Composite beams

The section of the beam consists of material 1 with elastic modulus E1and material 2 with elastic modulus E2.

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Figure 58: Composite beam section.

Step I

Assume material 1 (generally the with smaller E1) as reference mate-rial.

Define n1 = E1E1

= 1, n2 = E2E1

.

Step II

Estimate the position of the neutral axis Y using

Y =Âi ni Aiyi

Âi ni Ai

Step III

Calculate the moment of inertia of the cross-sectional area about theneutral axis (NA)

Ix = Âi

112

nibih3i + ni Aid2

i

Essentially the cross-sectional area is transformed into sectionshown here made up of only the reference material.

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Figure 59: Transformed beam section.

Step IV

Calculate the stress developed

s = �ni MyIx

The radius of curvature is given by

1r=

ME1 Ix

where E1 is the elastic modulus of the reference material.

Problem 6.

For the section shown here made of wood (E1 = 16 GPa) and steel(E2 = 200 GPa) calculate the bending stress at B and C when sub-jected to a moment of 1.5 kNm.

Subhayan De, USC

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Figure 60: Problem 6.

Step I

Assume wood with E1 = 10 GPa as reference material.Define n1 = E1

E1= 1, n2 = E2

E1= 200/16 = 12.5.

Step II

The distance is measured from bottom of the beam

ni Ai (mm2) yi (mm) ni Aiyi (mm3)1 20 ⇥ 103

120 2400 ⇥ 103

2 12.5 ⇥ 10310 125 ⇥ 103

S 32.5 ⇥ 103 2525 ⇥ 103

Estimate the position of the neutral axis Y using

Y =Âi ni Aiyi

Âi ni Ai= 77.7 mm

Step III

Moment of inertia of the cross-sectional area of the wood about theneutral axis (NA)

I1=1

12n1b1h3

1 + n1 A1d21

=1

12· (1) · (0.1 m) · (0.2 m)3 + (1) · (20 ⇥ 10�3 m2) · (0.120 m � 0.0777 m)2

= 102.5 ⇥ 10�6 m4

Subhayan De, USC

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Figure 61: Problem 6 (transformedsection).

Moment of inertia of the cross-sectional area of the steel plateabout the neutral axis (NA)

I2=1

12n2b2h3

2 + n2 A2d22

=1

12· (12.5) · (0.05 m) · (0.02 m)3 + (12.5) · (1 ⇥ 10�3 m2) · (0.015 m)2

= 57.7 ⇥ 10�6 m4

Hence, the moment of inertia of this composite beam is

Ix = I1 + I2 = 160.2 ⇥ 10�6 m4

Essentially the cross-sectional area is transformed into sectionshown below made up of only the reference material (wood here).

Step IV

The stress developed at point B

sB= �n1MyBIx

= � (1) · (1.5 ⇥ 103 Nm) · (0.22 m � 0.077 m)

160.2 ⇥ 10�6 m4

= �1.33 MPa = 1.33 MPa (C)

Subhayan De, USC

The stress developed at point C

sC= �n2MyCIx

= � (12.5) · (1.5 ⇥ 103 Nm) · (�0.077 m)

160.2 ⇥ 10�6 m4

= 9.09 MPa (T)

The radius of curvature is given by

1r=

ME1 Ix

=1.5 ⇥ 103 Nm

(16 ⇥ 109 Pa) · (160.2 ⇥ 10�6 m4)

= 0.585 ⇥ 10�3 m�1

) r = 1708.8 m

where E1 is the elastic modulus of the reference material (wood here).

Reinforced concrete sections

Reinforced concrete is made up of concrete and steel bars. Sinceconcrete can not take any tension and cracks appear in it only thearea of the concrete section above neutral axis and the steel barsshould be considered for the calculation of Ix.

Problem 7.

For the reinforced concrete section shown here (with 4 Re bars@20mm dia.) calculate the bending stress in the concrete at B (thetop) and in the steel when subjected to a moment of 20 kNm. Use20 GPa as the elastic modulus of concrete and 200 GPa as the elasticmodulus of steel.

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Figure 62: Problem 7.

Subhayan De, USC

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Figure 63: Problem 7 (transformedsection).

Step I

Assume concrete with E1 = 20 GPa as reference material.Define n1 = E1

E1= 1, n2 = E2

E1= 200/20 = 10.

Step II

Assume the position of the neutral axis as shown in the figure. De-note the distance from the bottom of the top flange to the neutral axisto be x.

The distance is measured from the assumed neutral axis of thebeam

Material ni Ai (mm2) yi (mm) ni Aiyi (mm3)

Concrete1 20000 20 + x 20000(20 + x)2 200x x

2 100x2

Steel 3

(10) · (4 · p4 · (20)2) �(180 � x) �12566(180 � x)

= 12566

S20000(20 + x) + 100x2

�12566(180 � x)

The position of the actual neutral axis Y from our assumed one is

Y =Âi ni Aiyi

Âi ni Ai

Subhayan De, USC

If our assumption of the neutral axis is true then

Y = 0

) Âi ni Aiyi

Âi ni Ai= 0

) Âi

ni Aiyi = 0

) 20000(20 + x) + 100x2 � 12566(180 � x) = 0

) x2 + 200(20 + x)� 125.66(180 � x) = 0

) x2 + 200x + 4000 � 22619 + 125.66x = 0

) x2 + 325.66x � 18619 = 0

) x ⇡ 50 mm

Step III

Moment of inertia of the cross-sectional area of the concrete partsabout the neutral axis (NA)

I1=1

12n1b1h3

1 + n1 A1d21

=112

· (1) · (0.5 m) · (0.04 m)3 + (1) · (20 ⇥ 10�3 m2) · (0.07 m)2

= 100.7 ⇥ 10�6 m4

I2=112

n1b2h32 + n1 A2d2

2

=1

12· (1) · (0.2 m) · (0.05 m)3 + (1) · (0.2 m ⇥ 0.05 m) ·

✓0.05

2m◆2

= 8.3 ⇥ 10�6 m4

Moment of inertia of the cross-sectional area of the steel about theneutral axis (NA)

Is= n2 Asd2s

= (10) · (4 · p

4· (0.02 m)2) · (0.13 m)2

= 212.4 ⇥ 10�6 m4

Note that we are ignoring the 1/12bh3 part for the transformed steelsection.

Hence, the moment of inertia of this composite beam is

Ix = I1 + I2 + Is = 321.4 ⇥ 10�6 m4

Essentially the cross-sectional area is transformed into sectionshown below made up of only the reference material.

Subhayan De, USC

Step IV

The stress developed at point B (i.e., the top fiber) in the concrete

sB= �n1MyBIx

= � (1) · (20 ⇥ 103 Nm) · (0.09 m)

321.4 ⇥ 10�6 m4

= �5.6 MPa = 5.6 MPa (C)

This is the maximum compressive stress in the concrete.The stress developed in the steel

sC= �n2MyCIx

= � (10) · (20 ⇥ 103 Nm) · (�0.13 m)

321.4 ⇥ 10�6 m4

= 80.9 MPa (T)

The radius of curvature is given by

1r=

ME1 Ix

=20 ⇥ 103 Nm

(20 ⇥ 109 Pa) · (321.4 ⇥ 10�6 m4)

= 3.111 ⇥ 10�3 m�1

) r = 321.4 m

where E1 is the elastic modulus of the reference material (concretehere).

Problem 8.

For the reinforced concrete section shown here (with 4 Re bars@20mm dia.) calculate the bending stress at B and C when sub-jected to a moment of 20 kNm. Use 20 GPa as the elastic modulus ofconcrete and 200 GPa as the elastic modulus of steel.

Step I

Assume concrete with E1 = 20 GPa as reference material.Define n1 = E1

E1= 1, n2 = E2

E1= 200/20 = 10.

Step II

Assume the position of the neutral axis as shown in the figure. De-note the distance from the bottom of the top flange to the neutral axisto be x.

Subhayan De, USC

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Figure 64: Problem 8.

The distance is measured from the assumed neutral axis of thebeam

Material ni Ai (mm2) yi (mm) ni Aiyi (mm3)Concrete 1 400x x

2 200x2

Steel 2

(10) · (4 · p4 · (20)2) �(570 � x) �12566(570 � x)

= 12566S 200x2 � 12566(570 � x)

The position of the actual neutral axis Y from our assumed one is

Y =Âi ni Aiyi

Âi ni Ai

If our assumption of the neutral axis is true then

Y = 0

) Âi ni Aiyi

Âi ni Ai= 0

) Âi

ni Aiyi = 0

) 200x2 � 12566(570 � x) = 0

) x2 + 62.83x � 35813 = 0

) x ⇡ 160 mm

Step III

Moment of inertia of the cross-sectional area of the concrete partsabout the neutral axis (NA)

Ic=1

12n1b1h3

1 + n1 A1d21

=1

12· (1) · (0.4 m) · (0.16 m)3 + (1) · (0.4 m ⇥ 0.16 m) ·

✓0.16 m

2

◆2

= 546 ⇥ 10�6 m4

Subhayan De, USC

Moment of inertia of the cross-sectional area of the steel about theneutral axis (NA)

Is= n2 Asd2s

= (10) · (4 · p

4· (0.02 m)2) · (0.41 m)2

= 2112 ⇥ 10�6 m4

Note that we are ignoring the 1/12bh3 part for the transformed steelsection.

Hence, the moment of inertia of this composite beam is

Ix = Ic + Is = 2658 ⇥ 10�6 m4

Essentially the cross-sectional area is transformed into sectionshown below made up of only the reference material.

Step IV

The stress developed at point B (i.e., the top fiber) in the concrete

sB= �n1MyBIx

= � (1) · (20 ⇥ 103 Nm) · (0.16 m)

2658 ⇥ 10�6 m4

= �1.2 MPa = 1.2 MPa (C)

This is the maximum compressive stress in the concrete.The stress developed in the steel

sC= �n2MyCIx

= � (10) · (20 ⇥ 103 Nm) · (�0.41 m)

2658 ⇥ 10�6 m4

= 30.85 MPa (T)

The radius of curvature is given by

1r=

ME1 Ix

=20 ⇥ 103 Nm

(20 ⇥ 109 Pa) · (2658 ⇥ 10�6 m4)

= 0.376 ⇥ 10�3 m�1

) r = 2658 m

where E1 is the elastic modulus of the reference material (concretehere).

Subhayan De, USC