Bending and shearing effect on deformations and stresses
It is known from the mechanics that the deformation of a member depends on the works done by bending moments and shear forces. The stiffness E ·I of earthquake resistant structural elements (cracked crosssections) is half of the respective stiffness of uncracked crosssections [EC8, §4.3.1(6),(7)]. The shear modulus G, on which the shear deformation depends, is related to the modulus of elasticity E and Poisson factor ν by the following expression, In EC2 it is assumed that for a cracked cross section ν=0 and for an uncracked crosssection ν=0.2 → Gvaries from 0.5E to 0.42E. In calculations for earthquake resistance it is recommended to assumeν =0 and therefore G=0.5E. A column fixed at both ends bearing at the top a mass m, will be now considered. Gravity acceleration causes a vertical load W=g·m. Due to seismic acceleration a, a horizontal forceH=a· m[=(a/g)·W is developed at the column top resulting its relative displacement by δ.
Figure 5.1.1: Fixedended column: K = (12 E I / h ^{3}) k _{va} Figure 5.1.1: Fixedended column: K = (12 E I / h ^{3}) k _{va}
The total displacement δ is due to: (a) the displacement δ_{1} due to bending caused by force H (b) the displacement δ_{2} due to shearing caused by force H where k_{z} is the correction factor of the shear effect, G is the concrete shear modulus and A the concrete crosssection. In conclusion, the horizontal seismic force Η is related to the corresponding displacement δ by the following expression: The term K denotes the stiffness of the fixedended column. k_{v}_{a} is the correction factor of the bending stiffness of the fixedended column due to shearing. Always k_{v}_{a} <1.0. It can be ignored for ordinary columns (i.e. it may be assumed that k_{v}_{a}=1.0). On thecontrary, in the case of to walls it must be taken into account since its value is significantly lower than 1.0. Indicatively for walls with length 1.0 m and 2.0 m, k_{v}_{a }is equal to 0.80 and 0.50 respectively. Consider a fixedended column with a cross section of 400m m /400 mm , height h =3.0 m , elasticity modulus E =32.8 GP , bearing a concentrated mass of m =80 t at the top. The design seismic acceleration factor at the top is a / g =0.10. Calculate the stress resultants and the relative displacement at the top. W =80 · 1000 kg · 10 m / sec ^{2} =800 kN , H =0.10·800=80 kN , V = H =80 kN M _{1} = Μ _{2} = H · h /2=80·3.0/2=120 kNm , I =0.40·0.40^{3}/12=21.33·10^{4 } m ^{4} Without shear effect, K=12 · EI / h ^{3} =12 · 32.8 · 10^{9} · Ν /m^{2} · 21.33 · 10^{4}m^{4}/(3.0^{3} · m^{3})=31.10·10^{6} N / m With shear effect, for G=0.5E, k_{z}=5/6 (rectangular crosssection) and A=b · d: This expression is independent of the crosssectional width and applies for all fixedended rectangular columns. K = (E· I / h^{3})· K_{va} =31.10·10^{6}·0.959 N/m =29.82·10^{6} N/m. Without shear effect, δ =H/K=80 · 10^{3}N/(31.10 · 10^{6}N/ m ) =2.57 mm . With shear effect, δ =H/K=80 · 10^{3}N/(29.82 · 10^{6}N/ m ) =2.68 mm , i.e. the displacement is 4% higher than the respective value without shear effect. Consider the same data and questions of the previous example for a wall with crosssection 2000/300. W =80 · 1000 kg · 10 m / sec ^{2} =800 kN , H =0.10·800=80 kN , V = H =80 kN M _{1} = Μ _{2} = H · h /2=80·3.0/2=120 kNm , I =0.30·2.00^{3}/12=0.20 m ^{4} Without shear effect, K=12 · EI / h ^{3} =12 · 32.8 · 10^{9} · Ν /m^{2} · 0.20m^{4}/(3.0^{3} · m^{3})=2915.56·10^{6} N / m therefore K = (E· I / h^{3})· K_{va} =2915.56·10^{6}·0.484 N / m =1410.75·10^{6} N / m Without shear effect, δ =H/K=80 · 10^{3}N/(2915.56 · 10^{6}N/ m ) =0.027 mm . With shear effect, δ =H/K=80 · 10^{3}N/(1410.75 · 10^{6}N/ m ) =0.057 mm , i.e. the displacement is 110% higher than the respective value without shear effect.
