STATICALLY INDETERMINATE MEMBERS &

THERMAL STRESSES

Chapter II

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.

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

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

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

COMPOUND BAR

Material(1)

Material(2)

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

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

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

W

L1 L2

(dL)1 = (dL)2

(σ1/ E1)L1 = (σ2 /E2)L2

σ1 = σ2 ×( E1/E2)(L1/L2) (1)

E1/E2 is called modular ratio

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

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.

250mm

250mm

Steel Bars 4No.

σc = 2.31 N/mm2

σs = 34.69N/mm2

Case (ii) As = 6250 mm2

(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

Steel Rod

Copper Tube

600mm

5mm

25mm

33mm

50KN

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:

600mm900mm

500mmA

B D E

C

F

20kN

σCD = 15.87N/mm2

σAB = 3.6 × 15.87 = 57.14N/mm2

σEF = 3 × 15.87 = 47.61N/mm2

(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

(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 x = 3.12m from A

(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.

Steel bar

80kN

Copper tube

0.15mm

1000mm

(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 a load of 50kN is applied at the position shown, find the stresses in the rods. Assume Ecu = 100 Gpa. Es= 200GPa Also find reaction at A.

Copper 200mmD

Steel 150mmE

A CB

1M 1.5M2M

50kN

Copper 200mmD

Steel 150mmE

A CB

C B

B'

Ps

(dL)s

(dL)cu

C'

Pcu

C

RA

A

C BRA

A

50kN

50kN

Temperature Stress

L

A B

L

AB

L

A B

B´P

αTL

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.

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= αTTemperature stress = αTE

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

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.

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

σ1A1 = σ2A2 (there is no external load)

σ1=( σ2A2)/A1 (1)

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.

(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

(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

Copper

40mm

60mm

40mm

1000mm

Solution:

(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.

Copper CopperSteel

PcuPsPcu

(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

Copper 200mmD

Steel 150mmE

A CB

200 300

(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

40kN

Aluminium

1m

Steel

3m

BA

(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=200mm2

A=400mm2A=600mm2

150mm200mm

150mm

Steel Copper Aluminium

Exercise problems

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

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

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 )

40

5) A 25mm diameter bar is fixed at both the ends as shown in the figure. Two collars one at CC and another at DD are provided. Find the stresses in portions AC, CD, & DB. Also find the distance through which Collars move and the reactions at A & B.

200

D

C

B

A

300

500

D

C

40

6) A compound bar is made up of a central steel plate 60mm wide 10mm thick, to which copper plates 40mm wide by 5mm thick are connected rigidly on each side. The length of the bar at normal temperature is 1meter. If the temperature is raised by 800C, determine the stresses in each metal and change in length.

Es=200GPa, Ec=100GPa, αs=12×10-6/0C, αc=17×10-6/0C

7) A Copper bar 40mm diameter is enclosed in a steel tube of external diameter 50mm. A pin of 15mm diameter is fitted transversely to the axis of the bar near each end to secure the bar to the tube as shown in the figure. Calculate the intensity of shear stress induced in the pin when temperature is raised by 600. Es=210GPa, Ec=120GPa, αs=12×10-6/0C, αc=17.5×10-6/0C. ANS:69.9MPa

PINcopper

steel