STATICALLY INDETERMINATE MEMBERS &

THERMAL STRESSES

Chapter II 1

STATICALLY INDETERMINATE MEMBERS

Structure for which equilibrium equations are sufficient to obtain the solution are classified as statically determinate. But for some combination of members subjected to axial loads, the solution cannot be obtained by merely using equilibrium equations.The structural problems with number of unknowns greater than the number independent equilibrium equations are called statically indeterminate.

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2) Equations based on geometric relations regarding elastic deformations, produced by the loads.

1) Equilibrium equations based on free body diagram of the structure or part of the structure.

The following equations are required to solve the problems on statically indeterminate structure.

COMPOUND BAR

(i) Change in length in all the materials are same.

(ii) Applied load is equal to sum of the loads carried by each bar.

Material(1)

Material(2)

W

L1 L2

3

A compound bar is one which is made of two or more than two materials rigidly fixed, so that they sustain together an externally applied load. In such cases

Total load = load carried by material 1

+ load carried by material 2

W = σ1 A1 + σ2 A2 (2)

From Equation (1) & (2) σ1 and σ2 can be calculated

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E1 L1

E2 L2 (1)σ1 = σ2 × ×

(dL)1 = (dL)2

σ1 .L1 σ2 .L2

E1 E2

=

E1/E2 is called modular ratio

Problems

(1) A load of 300KN is supported by a short concrete column 250mm square. The column is strengthened by 4 steel bars in corners with total c/s area of 4800mm2. If Es=15Ec, find the stress in steel and concrete.

If the stress in concrete not to exceed 4MPa, find the area of steel required so that the column can support a load of 600KN.

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250mm

250mm

Steel

Solution:

Deformation is same(dL)s = (dL)c

(σs / Es )× Ls = (σc / Ec)× Lc

σs / 15Ec= σc/Ec250mm

250mm

Steel

Case(i) : As = 4800mm2

Ac = (250×250) – 4800 = 57,700mm2

σs = 15σc (1)

W = σs As + σc Ac

300 × 103 = 15 σc × 4800 + σc× 57,700

σc = 2.31 N/mm2

σs = 15σc => 15 x 2.31 = 34.69N/mm2

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Case (ii)

W= σs As + σc Ac

600× 103 = 15 σc × As + σc Ac

600 × 103 = (15 × 4) As + 4 (250 × 250 – As)

As = 6250 mm2

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(2) A mild steel rod 5 mm diameter passes centrally through a copper tube of internal diameter 25mm and thickness 4mm. The composite section is 600mm long and their ends are rigidly connected. It is then acted upon by an axial tensile load of 50kN. Find the stresses & deformation in steel and copper. Take Ecu = 100GPa, Es = 200GPa

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600mm

Steel

Copper

5mm

25mm

33mm

50KN

Solution:

600mm

Steel

Copper

5mm

25mm

33mm

50KN

Since deformation are same(dL)s = (dL)cu

(σs / Es)×Ls =( σcu / Ecu )× Lcu

σs / (200 × 103 )= σcu / (100 × 103)σs = 2 σcu

50 × 103 = 2σcu ×( π/4) (5)2 + σcu × π/4 [(33)2 – (25)2]

(dL)s = [247.72/(200 ×103)] ×600 0.74mm = (dL)cu

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W = σs As+ σcu Acu

(dL)s = (σs / Es )× Ls

σcu = 123.86N/mm2 σs = 247.72 N/mm2

(3) Three vertical rods AB, CD, EF are hung from rigid supports and connected at their ends by a rigid horizontal bar. Rigid bar carries a vertical load of 20kN. Details of the bar are as follows:

(i) Bar AB :- L=500mm, A=100mm2, E=200GPa

(ii) Bar CD:- L=900mm, A=300mm2, E=100GPa

(iii) Bar EF:- L=600mm, A=200mm2, E=200kN/mm2

If the rigid bar remains horizontal even after loading, determine the stress and elongation in each bar.

Solution:

600mm900mm500mm

A

B D E

C

F

20kN

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W = (σAB× AAB ) + (σCD ×ACD) + (σEF ×AEF)

20 × 103 = (3.6 × σCD × 100) + (σCD × 300) + (3σCD × 200)

σCD = 15.87N/mm2

σAB = 3.6 × 15.87 = 57.14N/mm2

σEF = 3 × 15.87 = 47.61N/mm2

dLAB = (σAB / EAB) × LAB = [57.14/(200 × 103)] ×500

dLAB = 0.14 = (dL)CD = (dL)EF

Deformations are same(dL)AB = (dL)CD = (dL)EF

(σAB / EAB) × LAB = (σCD / ECD) × LCD = (σEF / EEF) × LEF

[σAB/(200 ×103)]× 500 =[σCD /(100 ×103)] ×900= [σEF /(200× 103 )] × 600

σAB = 3.6× σCD ; σEF = 3× σCD

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(4) Two copper rods and one steel rod together supports as shown in figure. The stress in copper and steel not to exceed 60MPa and 120MPa respectively. Find the safe load that can be supported. Take Es = 2Ecu

W

Copper

(30mm×30mm)

Copper

(30mm×30mm)

Steel

(40mm×40mm)

120mm

80mm

12

Solution:

Deformations are same i.e., (dL)s = (dL)cu

(σs / Es) × Ls = (σcu / Ecu )× Lcu

(σs / 2Ecu )× 200 =( σcu / Ecu )× 120=> σ s = 1.2 σcu

Let σcu=60MPa=60N/mm2,

σs=1.2x60 = 72N/mm2 < 120N/mm2 (safe)

Safe load = W = σs× As + 2( σcu ×Acu )

= 72(40× 40) + 2 ×[60×(30 × 30)]

Safe load = W = 223.2 × 103 N = 223.2 kN

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(5)A rigid bar AB 9m long is suspended by two vertical rods at its end A and B and hangs in horizontal position by its own weight. The rod at A is brass, 3m long, 1000mm2 c/s and Eb = 105N/mm2. The rod at B is steel, length 5m, 445mm2 c/s and Es = 200GPa. At what distance x from A, if a vertical load P = 3000N be applied if the bar remains horizontal after the load is applied.

9m

5m

3m

Steel

A B

3000N

Brass

x

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W= σsAs + σbAb

3000= (1.2 σb × 445) + (σb × 1000)

σb = 1.95N/mm2 σs = 2.34N/mm2

Deformations are same i.e., (dL)b = (dL)s

(σb / Eb )× Lb = (σs / Es )× Ls

(σb / 105) ×(3× 103) = [σs / (200 × 103)] × [5 × 103]=> σs = 1.2 σb

+ve ΣMA= 0 => -(3000) (x) + (2.34 × 445) × 9000 = 0

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x = 3123.9 mm = 3.12m from A

If the load of 3000N is kept at a distance of 3.12m from A, bar AB will remain horizontal.

(6) A mild steel bar of c/s 490mm2 is surrounded by a copper tube of 210mm2 as shown. When they are placed centrally over a rigid bar, it is found that steel bar is 0.15 mm longer. Over this unit a rigid plate carrying a load of 80 kN is placed. Find the stress in each bar, if the length of the compound bar is 1m. Take Es = 200 GPa, Ecu = 100 GPa.

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Steel bar

80kN

Copper tube

0.15mm

1000mm

(dL)s = (dL)cu + 0.15

(σs / Es ) × Ls =( σcu / Ecu )× Lcu + 0.15

[σs / (200 × 103)] × 1000.15 = {[σcu / (100 × 103)] ×1000} + 0.15

σ s = 2 σcu + 30

W = σs ×As + σcu ×Acu

σcu = 54.87N/mm2 σs = (2×54.87) + 30 = 139.84N/mm2

Solution: 17

80× 103 = [( 2σcu + 30)× 490] + (σcu × 210)

Temperature Stress

Any material is capable of expanding or contracting freely due to rise or fall in temperature. If it is subjected to rise in temperature of T˚C, it expands freely by an amount ‘αTL’ as shown in figure. Where α is the coefficient of linear expansion, T˚C = rise in temperature and L = original length.

L

A B

L

AB

L

A B

B´P

αTL

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From the above figure it is seen that ‘B’ shifts to B' by an amount ‘αTL’. If this expansion is to be prevented a compressive force is required at B'.

Temperature strain = αTL/(L + αTL) ≈ αTL/L= αT

Hence the temperature strain is the ratio of expansion or contraction prevented to its original length.

If a gap δ is provided for expansion then

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Temperature stress = αTE

Temperature stress = [(αTL – δ)/L] E

Temperature strain = (αTL – δ) / L

Temperature stress in compound bars:-

Material(2)

Material(1)

α2TL

∆

α1TL

(dL)1

P1

(dL)2

P2

x

xWhen a compound bar is subjected to change in temperature, both the materials will experience stresses of opposite nature.

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σ1A1 = σ2A2 (there is no external load)

σ1 = ( σ2A2)/A1 (1)

Compressive force on material (1) = tensile force on material (2)

As the two bars are connected together, the actual position of the bars will be at XX.

Actual expansion in material (1) = actual expansion in material (2)

α1TL – (dL)1 = α2TL + (dL)2

α1TL – (σ1 / E1) L =α2TL + (σ2 / E2) L

αT – (σ1 / E1) = α2T + σ2 / E2 --------------------------(2)

From (1) and (2) magnitude of σ1 and σ2 can be found out.

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(1) A steel rail 30m long is at a temperature of 24˚C. Estimate the elongation when temperature increases to 44˚C. (1) Calculate the thermal stress in the rail under the following two conditions :

(i) No expansion gap provided

(ii) If a 6mm gap is provided for expansion

(2) If the stress developed is 60MPa , what is the gap left between the rails?

Take E = 200GPa, α = 18 x 10-6 /˚C

PROBLEMS 22

Free expansion αTL = 18 × 10-6 ×(44-24) × 30 × 103 = 10.8mm

i) No expansion joints provided:-

Temperature stress = αTE

= 18 × 10-6 × 20× 200 × 103

= 72N/mm2

ii) 6mm gap is provided for expansion

Temperature stress = [( αTL – δ) / L] E

= [(10.8 – 6)/(30 × 103 )] ×200 × 103

= 32N/mm2

When stress = 60MPa

Temperature stress = [( αTL – δ) / L] E

= [(10.8 – δ ) / (30 × 103 )] × 200 × 103

Solution:

δ = 1.8mm

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(2) A steel bar is placed between two copper bars. Steel bar and copper bar has c/s 60mm × 10mm and 40mm × 5mm respectively connected rigidly on each side. If the temperature is raised by 80°C, find stress in each metal and change in length. The length of bar at normal temperature is 1m. Es = 200GPa, Ecu= 100GPa, αs = 12 x 10-6/° C, αcu = 17x10-6/ ° C

Steel

Copper

αcuTLcu(dL)cu

Pcu

(dL)s

Ps

Copper

40mm

60mm

40mm

1000mm

Pcu

x

αsTLs

x∆

Solution:

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Compressive force on copper bar = tensile force on steel bar

2σcu ×Acu = σs ×As

2σcu ( 40 × 5) = σs ( 60 × 10)

αcuTLcu - (dL)cu = αsTLs + (dL)s

αcuTLcu - (σcu / Ecu) Lcu = αsTLs + (σs / Es) Ls

=> σcu = 1.5σs

Actual expansion in copper = Actual expansion in steel

Since Lcu = Ls

(17 × 10-6 × 80 )– 1.5σs /(100 × 103) = (12× 10-6 × 80) + σs /(200 × 103)

σs = 20N/mm2(T) σcu=1.5 × 20 = 30N/mm2 (C)

Δ = Change in length = αcu ×T×Lcu – (σcu / Ecu) Lcu

= 17×10-6 × 80 × 1000 – [30/(100 × 103)]× 1000

Δ = 1.06mm

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(3) A horizontal rigid bar weighing 200 kN is hung by three vertical rods each of 1m length and 500mm2 c/s symmetrically as shown. Central rod is steel and the outer rods are copper. Temperature rise is 40ºC. (1) Determine the load carried by each rod and by how much the horizontal bar descend? Given Es = 200GPa. Ecu=100GPa. αs =1.2 x 10-

5/ºC. αcu=1.8x 10-5/ºC. (2) What should be the temperature rise if the entire load of 200kN is to be carried by steel alone.

(dL)T

(dL)L

(dL)T

(dL)L

(dL)T

(dL)L

PcuPsPcu

Copper CopperSteel

200kN

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[ (dL) T + (dL) L]cu = [ (dL)T + (dL)L]st

[αcuTLcu +( PcuLcu / Acu Ecu ) ] = [αsTLs +( PsLs / AsEs )] -----(1)

For equilibrium condition ∑Fv = 0 => Ps + 2Pcu = 200 × 103

Ps = 200 × 103 – 2Pcu

Substituting in (1) [1.8 × 10-5 × 40 + Pcu/ (500 × 100 × 103)]

={ 1.2 × 10-5 × 40 +[ (200 × 103 – 2Pcu ) / (500 × 200 × 103)]}

Pcu = 44,000N Ps = 112 × 103NPs = 200 × 103 – 2 × 44,000

Elongation = αcuTLcu + (Pcu ×Lcu )/(Acu ×Ecu)

= 1.8 ×10-5 × 40 ×1000 +[ 44000 × 1000/(500 × 100 × 103)]

dL=1.6mm

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Case (ii)

[ (dL)T ]cu = [ (dL)L + (dL)T]s => αcuTLcu = PsLs / AsEs +αsTLs

=> [1.8 × 10-5 ×T] = [(200 × 103) / (500 × 200 × 103) +1.2 × 10-5 × T]

=> T = 333.33ºC

PcuPsPcu

(dL)T

(dL)L(dL)T

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CopperSteel

(dL)T 200kN

Copper

(4) A rigid bar AB is hinged at A and is supported by copper and steel bars as shown each having c/s area 500mm2. If temperature is raised by 50ºC, find stresses in each bar. Assume Ecu = 100 Gpa. Es= 200GPa, αs = 1.2 x 10-5/ºC αcu = 18 x 10-6/ºC

C B

B'

αsTLs

Ps(dL)s

(dL)cu

C'αcuTLcu

Pcu

C"

B"

RA

A

29

Copper 200mmD

Steel 150mmE

A CB

200mm 300mm

Copper:- Temp.Stress = [(αcuTLcu)– (dL)cu / 200] Ecu

Pcu= {[(αcuTLcu)– (dL)cu ]/ 200}× (Ecu × Acu)

= [(18 × 10-6× 50 × 200) – (dL)cu) /200] 100 ×103 × 500

Pcu= [45 × 103 - 250 ×103(dL)cu]-----------(1)

Steel:- Ps = {[(dL)s – αsTLs] / 150} EsAs

Ps = {[(dL)s – 12 x 10-6 × 50 × 150]/150 ]} × (200× 103 × 500)

Ps = 6.66 ×105 (dL)s – 6 × 104 ----------------(2)

From similar Δle (dL)cu/200 = (dL)s/500 => (dL)s = 2.5(dL)cu --- (3)

Pcu = 2.5Ps (4)

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+ ve Σ MA = 0 => -Ps × 500 +Pcu × 200 = 0

Substituting (1) & (2) in (4)

(45 × 103)– [250 ×103 (dL)cu]= 2.5 {[ 6.66× 105 (dL)s] – (6×104)}

45 ×103 – 250 ×103 (dL)cu= 2.5 {[ 6.66 × 105× 2.5(dL)cu]– (6×104)}

(dL)cu= 0.044mm and (dL)s=0.11mm

Substituting in (1) & (2)

Pcu = (45 × 103) – (250 × 103 × 0.044) = 34,000N

Ps = 34,000/2.5 = 13,600N

σs = Ps / As =13,600/500 = 27.2N/mm2 (T)

σcu = Pcu / Acu = 34,000/500 = 68N/mm2 (C)

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(5) A composite bar is rigidly fixed at A and B.Determine the reaction at the support when the temperature is raised by 20ºC. Take EAl = 70GPa, Es = 200GPa, αAl = 11 x 10-6/ºC, αs = 12 x 10-6/ºC.

A = 600mm2A = 300mm2

40kNAluminium

1m

Steel3m

BA

32

A = 600mm2

A = 300mm2

40kNAL

1m

Steel3m

RB

RA

Solution:

RA + RB = 40 ×103

=> RA = 40× 103 – RB

[ (dL)T + (dL)L ]al + [ (dL)T + (dL)L ]s = 0

{[(α T L)– [(RA x L) /(A × E )]}al+ {[(αTL)+[( RB×L )/(A×E)]s= 0

Aluminium:-

RA RA

Steel:-

RB RB

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40kN

(11 × 10-6 × 20 × 1000) –[{(40 × 103 – RB)×1000}/(600 × 70 × 103)] + (12 ×10-6 × 20 ×3000 )+ (RB × 3000)/(300 × 200 × 103)= 0

0.22 – 0.95 + (2.38 × 10-5 RB) + 0.072 +( 5 ×10-5 RB )= 0

7.38 × 10-5 RB = 0.658

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RA = 8915.99 ≈ 8916N ( )

RB = 31,084N ( )

(6) A bar is composed of 3 segments as shown in figure. Find the stress developed in each material when the temperature is raised by 50˚C under two conditions

i)Supports are perfectly rigid

ii) Right hand support yields by 0.2mm

Take Es = 200GPa, Ecu =100GPa, Eal = 70GPa, αs = 12 x 10-6/ºC,

αcu = 18 x 10-6/ºC, αal = 24 x 10-6/ºC.

A=200mm2A=400mm2

A=600mm2

150mm200mm

150mm

Steel Copper Aluminium

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Case(i): Supports are perfectly rigid

(dL)s + (dL)cu + (dL)al = αsTLs + αcuTLcu + αalTLal

= (12× 10-6 ×50 × 150 ) +(18 × 10 -6 × 50 × 200) +(24 × 10-6 × 50 ×150)

= 0.45mm

(σs/Es) Ls + (σcu/Ecu) Lcu + (σal/Eal) Lal = 0.45mm

[σs/(200× 103)]×150+[σcu/(100×103)]×200+[σal/(70× 103)]×150=0.45 -(1)

From principle of compound bars

σsAs = σcuAcu = σalAal => σs × 200 = σcu × 400 = σal × 600

σs=2σcu σal = 0.67σcu

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Substituting in (1)

[2σcu/(200 × 103)]×150+[σcu/(100 × 103)]×200+[0.67σcu/(70×103)] ×150=0.45

σcu = 91.27N/mm2

σs = 2σcu = 182.54N/mm2, σal = 0.67 σcu = 61.15N/mm2

Case (ii) Right hand support yield by 0.2mm

(σs/Es) Ls + (σcu/Ecu) Lcu + (σal/Eal) Lal = 0.45 – 0.2=0.25

[2σcu/(200 × 103)]×150 + [σcu/(100 × 103)]×200 + [0.67σcu/(70×103)] ×150 = 0.25

σcu = 50.61N/mm2

σs = 2σcu = 101.22N/mm2 ; σal = 0.67 σcu = 33.91N/mm2

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Exercise problems 38

1) A circular concrete pillar consists of six steel rods of 22mm diameter each reinforced into it. Determine the diameter of pillar required when it has to carry a load of 1000kN. Take allowable stresses for steel & concrete as 140Mpa & 8Mpa respectively. The modular ratio is 15 ANS: D=344.3mm

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2) Determine the stresses & deformation induced in Bronze & steel as shown in figure. Given As=1000mm2, Ab=600mm2, Es= 200Gpa, Eb= 83Gpa ANS: ( σb=55Mpa, σs=93.5Mpa, dLs=dLb=0.093mm)

160kN

Bronze BronzeSteel

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3) A cart wheel of 1.2m diameter is to be provided with steel tyre. Assume the wheel to be rigid. If the stress in steel does not exceed 140MPa, calculate minimum diameter of steel tyre & minimum temperature to which it should be heated before on to the wheel.

ANS: d=1199.16mm T=58.330C

4) A brass rod 20mm diameter enclosed in a steel tube of 25mm internal diameter & 10mm thick. The bar & the tube are initially 2m long & rigidly fastened at both the ends. The temperature is raised from 200C to 800C. Find the stresses in both the materials.

If the composite bar is then subjected to an axial pull of 50kN, find the total stress. Es=200GPa, Eb=80GPa, αs=12×10-6/0C, αb=19×10-6/0C.

ANS: σb=8.81N/mm2 ( C ) , σs=47.99N/mm2( T )