Problem 5.5. Shrinkage with additional moment

The cross section given in Figure a) of Problem 5.4 is subjected to an additional moment, its value is M = 36 kNm. Determine the stresses in the steel and in the extreme compressed concrete fibre. Final value of the shrinkage is εcs = 5×10-4. Elastic modulus of concrete is Ec = 31 GPa, elastic modulus of steel is Es = 200 GPa. Assume uncracked concrete. Diameter of steel bars is Φ = 12 mm.

Solve Problem

Solve

Maximum stress in the concrete, σc [N/mm2]=

Maximum stress in the steel, σs [N/mm2]=

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Steps

Step by step

Step 1. Calculate the section properties.

Check section properties

Cross sectional properties of the replacement homogeneous cross section are the following

Eqs.(4-4)-(4-8)

Ac=300×400=1.2×105 mm2As=4ϕ2π4=122π=452 mm2Ae=Ac+EsEcAs=120000+20031452=1.23×105 mm2se=h2=200 mmIe=bh312+2EsEcAs2dse2=300×400312+452200313552002=1.670×109 mm4

Holes in the concrete and the moment of inertia of the steel bars about their centre of gravity are neglected.

Step 2. Determine the kinematic load and deformations of the beam.

See also in Problem 5.4 c)


Check deformations

The kinematic load is

NK=AcEcεcs

The cross section is symmetric, thus the curvature from the axial load is zero, the elongation of the beam is

Eq.(5-36)

εoK=εcsAcAe=5×1041.2×1051.23×105=4.888×104

Step 3. Determine deformation from the bending moment.

Check bending deformation

From the moment the curvature of the beam’s axis is:

Eqs.(3-19) and (4-7)

κM=MEIe=36×10631×1031.67×109=6.953×1071mm

Step 4. Summarize stresses in the concrete and in the steel from the bending and from the shrinkage.

Check stresses

Eq.(5-37)


σc1=σK,Rc+σMc=EcεoK+εcs+EcκMh2=EcεoK+εcs+MIeh2=    =31×1034.88×104+5×104+6.95×107×200=4.68Nmm2 (tension)σc2=σK,RcσMc=EcεoK+εcsEcκMh2=3.94Nmm2 (compression)σK,Rs2=EsεoKEsκMdse=               =200×1034.88×1046.95×107355200=119.2Nmm2(compression)σK,Rs1=EsεoK+EsκMdse=76.1Nmm2(compression)

Results

Worked out solution

Cross sectional properties of the replacement homogeneous cross section are the following

Eqs.(4-4)-(4-8)

Ac=300×400=1.2×105 mm2As=4ϕ2π4=122π=452 mm2Ae=Ac+EsEcAs=120000+20031452=1.23×105 mm2se=h2=200 mmIe=bh312+2EsEcAs2dse2=300×400312+452200313552002=1.670×109 mm4

Holes in the concrete and the moment of inertia of the steel bars about their centre of gravity are neglected.

The kinematic load is

NK=AcEcεcs

See also in Problem 5.4 c)

The cross section is symmetric, thus the curvature from the axial load is zero, the elongation of the beam is

Eq.(5-36)

εoK=εcsAcAe=5×1041.2×1051.23×105=4.888×104

The beam’s axis curves also from the bending moment.

Eqs.(3-19) and (4-7)

κM=MEIe=36×10631×1031.67×109=6.953×1071mm

Stresses in the concrete and in the steel from the bending and from the shrinkage are summarized as follows

Eq.(5-37)


σc1=σK,Rc+σMc=EcεoK+εcs+EcκMh2=EcεoK+εcs+MIeh2=    =31×1034.88×104+5×104+6.95×107×200=4.68Nmm2 (tension)σc2=σK,RcσMc=EcεoK+εcsEcκMh2=3.94Nmm2 (compression)σK,Rs2=EsεoKEsκMdse=               =200×1034.88×1046.95×107355200=119.2Nmm2(compression)σK,Rs1=EsεoK+EsκMdse=76.1Nmm2(compression)