Problem 4.3. Composite cross section

Determine the bending resistance of the given composite crosssection. Perform

a) elastic analysis,
b) plastic analysis.

Compressive strength of concrete is fc =13.33 N/mm2, elastic modulus is Ec = 18.3 GPa. Section properties of the steel cross section are: A = 363.5 cm2, Iz= 481200 cm4, material properties are: fy =235 N/mm2, Es= 210 GPa. (In plastic analysis assume perfectly plastic material. The thicknesses of the web and the flanges are t w = 18.5 mm, t f = 35 mm.)

Solve Problem

Solve

Problem a)

Bending resistance, MR,e [kNm]=

Problem b)

Bending resistance, MR,p [kNm]=

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Steps

Step by step

Problem a)

Step 1. Calculate the section properties.

Check section properties

The inhomogeneous cross section is replaced by an equivalent homogeneous one. The reference material is the concrete, the ratio of the elastic moduli is

α=EsEc=11.46

Tensile stiffness, position of the centroid and bending stiffness of the replacement homogeneous cross section are:

EA¯=Echcbc+EsAs=1.24×1010 Nyc=1EA¯Echcbchc2+EsAshc+hs2=438.8 mmEI¯=Ecbchc312+Echcbcychc22+EsIs+EsAshs2yc2=1.91×1015 mm4

Step 2. Determine elastic moment resistance.

Check elastic resistance

Failure occurs when the maximum stress of concrete reaches the its compressive strength, or the maximum stress of steel reaches its yield strength. The resistance can be expressed from the maximum stresses:

σc=MEI¯ycEc=fc       MRc=fcEI¯Ecyc=13.33×1.91×101518.3×103×438.8=3175×106 Nmm=3175 kNmσs=MEI¯hycEs=fy       MRs=fsEI¯Eshs+hcyc=235×1.91×1015210×103×(900+200438.8)=3234×106 Nmm=3234 kNm

The maximum allowed stress is reached in the steel first, the elastic moment resistance is

MRe=min (MRc,MRs)=MRc=3175 kNm

Problem b)

Step 1. Assume the location of the neutral axis. Draw the normal stress diagram.

Check diagram

At plastic failure stresses reach the strength in each points of the cross section. Let us assume that stresses change sign in the web of the steel girder.

Step 2. Calculate the location of the neutral axis. Check the assumption.

Check neutral axis

The location of the neutral axis can be determined from the normal force equilibrium of the cross section. To check the assumption we compare the stress resultants of the cross sectional parts:

Nc=fcbchc=3.46×106 NNf=fybftf=2.47×106 NNw=fytwhw=3.61×106 NNc+Nf< Nf+Nw

which means that the web must be partially tensioned, our assumption is true, the neutral axis passes through the web.

N=0Nc+Nf+fytwx Nffytwhwx=0x=Nc+fytwhw2fytw=16.4 mm

Step 3. Determine the plastic moment resistance

Check plastic resistance

Plastic moment resistance is determined from a moment equilibrium equation. Let us write the moment equilibrium about the neutral axis:

MR,p=Ncx+hc2+tf+Nfx+tf2+Nfhsxtf2tf+fytwx22+hwx22=4099 kNm

Results

Worked out results

Problem a)

The inhomogeneous cross section is replaced by an equivalent homogeneous one. The reference material is the concrete, the ratio of the elastic moduli is

α=EsEc=11.48

Tensile stiffness, position of the centroid and bending stiffness of the replacement homogeneous cross section are:

EA¯=Echcbc+EsAs=1.26×1010 Nyc=1EA¯Echcbchc2+EsAshc+hs2=438.8 mmEI¯=Ecbchc312+Echcbcychc22+EsIs+EsAshs2yc2=1.91×1015 mm4

Failure occurs when the maximum stress of concrete reaches the its compressive strength, or the maximum stress of steel reaches its yield strength. The resistance can be expressed from the maximum stresses:

σc=MEI¯ycEc=fc       MRc=fcEI¯Ecyc=13.33×1.91×101518.3×103×438.8=3175×106 Nmm=3175 kNmσs=MEI¯hycEs=fy       MRs=fsEI¯Eshs+hcyc=235×1.91×1015210×103×(900+200438.8)=3234×106 Nmm=3234 kNm

The maximum allowed stress is reached in the steel first, the elastic moment resistance is

MRe=min (MRc,MRs)=MRc=3175 kNm

Problem b)

At plastic failure stresses reach the strength in each points of the cross section. Let us assume that stresses change sign in the web of the steel girder.

The location of the neutral axis can be determined from the normal force equilibrium of the cross section. To check the assumption we compare the stress resultants of the cross sectional parts:

Nc=fcbchc=3.46×106 NNf=fybftf=2.47×106 NNw=fytwhw=3.61×106 NNc+Nf< Nf+Nw

which means that the web must be partially tensioned, our assumption is true, the neutral axis passes through the web.

N=0Nc+Nf+fytwx Nffytwhwx=0x=Nc+fytwhw2fytw=16.4 mm

Plastic moment resistance is determined from a moment equilibrium equation. Let us write the moment equilibrium about the neutral axis:

MR,p=Ncx+tf+hc2+Nfx+tf2+Nfhsxtf2tf+fytwx22+hwx22=4099 kNm