Design of structural elements

Once the design loads acting on the structure have
 been estimated it is then possible to calculate the
 design loads acting on individual elements. As was
 pointed out at the beginning of this chapter, this
 usually requires the designer to make assumptions
 regarding the support conditions and how the loads
 will eventually be transmitted down to the ground.
 Figures 2.5(a) and (b) illustrate some of the more
 Fig. 2.5 Typical beams and column support conditions.
 In design it is common to assume that all the joints
 in the structure are pinned and that the sequence of
 load transfer occurs in the order: ceiling/floor loads to
 beams to columns to foundations to ground. These
 assumptions will considerably simplify calculations
 and lead to conservative estimates of the design
 loads acting on individual elements of the struc
ture. The actual calculations to determine the forces
 acting on the elements are best illustrated by a
 number of worked examples as follows.
 13
Basic structural concepts and material properties
 Example 2.2 Design loads on a floor beam
 A composite floor consisting of a 150 mm thick reinforced concrete slab supported on steel beams spanning 5 m and
 spaced at 3 m centres is to be designed to carry an imposed load of 3.5 kN m−2. Assuming that the unit mass of the
 steel beams is 50 kg m−1 run, calculate the design loads on a typical internal beam.
 UNIT WEIGHTS OF MATERIALS
 Reinforced concrete
 From Table 2.1, unit mass of reinforced concrete is 2400 kg m−3. Assuming the gravitational constant is 10 m s−2, the
 unit weight of reinforced concrete is
 2400 × 10 = 24 000 N m−3 = 24 kN m−3
 Steel beams
 Unit mass of beam = 50 kg m−1 run
 Unit weight of beam = 50 × 10 = 500 N m−1 run = 0.5 kN m−1 run
 LOADING
 Slab
 Slab dead load (gk
 )
 = self-weight = 0.15 × 24 = 3.6 kN m−2
 Slab imposed load (qk
 ) = 3.5 kN m−2
 Slab ultimate load
 = 1.4gk
 + 1.6qk
 = 1.4 × 3.6 + 1.6 × 3.5
 = 10.64 kN m−2
 Beam
 Beam dead load (gk
 ) = self-weight = 0.5 kN m−1 run
 Beam ultimate load = 1.4gk
 DESIGN LOAD
 = 1.4 × 0.5 = 0.7 kN m−1 run
 Each internal beam supports a uniformly distributed load from a 3 m width of slab (hatched \\\\\\\) plus self-weight.
 Hence
 Design load on beam = slab load + self-weight of beam
 = 10.64 × 5 × 3 + 0.7 × 5
 = 159.6 + 3.5 = 163.1 kN
 14
Design loads acting on elements
 Example 2.3 Design loads on floor beams and columns
 The floor shown below with an overall depth of 225 mm is to be designed to carry an imposed load of 3 kN m−2 plus
 f
 loor finishes and ceiling loads of 1 kN m−2. Calculate the design loads acting on beams B1–C1, B2–C2 and B1–B3 and
 columns B1 and C1. Assume that all the column heights are 3 m and that the beam and column weights are 70 and
 60 kg m−1 run respectively.
 UNIT WEIGHTS OF MATERIALS
 Reinforced concrete
 From Table 2.1, unit mass of reinforced concrete is 2400 kg m−3. Assuming the gravitational constant is 10 m s−2, the
 unit weight of reinforced concrete is
 2400 × 10 = 24 000 N m−3 = 24 kN m−3
 Steel beams
 Unit mass of beam = 70 kg m−1 run
 Unit weight of beam = 70 × 10 = 700 N m−1 run = 0.7 kN m−1 run
 Steel columns
 Unit mass of column = 60 kg m−1 run
 Unit weight of column = 60 × 10 = 600 N m−1 run = 0.6 kN m−1 run
 LOADING
 Slab
 Slab dead load (gk
 )
 = self-weight + finishes
 = 0.225 × 24 + 1 = 6.4 kN m−2
 Slab imposed load (qk
 ) = 3 kN m−2
 Slab ultimate load
 Beam
 = 1.4gk
 + 1.6qk
 = 1.4 × 6.4 + 1.6 × 3 = 13.76 kN m−2
 Beam dead load (gk
 ) = self-weight = 0.7 kN m−1 run
 Beam ultimate load = 1.4gk
 = 1.4 × 0.7 = 0.98 kN m−1 run
 Column
 Column dead load (gk
 ) = 0.6 kN m−1 run
 Column ultimate load = 1.4gk
 = 1.4 × 0.6 = 0.84 kN m−1 run
 15
Basic structural concepts and material properties
 Example 2.3 continued
 DESIGN LOADS
 Beam B1–C1
 Assuming that the slab is simply supported, beam B1–C1 supports a uniformly distributed load from a 1.5 m width of
 slab (hatched //////) plus self-weight of beam. Hence
 Design load on beam B1–C1 = slab load + self-weight of beam
 = 13.76 × 6 × 1.5 + 0.98 × 6
 = 123.84 + 5.88 = 129.72 kN
 Since the beam is symmetrically loaded,
 RB1
 = RC1
 = 129.72/2 = 64.86 kN
 Beam B2–C2
 Assuming that the slab is simply supported, beam B2–C2 supports a uniformly distributed load from a 3 m width of
 slab (hatched \\\\\) plus its self-weight. Hence
 Design load on beam B2–C2 = slab load + self-weight of beam
 = 13.76 × 6 × 3 + 0.98 × 6
 = 247.68 + 5.88 = 253.56 kN
 Since the beam is symmetrically loaded, RB2
 and RC2
 are the same and equal to 253.56/2 = 126.78 kN.
 Beam B1–B3
 Assuming that the slab is simply supported, beam B1–B3 supports a uniformly distributed load from a 1.5 m width of
 slab (shown cross-hatched) plus the self-weight of the beam and the reaction transmitted from beam B2–C2 which
 acts as a point load at mid-span. Hence
 Design load on beam B1–B3 = uniformly distributed load from slab plus self-weight of beam
 + point load from reaction RB2
 = (13.76 × 1.5 × 6 + 0.98 × 6) + 126.78
 = 129.72 + 126.78 = 256.5 kN
 16
Example 2.3 continued
 Since the beam is symmetrically loaded,
 Column B1
 RB1
 = RB3
 = 256.5/2 = 128.25 kN
 Structural analysis
 Column B1 supports the reactions from beams A1–B1, B1–C1 and B1–B3 and its self-weight. From the above, the
 reaction at B1 due to beam B1–C1 is 64.86 kN and from beam B1–B3 is 128.25 kN. Beam A1–B1 supports only its
 self-weight = 0.98 × 3 = 2.94 kN. Hence reaction at B1 due to A1–B1 is 2.94/2 = 1.47 kN. Since the column height is
 3 m, self-weight of column = 0.84 × 3 = 2.52 kN. Hence
 Design load on column B1 = 64.86 + 128.25 + 1.47 + 2.52
 = 197.1 kN
 Column C1
 Column C1 supports the reactions from beams B1–C1 and C1–C3 and its self-weight. From the above, the reaction at
 C1 due to beam B1–C1 is 64.86 kN. Beam C1–C3 supports the reactions from B2–C2 (= 126.78 kN) and its self
weight (= 0.98 × 6) = 5.88 kN. Hence the reaction at C1 is (126.78 + 5.88)/2 = 66.33 kN. Since the column height
 is 3 m, self-weight = 0.84 × 3 = 2.52 kN. Hence
 Design load on column C1 = 64.86 + 66.33 + 2.52 = 133.71 kN