Design of structural elements

 The design axial loads can be used directly to size col
umns. Column design will be discussed more fully
 in section 2.5. However, before flexural members
 such as beams can be sized, the design bending
 moments and shear forces must be evaluated.
 Such calculations can be performed by a variety of
 methods as noted below, depending upon the com
plexity of the loading and support conditions:
 1. equilibrium equations
 2. formulae
 3. computer methods.
 Hand calculations are suitable for analysing
 statically determinate structures such as simply sup
ported beams and slabs (section 2.4.1). For various
 standard load cases, formulae for calculating the
 maximum bending moments, shear forces and
 deflections are available which can be used to rapidly
 17
Basic structural concepts and material properties
 analyse beams, as will be discussed in section 2.4.2.
 Alternatively, the designer may resort to using vari
ous commercially available computer packages, e.g.
 SAND. Their use is not considered in this book.
 2.4.1 EQUILIBRIUM EQUATIONS
 It can be demonstrated that if a body is in equilib
rium under the action of a system of external forces,
 all parts of the body must also be in equilibrium.
 This principle can be used to determine the bend
ing moments and shear forces along a beam. The
 actual procedure simply involves making fictitious
 ‘cuts’ at intervals along the beam and applying the
 equilibrium equations given below to the cut por
tions of the beam.
 Σ moments (M) = 0
 Σ vertical forces (V ) = 0
 Example 2.4 Design moments and shear forces in beams using
 equilibrium equations
 Calculate the design bending moments and shear forces in beams B2–C2 and B1–B3 of Example 2.3.
 BEAM B2–C2
 Let the longitudinal centroidal axis of the beam be the x axis and x = 0 at support B2.
 x = 0
 By inspection,
 x = 1
 Moment at x = 0 (Mx=0
 ) = 0
 Shear force at x = 0 (Vx=0
 ) = RB2
 = 126.78 kN
 (2.1)
 (2.2)
 Assuming that the beam is cut 1 m from support B2, i.e. x = 1 m, the moments and shear forces acting on the cut
 portion of the beam will be those shown in the free body diagram below:
 From equation 2.1, taking moments about Z gives
 126.78 × 1 − 42.26 × 1 × 0.5 − Mx=1
 = 0
 Hence
 Mx=1
 = 105.65 kN m
 From equation 2.2, summing the vertical forces gives
 126.78 − 42.26 × 1 − Vx=1
 = 0
 18
Example 2.4 continued
 Hence
 x = 2
 Vx=1
 = 84.52 kN
 Structural analysis
 The free body diagram for the beam, assuming that it has been cut 2 m from support B2, is shown below:
 From equation 2.1, taking moments about Z gives
 126.78 × 2 − 42.26 × 2 × 1 − Mx=2
 = 0
 Hence
 Mx=2
 = 169.04 kN m
 From equation 2.2, summing the vertical forces gives
 126.78 − 42.26 × 2 − Vx=2
 = 0
 Hence
 Vx=2
 = 42.26 kN
 If this process is repeated for values of x equal to 3, 4, 5 and 6 m, the following values of the moments and shear
 forces in the beam will result:
 x(
 m) 0 1 2 3 4 5 6
 M(kN m)
 V(kN)
 0
 126.78
 105.65
 84.52
 169.04
 42.26
 190.17
 0
 169.04
 −42.26
 105.65
 −84.52
 0
 −126.78
 This information is better presented diagrammatically as shown below:
 19
Basic structural concepts and material properties
 Example 2.4 continued
 Hence, the design moment (M) is 190.17 kN m. Note that this occurs at mid-span and coincides with the point of
 zero shear force. The design shear force (V ) is 126.78 kN and occurs at the supports.
 BEAM B1–B3
 Again, let the longitudinal centroidal axis of the beam be the x axis of the beam and set x = 0 at support B1.
 The steps outlined earlier can be used to determine the bending moments and shear forces at x = 0, 1 and 2 m and
 since the beam is symmetrically loaded and supported, these values of bending moment and shear forces will apply
 at x = 4, 5 and 6 m respectively.
 The bending moment at x = 3 m can be calculated by considering all the loading immediately to the left of the
 point load as shown below:
 From equation 2.1, taking moments about Z gives
 128.25 × 3 − 64.86 × 3/2 − Mx=3
 = 0
 Hence
 Mx=3
 = 287.46 kN m
 In order to determine the shear force at x = 3 the following two load cases need to be considered:
 In (a) it is assumed that the beam is cut immediately to the left of the point load. In (b) the beam is cut immediately
 to the right of the point load. In (a), from equation 2.2, the shear force to the left of the cut, Vx=3,L
 , is given by
 128.25 − 64.86 − Vx=3,L
 = 0
 Hence
 20
 Vx=3,L
 = 63.39 kN
Example 2.4 continued
 In (b), from equation 2.2, the shear force to the right of the cut, Vx=3,R
 , is given by
 128.25 − 64.86 − 126.78 − Vx=3,R
 = 0
 Hence
 Vx=3,R
 = −63.39 kN
 Summarising the results in tabular and graphical form gives
 Structural analysis
 x(m)
 0
 1
 2
 3
 4
 5
 6
 M(kN m)
 V(kN)
 0
 128.25
 117.44
 106.63
 213.26
 85.01
 287.46
 63.39|−63.39
 213.26
 −85.01
 117.44
 −106.63
 0
 −128.25
 Hence, the design moment for beam B1–B3 is 287.46 kN m and occurs at mid-span and the design shear force is
 128.25 kN and occurs at the supports.
 2.4.2 FORMULAE
 An alternative method of determining the design
 bending moments and shear forces in beams in
volves using the formulae quoted in Table 2.3. The
 table also includes formulae for calculating the
 maximum deflections for each load case. The for
mulae can be derived using a variety of methods
 for analysing statically indeterminate structures, e.g.
 slope deflection and virtual work, and the inter
ested reader is referred to any standard work on
 this subject for background information. There will
 be many instances in practical design where the
 use of standard formulae is not convenient and
 equilibrium methods are preferable.
 21
Basic structural concepts and material properties
 22
 Table 2.3 Bending moments, shear forces and deflections for
 various standard load cases
 Loading Maximum Maximum Maximum
 bending shearing deflection
 moment force
    
WL
 EI
 3
 8    
WL
 2 W
    
W
 2    
WL
 EI
 3
 48    
WL
 4
    
23
 1296
 3 WL
 EI    
WL
 6    
W
 2
    
11
 768
 3 WL
 EI    
W
 2    
WL
 8
    
5
 384
 3 WL
 EI    
W
 2    
WL
 8
 WL W
    
WL
 EI
 3
 3
    
WL
 EI
 3
 60    
W
 2    
WL
 6
    
WL
 EI
 3
 192
 (at supports
 and at midspan)
    
WL
 8    
W
 2
    
WL
 12    
W
 2    
WL
 EI
 3
 384
 at supports
    
WL
 24
 at midspan
23
 Structural analysis
 This load case can be solved using the principle of superposition which can be stated in general terms as follows: ‘The
 effect of several actions taking place simultaneously can be reproduced exactly by adding the effects of each case
 separately.’ Thus, the loading on beam B1–B3 can be considered to be the sum of a uniformly distributed load (Wudl
 )
 of 129.72 kN and a point load (Wpl
 ) at mid-span of 126.78 kN.
 By inspection, maximum moment and maximum shear force occur at beam mid-span and supports respectively. From
 Table 2.3, design moment, M, is given by
    
M Wl    .  = = ×
 8
 25356 6
 8 = 190.17 kN m
 and design shear force, V, is given by
    
V W    .  . = = = 2
 25356
 2 12678kN
 BEAM B1–B3
 By inspection, the maximum bending moment and shear force for both load cases occur at beam mid-span and
 supports respectively. Thus, the design moment, M, is given by
    
M W l Wl      .    .    . = + = × + × = udl pl kN m 8 4
 12972 6
 8
 12678 6
 4 28746
 and the design shear force, V, is given by
    
V W W      .  .  . = + = + = udl pl kN 2 2
 12972
 2
 12678
 2 1282