ACTIONS/LOADS AND DISTRIBUTION CHAPTER 3
Actions / Loads • Actions / loads are a primary consideration in any building design. • It can be define as an external forces that a building must resist to provide reasonable performance (i.e., safety and serviceability) throughout the structure’s design working life. • The anticipated loads are influenced by a building’s intended use (occupancy and function), configuration (size and shape), and location (climate and site conditions). • Ultimately, the type and magnitude of design loads affect critical decisions such as material selection, construction details, and architectural configuration. • Thus, to optimize the value (i.e., performance versus economy) of the finished product, it is essential to estimate the design loads realistically.
Loading estimation Elements Type of actions Unit Slab ✔ Permanent action: (i) Selfweight of slab, (ii) Finishes and services, and (iii) Ceiling ✔ Variable action (depend on function of floor) kN/m2 Beam ✔ Permanent action: (i) Distribution from slab / roof, (ii) Selfweight of beam, and (iii) Brickwall ✔ Variable action: (i) Distribution from slab / roof kN/m Column ✔ Permanent action: (i) Distribution from beam, (ii) Selfweight of column ✔ Variable action: (i) Distribution from beam kN Foundation ✔ Permanent action: (i) Distribution from columns, (ii) Selfweight of footing ✔ Variable action: (i) Distribution from column kN
Loading estimation Permanent action of slab • Selfweight of slab = Slab thickness,h x unit weight of RC (25 kN/m3 ) • In common practices, minimum slab thickness = 100 mm and maximum slab span (short / long span) = 36h. • Finishes – Depending on floor finishes assigned by architect – refer to manufacture catalogue. (For Example) • Ceramic tile => [(31.68 kg x 9.81)/1000] / 1.44 m2 = 0.22 kN/m2 • 30 mm thk. cement render => [(1900 kg/m3 x (9.81/1000) x 0.030] = 0.56 kN/m2 • Total finishing load = 0.22 + 0.56 = 0.78 kN/m2 • Ceiling weight normally range = 0.3 – 0.5 kN/m2 (Depend on ceiling type)
Loading estimation Variable action of slab • Characteristic variable action, Qk acting on slab is caused by people, furniture, equipment etc, which the variation in magnitude time dependent is considered. • Examples of variable action as given in EC1 are shown in the table:
Loading estimation Selfweight of beam h b Rectangular beam Selfweight = b x h x Unit weight of concrete Flange beam beff beff bw bw hf h Selfweight = bw x [ h – hf ] x Unit weight of concrete
Loading estimation b l Selfweight of column s/w column = b x h x l x u/weight of concrete s/w footing = Bf x Lf x tf x u/weight of concrete Selfweight of foundation Brickwall load h B/wall load = 2.6 kN/m2 x wall height (h)
Loading distribution
Loading distribution (Floor) ▪ Actions that applied on a beam may consist of beams self- weight, permanent and variable actions from slabs, actions from secondary beams and other structural or non-structural members supported by the beam. ▪ The distribution of slab actions on beams depends on the slab dimension, supporting system and boundary condition. ▪ It is important to determine the type of slab using the following criteria:
Loading distribution (Floor)
Loading distribution (Floor) • There are alternatives methods which consider various support conditions and slab continuity. The methods are (i). Slab shear coefficient from Table 3.15 BS 8110, (ii). Yield line analysis and (iii). Table 63 Reinforce Concrete Designer’s Handbook by Reynold.
Loading distribution (Floor) Case 1 Case 2 Case 3 Case 4 Case 5 Case 6 Case 7 Case 8 Case 9
Example 3.1 (Loading distribution) Determine the characteristic permanent and variable actions on beam FB1 as shown in the above figure. A brickwall of 4 m height is sitting on the beam. Unit weight of concrete = 25 kN/m3 ; finishes = 1.0 kN/m2 ; brickwall = 2.6 KN/m2 ; variable action = 4.0 kN/m2 .
Solution of Example 3.1 Permanent action (Gk): (i) self‐weight = 0.25 x 0.7 x 25 = 4.38 kN/m (ii) Brickwall = 2.6 x 4 = 10.4 kN/m (iii) Slab: Gk = (0.15 x 25) + 1.0 = 4.75 kN/m2 Slab distribution = 0.5 x 4.75 x 6 = 14.25 kN/m Total Gk = 4.38 + 10.4 + 14.25 = 29.0 kN/m Variable action (Qk ) = 0.5 x 4.0 x 6 = 12 kN/m
Example 3.2 (Loading distribution) Determine the characteristic permanent and variable actions on beam B/1-3. Given the following data: Unit weight of concrete = 25 kN/m3 ; Finishes, ceiling and services = 2.0 kN/m2 ; Variable action (all slabs) = 3.0 kN/m2 . Solution Example 3.2
Loading distribution (Secondary beam) • Secondary beam will distributed load acting on it to the primary or main beam via it’s support reactions. R (kN) R1 R2
Loading distribution (Secondary beam) Determine the characteristic permanent and variable actions on beam B/1-3. Given the following data: Unit weight of concrete = 25 kN/m3 ; Finishes, ceiling and services = 2.0 kN/m2 ; Variable action (all slabs) = 3.0 kN/m2 . Consider the column 2/B is not exist and beam 2/B-C (200 x 450 mm) is a secondary beam. Column 2/B have been remove #Assignment
Loading distribution (Roof) • Typical Malaysian building roof type, • Load transfer mechanism of flat roof is similarly as slab load distribution. • For hip and gable roof (pitch roof), the roof load will transfer to the beam via roof truss system.
Loading distribution (Roof) Roof truss system • Permanent action, Gk • Roof tile / metal deck • Insulation (if applicable) • Selfweight of truss • Batten / purlin • Ceiling • Services • Variable action, Qk • 0.75 kN/m2 – 1.25 kN/m2 (For maintenance purposes) The weight of roofing material can be refer to manufacture catalogue depend on architectural design.
Loading distribution (Roof) • Type of roof truss system Cold-formed steel truss system => Span < 13 m Design by fabricator – roof load can be distribute to the four side of roof beam by using simplified approach. Load in x-x direction = 0.5 x n x Lx Load in y-y direction = 0.5 x n x Ly Lx Ly
Loading distribution (Roof) Hot rolled steel truss system => Span > 13 m Design by engineer – roof load can be distribute to the roof beam via truss support reaction at the respective side.
Loading distribution (Axial load) • The beam will distribute load into column in the form of axial load (kN) via summation of support reaction at particular beam column joint. Permanent & Variable actions Axial load in Column A = R1 + R3 + Column s/weight
Structural Analysis • The primary purpose of structural analysis is to establish the distribution of internal forces and moments over the whole part of a structure and to identify the critical design conditions at all sections. • In practice there are many different loads will act together and this has to be considered in the structure design. • Various combinations of the characteristic loads such as Gk , Qk and Wk ; and their partial factors of safety must be considered for the loading of the structure. • The RC beam must be designed to resist the ultimate bending moments and shear forces. In some cases, the beam is also required to resist torsional moments and axial forces. • The beam is required to achieve serviceability requirements such as deflection and cracking.
Loading arrangement (Load combination) ▪ In considering the combinations of actions, the relevant cases shall be considered to enable the critical design conditions to be established at all sections, within the structure or part of the structure considered. ▪ For simply supported beam, the analysis for bending and shear force can be carried out using statically determinate approach. For the ultimate limit state we need only consider the maximum load of 1.35Gk + 1.5Qk on the span. Clause 5.1.3 : MS EN 1992-1-1
Loading arrangement (Load combination) • For continuous beam, the following simplified load arrangements (based on National Annex) are recommended: • Load set 1: Alternate or adjacent spans loaded • Load set 2: All or alternate spans loaded Load set 1: Alternate or adjacent spans loaded (Continuous Beam) ▪ Alternate span carrying the design permanent and variable load (1.35Gk + 1.5Qk ), other spans carrying only the design permanent loads (1.35Gk ) ▪ Any two adjacent spans carrying the design permanent and variable loads (1.35Gk + 1.5Qk), all other spans carrying only the design permanent load (1.35Gk ) Section 5.1.3 : MS EN 1992-1-1
Loading arrangement (Load combination) Alternate spans loaded 1.35Gk + 1.5Qk 1.35Gk 1.35Gk + 1.5Qk 1.35Gk 1.35Gk + 1.5Qk 1.35Gk 1.35Gk + 1.5Qk 1.35Gk
Loading arrangement (Load combination) Adjacent Span Loaded 1.35Gk + 1.5Qk 1.35Gk 1.35Gk 1.35Gk + 1.5Qk 1.35Gk + 1.5Qk 1.35Gk + 1.5Qk 1.35Gk 1.35Gk 1.35Gk + 1.5Qk 1.35Gk + 1.5Qk 1.35Gk 1.35Gk
Loading arrangement (Load combination) Load set 2: All or alternate spans loaded (Continuous Beam) ▪ All span carrying the design permanent and variable loads (1.35Gk+ 1.5Qk ) ▪ Alternate span carrying the design permanent and variable load (1.35Gk+ 1.5Qk ), other spans carrying only the design permanent loads (1.35Gk ) 1.35Gk + 1.5Qk 1.35Gk + 1.5Qk 1.35Gk + 1.5Qk 1.35Gk + 1.5Qk 1.35Gk + 1.5Qk 1.35Gk + 1.5Qk 1.35Gk 1.35Gk 1.35Gk 1.35Gk + 1.5Qk 1.35Gk + 1.5Qk 1.35Gk
Loading arrangement (Load combination) ▪ The shear force and bending moment diagrams can be drawn for each of the load cases required in the patterns of loading. ▪ A composite diagram comprising a profile indicating the maximum values including all possible load cases can be drawn; this is known as an envelope. ▪ In this subject, the load set 2 will be further considered in structural analysis. ▪ All span carrying the design permanent and variable loads (1.35Gk+ 1.5Qk ) ▪ Alternate span carrying the design permanent and variable load (1.35Gk+ 1.5Qk ), other spans carrying only the design permanent loads (1.35Gk )
Loading arrangement (Load combination) Load Case 1 Load Case 2 Load Case 3 SHEAR FORCE DIAGRAM ENVELOPE BENDING MOMENT DIAGRAM ENVELOPE
Method of structural analysis ▪ Three analysis methods may be used in order to obtain shear force and bending moment for design purposes. There are; Elastic analysis using moment distribution method (Modified Stiffness Method) Simplified method using shear and moment coefficient from Table 3.5: BS 8110: Part 1. Using commercial analysis software such as Staad Pro, Esteem, Orion, Tekla and etc. 1 2 3
Moment distribution method ▪ Moment distribution method is only involving distribution moments to joint repetitively. ▪ The accuracy of moment distribution method is dependent to the number repeat which does and usually more than 5 repeat real enough. Right value will be acquired when no more moments that need distributed. ▪ In general the value is dependent to several factor as : • Fixed end moment, FEM • Carry over factor, COF • Member Stiffness Factor (distribution factor), K
Moment distribution method Fixed end moment (FEM) ▪ The moment at the fixed joints of a loaded member are called fixed-end moment. ▪ This moment can be determined from table below, depending upon the type of loading on the member.
Moment distribution method Carry over factor (COF) ▪ The carry-over factor to a fixed end is always 0.5, otherwise it is 0.0.
Moment distribution method
Example 3.3 A series of 250 x 450 mm reinforced concrete beams spaced at 3 m centres and spanning 7.5 m support a 175 mm thick reinforced concrete slab as shown in Figure 2.1. If the variable floor action is 3 kN/m2 and the load induced by the weight of concrete is 25 kN/m3 , calculate the maximum shear force and bending moment of beam B/1-2.
Solution of Example 3.3 Slab Permanent action, Gk from 175 mm slab : 25 x 0.175 = 4.38 kN/m2 Variable action, Qk : 3.0 kN/m2 Total ultimate load : 1.35 (4.38) + 1.5(3.0) = 10.41 kN/m2 Beam Ultimate load from slab : 10.41 x 3 = 31.23 kN/m Beam self-weight, Gk : 25 x 0.4 x 0.25 = 2.5 kN/m Ultimate load for beam : 1.35 x 2.5 = 3.38 kN/m Total ultimate load, w : 31.23 + 3.38 = 34.61 kN/m Maximum shear force = wL/2 = 34.61 x 7.5/2 = 129.79 kN Maximum bending moment = wL2 /8 = 34.61 x 7.52 /8 = 243.35 kNm
Example 3.4 Figure below shows the first floor layout plan of commercial building. If all beams size are 300 x 500 mm, determine the following; 1. Characteristic permanent and variable action act on the beam 1/A-E if all slab thickness are 150 mm and the brickwall heights is 3 m. 2. Shear force and bending moment envelope of beam 1/A-E. Given the following data; Variable load on slab = 4.0 kN/m2 Finishes, ceiling & services = 1.5 kN/m2 Unit weight of concrete = 25 kN/m3
Example 3.4
Solution of Example 3.4 1) Characteristic permanent and variable action act on the beam 1/A-E Action on slab Permanent action, Gk Selfweight of slab = 0.15 x 25 = 3.75 kN/m2 Finishes, ceiling and service = 1.5 kN/m2 Total permanent action on slab = 5.25 kN/m2 Variable action, Qk = 4.0 kN/m2 Action on beam Permanent action, Gk Load from slab = 0.5 x 5.25 x 3 = 7.88 kN/m Beam selfweight = (0.3 x (0.5 – 0.15) x 25) = 2.63 kN/m Brickwall = 2.6 x 3 = 7.8 kN/m Total permanent action on beam = 18.31 kN/m Variable action, Qk Load from slab = 0.5 x 4.0 x 3.0 = 6.00 kN/m
Solution of Example 3.4 2)Shear force and bending moment envelope of beam 1/A-E. Loading
Solution of Example 3.4 Moment of Inertia, I I = bh3 /12 = 300 x 5003 /12 = 3.125 x 109 mm4 Stiffness, K A-B KAB = KBA= 3I/L = 3 x 3.125 x 109 /8000 = 1.17 x 106 mm3 B-C KBC = KCB = 4I/L = 4 x 3.125 x 109 /8000 = 1.56 x 106 mm3 C-D KCD = KDC = 4I/L = 4 x 3.125 x 109 /8000 = 1.56 x 106 mm3 D-E KDE = KED= 3I/L = 3 x 3.125 x 109 /8000 = 1.17 x 106 mm3 Distribution Factor, DF Joint A & E ~ DFAB & DFED = KAB / (KAB + 0) = 1.17 / (1.17 + 0) = 1.0 Joint B ~ DFBA = KBA / (KBA + KBC) = 1.17 / (1.17 + 1.56) = 0.43 DFBC = KBC / (KBA + KBC) = 1.56 / (1.17 + 1.56) = 0.57 Joint C ~ DFCB = KCB / (KCB + KCD) = 1.56 / (1.56 + 1.56) = 0.50 DFCD = KCD / (KCB + KCD) = 1.56 / (1.56 + 1.56) = 0.50 Joint D ~ DFDC = KDC / (KDC + KDE) = 1.56 / (1.17 + 1.56) = 0.57 DFDE = KDE / (KDC + KDE) = 1.17 / (1.17 + 1.56) = 0.43
Fix End Moment, FEM - MAB = MBA = wL2 /12 = 33.72 x 8 2 /12 = 179.84 kNm - MBC = MCB = wL2 /12 = 33.72 x 8 2 /12 = 179.84 kNm - MCD = MDC = wL2 /12 = 33.72 x 8 2 /12 = 179.84 kNm - MDE = MED = wL2 /12 = 33.72 x 8 2 /12 = 179.84 kNm Load Case 1
SFD BMD
Load Case 2 Fix End Moment, FEM - MAB = MBA = wL2 /12 = 33.72 x 8 2 /12 = 179.84 kNm - MBC = MCB = wL2 /12 = 24.72 x 8 2 /12 = 131.84 kNm - MCD = MDC = wL2 /12 = 33.72 x 8 2 /12 = 179.84 kNm - MDE = MED = wL2 /12 = 24.72 x 8 2 /12 = 131.84 kNm
SFD BMD
Fix End Moment, FEM - MAB = MBA = wL2 /12 = 24.72 x 82 /12 = 131.84 kNm - MBC = MCB = wL2 /12 = 33.72 x 82 /12 = 179.84 kNm - MCD = MDC = wL2 /12 = 24.72 x 82 /12 = 131.84 kNm - MDE = MED = wL2 /12 = 33.72 x 82 /12 = 179.84 kNm Load Case 3