams
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.
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.
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.
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.
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.
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.
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 1+ 1+ 6.76 0.7F 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
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
where F cr = C π b L r b ts 2 E 2 1+ 0.078 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:
h For 2.24 E / (= 63.4 for A36 steel) t w C v = 1.0 F yw 0.9 0.6 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
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.
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 20 3- For roof purlins, DEFLECTIONS L /d 6900 / F y (27.5 for A36 steel, sometimes relaxed to a value equal to 30)
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
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
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
5. For the above case, with UDL also present within supports, wa max ( 2 3 3 4a 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
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 4.18. 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.
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 = 1.2 6.6 + 1.6 5 = 15.92 kn/m
15.92 kn/m 55.72 kn 7 m 55.72 kn 55.72 kn 55.72 kn S.F.D. 97.51 kn-m B.M.D.
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 = 97.51 kn-m V u = V max = 55.72 kn Compression flange continuously braced. (a) A36 Steel Assuming the section to be internally compact and knowing that there is no LTB,
6 6 M u 10 97.51 10 (Z x ) req = = = 433.4 10 3 mm 3 φ F 0.9 250 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: W310 32.7 provides sufficient strength but depth is little lesser.
Select W360 32.9; Z x = 544 10 3 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
φ b M n = φ b M p = 0.9 544 10 3 250 / 10 6 121 kn-m M u = 97.51 kn-m < φ b M p OK h/t w =53.3 < 63.4 shear yield formula is applicable 0.9 0.6 1000 0.9 0.6 1000 φ v V n = F yw A w C v = 250 349 5.8 1.0 = 273.27 kn
V u = 55.72 (kn) < φ v V n OK act due to service live load = 5 w l 4 l 384 EI ( ) 4 5 5 7000 = = 9.44 mm 4 384 200000 8280 10 all = L/360 = 7000 / 360 = 19.44 mm act < all OK Self-weight = 32.9 9.81 1000 = 0.32 kn/m
This is lesser than assumed self-weight of 0.6 kn/m in the start. OK Final Selection: W360 32.9 (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 10 97.51 10 (Z x ) req = = = 314.0 10 3 mm 3 φb Fy 0.9 345 7000 d min = = 438 mm 16
This d min is much larger, so the direct deflection criteria will be used. I req = l 5 = 450 384 3 5 450 wll 384 E w L 4 = l EI 5 450 384 5 7000 200000 = 5024 10 4 mm 4 3 Z sel Z req Selection of section Min. weight section I I min
Consulting beam selection tables (Reference-1), following result is obtained: W310 28.3 provides sufficient strength and moment of inertia. I x = 5410 10 4 mm 4 ; Z x = 405 10 3 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.
Continuously braced conditions imply: and C b = N.A. L b < L p φ b M n = φ b M p = 0.9 405 10 3 345 / 10 6 125.8 kn-m M u = 97.51 kn-m < φ b M p OK 2.24 E / F y h/t w =53.3 < = 53.9 shear yield formula is applicable
0.9 0.6 1000 φ v V n = F yw A w C v 0.9 0.6 1000 = 345 309 6.0 1.0 = 345.40 kn V u = 55.72 (kn) < φ v V n OK act < all Already satisfied OK Self-weight = 28.3 9.81 1000 = 0.28 kn/m
This is lesser than assumed self-weight of 0.6 kn/m in the start. OK Final Selection: W310 28.3 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.
P D = 60 kn P L = 50 kn w D = 8 kn/m w L = 10 kn/m 3.5m 3.5m
Solution: Self load = 0.1 8 + 7 60 = 1.7kN/m w u = 1.2 9.7 + 1.6 10 = 27.64 kn/m P u = 1.2 60 + 1.6 50 = 152 kn The factored loading and the shear force and bending moment diagrams are shown in Figure 4.20.
P u = 152 kn w u = 27.64 kn/m 172.74 kn 3.5m 3.5m 172.74 kn 76 172.74 kn S.F.D. 76 172.74 kn 169.30 kn-m B.M.D. 266.00 kn-m
Calculation of C b : M B = M max = 435.3 kn-m M A = M C = 172.4 1.75 27.64 1.75 2 /2 = 259.97 kn-m C b = 12.5 435.30 R m = 1.24 6.5 435.30 + 6 259.97 3.0 (where R m = 1.0) M max = 435.3 kn-m : V max = 172.74 kn : L b = 7 m
Assuming the section to be internally compact with no LTB, (Z x ) req = d min. M u φ b 6 10 F y 435.3 10 0.9 250 6 = = 1935 10 3 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,
Trial section: W610 82 M p = 549.00 kn-m, M r = 327.25 kn-m, BF = 64.99, L r = 5.10 m 549.00 1.24 1 64.99 1.24 L p = 1.67 m : L m = 1.67 + = 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 = 435.3 / 1.24 = 351.05 knm, from design charts of Reference-1,
Select W410 100 φ b M p = 479.25 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.
L p L r BF M p = 3.11 m = 10.00 m = 28.53 kn = 532.50 kn-m M C 1 BF Cb 532.50 1.24 1 28.53 1.24 p b L m = L p + L r L m = 3.11 + = 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 = 1.24 0.9 [532.50 28.53(7.00 3.11)] = 470.44 kn-m
M u = 435.3 kn-m φ b M n OK h/t w = 35.9 < 63.4 0.9 0.6 1000 φ v V n = F yw A w C v 0.9 0.6 1000 = 250 415 10.0 1.0 = 560.25 kn V u = 172.74 kn < φ v V n OK
act = 5 384 ( 7000) 10 200000 39700 10 = 3.94 + 4.50 = 8.44 mm L/360 = 19.44 mm 4 4 50000 3500 + 12 200000 39700 10 4 2 2 ( 0.75 ( 7000) ( 3500) ) act < L/360 OK 100 9.81 Self weight = = 0.981 kn/m 1000 < 1.7 kn/m assumed in the start OK Final Selection: W410 100
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.
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
w u =6.8 kn/m P u =152 kn P u =90 kn 81.56 kn 208.04 kn 81.56 33.96 90 90 S.F.D. 118.04 404.32 kn-m 118.04 (+) ( ) B.M.D. 540.00 kn-m
M u = M max = 540 kn-m M u = 404.32 kn-m for portion AC M u = 540.00 kn-m for portions CB and BD V u = V max = 118.04 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
M = 81.56 x 6.8 x 2 /2 M max = 404.32 kn-m M A = 132.32 kn-m (x = 1.75 m) M B = 243.81 kn-m (x = 3.50 m) M C = 334.48 kn-m (x = 5.25 m) 12.5 404.32 3 132.32 + 4 243 C b = R m 2.5 404.32 3.0 (where R m = 1.0) = 1.49 ii) Portion CB + M = 540 118.04 x (x starts from the end B).81 + 3 334.48
M max = 540.00 kn-m M A = 303.92 kn-m (x = 2 m) M B = 67.84 kn-m (x = 4 m) M C = 168.24 kn-m (x = 6 m) C b = R m 2.5 540.00 12.5 540.00 + 3 303.92 + 4 67.84 3.0 (where R m = 1.0) = 2.22 iii) Portion BD C b = 1.0 + 3 168.24
Portion (M u ) max (kn-m) (V u ) max (kn) L b (m) C b AC 404.32 81.56 7 1.49 CB 540.00 118.04 8 2.22 BD 540.00 90.00 6 1.00
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.
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.
l 15000 d min = = 682 mm 22 22 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 = 2400 10 3 mm 3 M 10 6 6 540 10 250
From beam selection tables, the trial section is: W610 92 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 W610 125 is the first choice but select W690 125 to satisfy d min.
Trial Section: W690 125 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 = 999.50 kn-m ; M r = 610.75 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.
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 = 1.0 0.9 [999.50 77.0(6 2.62)] = 665.22 kn-m 899.55 kn-m M u = 540 kn-m < φ b M n OK h/t w = 52.7 63.4 0.9 0.6 1000 φ v V n = 250 678 11.7 1.0 = 1070.9 kn
V u = 118.04 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 = 1.49 0.9 [999.50 77.0(7 2.62)] = 887.92 kn-m 899.55 kn-m
φ b M n = 887.92 kn-m M u = 404.32 kn-m < φ b M n OK Check For Portion CB For W690 125: S x = 3490 10 3 mm 3 M p = 999.50 kn-m ; M r = 610.75 kn-m
φ b M p = 899.55 kn-m J = 117 10 4 mm 4 ; r y = 52.6 mm L b = 8m ; C b = 2.22 I y = 4410 10 4 mm 4 ; r y = 52.6 mm d = 678 mm ; t f = 16.3 mm h o r ts = d t f = 678 16.3 = 661.7 mm = I y h 2S o x 4410 10 4 661.7 = = 64.66 mm 2 3490 10 3
L b >L r L r b ts = φ b M n = φ b 8000 64.59 C π b L r b ts 2 E 2 = 123.86 J c 1+ 0.078 S h x o L r b ts 2 S x / 10 6 φ b M p = 2.22 π 1170000 1 2 2 0.9 1+ 0.078 123.86 3490 2 3 123.86 200000 3490 10 661.7 10 3 /10 6 φ b M n = 1137.10 kn-m 899.55 kn-m Hence φ b M n = φ b M p = 899.55 kn-m and portion BD is most critical as expected earlier.
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 + 6,000) 3 200,000 119,000 10 4
max = 31.76 mm (will be reduced by loads on span l ) Also max considering one end of the overhang as fixed = 3 Pa 3EI 30,000 6,000 3 200,000 119,000 10 3 4 = 9.08 mm l/360 = 6000/360 = 16.67mm Deflection may be critical, detailed calculations are recommended.
Self weight = 125 9.81 1000 = 1.23 kn/m (asked to be ignored in the problem statement) Final Selection: W690 125 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.
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 = 1.2 4.8 + 1.6 10 = 21.76 kn/m 21.76 8 M u = M max = = 174.08 kn-m 8 2
Assuming the section to be internally compact and knowing that there is no LTB, M u (Z x ) req = = φ b 10 F y = 773.7 10 3 mm 3 8000 d min = = 364 mm 22 act due to service live load = 5 384 ( 8,000) 10 200,000 I 6 4 min 174.08 10 0.9 250 6 w 4 l 5 l 384 EI = 8,000 / 1500
I min = 50,000 10 4 mm 4 Z sel 773.7 10 3 mm 3 Selection of section Minimum weight section d 364 mm From beam selection tables, Trial section no.1: W360 44 I min = 50,000 10 4 mm 4 d = 352 mm < d min I x = 12,100 10 4 mm 4
Moving upwards in the same tables, Trial section no.2: W410 46.1 d = 403 mm > d min I x = 15,600 10 4 mm 4 Trial section no.3: W360 51 d = 355 mm < d min I x = 14,200 10 4 mm 4 Trial section no.4: W460 52 Ix = 21,200 10 4 mm 4
Skipping W410 60 and W460 60, Trial section no.5: W530 66 I x = 35,100 10 4 mm 4 Trial section no.6: W530 74 I x = 41,000 10 4 mm 4 Trial section no.7: W610 82 I x = 56,200 10 4 mm 4 Trial Section: W610 82 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.
End of Chapter 4