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STATICALLY INDETERMINATE MEMBERS & THERMAL STRESSES Chapter II 1.

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1 STATICALLY INDETERMINATE MEMBERS & THERMAL STRESSES Chapter II 1

2 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. 2 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.

3 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 L1L1 L2L2 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

4 Total load = load carried by material 1 + load carried by material 2 W = σ 1 A 1 + σ 2 A 2 (2) From Equation (1) & (2) σ 1 and σ 2 can be calculated 4 E 1 L 1 E 2 L 2 (1) σ 1 =σ 2 ×× (dL) 1 = (dL) 2 σ 1.L 1 σ 2.L 2 E 1 E 2 = E 1 /E 2 is called modular ratio

5 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 4800mm 2. If E s =15E c, 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 mm Steel

6 Solution: Deformation is same (dL) s = (dL) c (σ s / E s )× L s = (σ c / E c )× L c σ s / 15E c = σ c /E c 250mm Steel Case(i) : A s = 4800mm 2 A c = (250×250) – 4800 = 57,700mm 2 σ s = 15σ c (1) W = σ s A s + σ c A c 300 × 10 3 = 15 σ c × σ c × 57,700 σ c = 2.31 N/mm 2 σ s = 15σ c => 15 x 2.31 = 34.69N/mm 2 6

7 Case (ii) W= σ s A s + σ c A c 600× 10 3 = 15 σ c × A s + σ c A c 600 × 10 3 = (15 × 4) A s + 4 (250 × 250 – A s ) A s = 6250 mm 2 7

8 ( 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 E cu = 100GPa, E s = 200GPa 8 600mm Steel Copper 5mm 25mm 33mm 50KN

9 Solution: 600mm Steel Copper 5mm 25mm 33mm 50KN Since deformation are same (dL) s = (dL) cu (σ s / E s )×L s =( σ cu / E cu )× L cu σ s / (200 × 10 3 )= σ cu / (100 × 10 3 ) σ s = 2 σ cu 50 × 10 3 = 2σ cu ×( π/4) (5) 2 + σ cu × π/4 [(33) 2 – (25) 2 ] (dL) s = [247.72/(200 ×10 3 )] × mm = (dL) cu 9 W = σ s A s + σ cu A cu (dL) s = (σ s / E s )× L s σ cu = N/mm 2 σ s = N/mm 2

10 (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=100mm 2, E=200GPa (ii)Bar CD:- L=900mm, A=300mm 2, E=100GPa (iii)Bar EF:- L=600mm, A=200mm 2, E=200kN/mm 2 If the rigid bar remains horizontal even after loading, determine the stress and elongation in each bar. Solution: 600mm 900mm 500mm A B D E C F 20kN 10

11 W = (σ AB × A AB ) + (σ CD ×A CD ) + (σ EF ×A EF ) 20 × 10 3 = (3.6 × σ CD × 100) + (σ CD × 300) + (3σ CD × 200) σ CD = 15.87N/mm 2 σ AB = 3.6 × = 57.14N/mm 2 σ EF = 3 × = 47.61N/mm 2 dL AB = (σ AB / E AB ) × L AB = [57.14/(200 × 10 3 )] ×500 dL AB = 0.14 = (dL) CD = (dL) EF Deformations are same (dL) AB = (dL) CD = (dL) EF (σ AB / E AB ) × L AB = (σ CD / E CD ) × L CD = (σ EF / E EF ) × L EF [σ AB /(200 ×10 3 )]× 500 =[σ CD /(100 ×10 3 )] ×900= [σ EF /(200× 10 3 )] × 600 σ AB = 3.6× σ CD ; σ EF = 3× σ CD 11

12 (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 E s = 2E cu W Copper (30mm×30mm) Copper (30mm×30mm) Steel (40mm×40mm) 120mm 80mm 12

13 Solution: Deformations are same i.e., (dL) s = (dL) cu (σ s / E s ) × L s = (σ cu / E cu )× L cu (σ s / 2E cu )× 200 =( σ cu / E cu )× 120 => σ s = 1.2 σ cu Let σ cu =60MPa=60N/mm 2, σ s =1.2x60 = 72N/mm 2 < 120N/mm 2 (safe) Safe load = W = σ s × A s + 2( σ cu ×A cu ) = 72(40× 40) + 2 ×[60×(30 × 30)] Safe load = W = × 10 3 N = kN 13

14 ( 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, 1000mm 2 c/s and E b = 10 5 N/mm 2. The rod at B is steel, length 5m, 445mm 2 c/s and E s = 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 AB 3000N Brass x 14

15 W= σ s A s + σ b A b 3000= (1.2 σ b × 445) + (σ b × 1000) σ b = 1.95N/mm 2 σ s = 2.34N/mm 2 Deformations are same i.e., (dL) b = (dL) s (σ b / E b )× L b = (σ s / E s )× L s (σ b / 10 5 ) ×(3× 10 3 ) = [σ s / (200 × 10 3 )] × [5 × 10 3 ] => σ s = 1.2 σ b +ve ΣM A = 0 => -(3000) (x) + (2.34 × 445) × 9000 = 0 15 x = 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.

16 (6) A mild steel bar of c/s 490mm 2 is surrounded by a copper tube of 210mm 2 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 E s = 200 GPa, E cu = 100 GPa. 16 Steel bar 80kN Copper tube 0.15mm 1000mm

17 (dL) s = (dL) cu (σ s / E s ) × L s =( σ cu / E cu )× L cu [σ s / (200 × 10 3 )] × = {[σ cu / (100 × 10 3 )] ×1000} σ s = 2 σ cu + 30 W = σ s ×A s + σ cu ×A cu σ cu = 54.87N/mm 2 σ s = (2×54.87) + 30 = N/mm 2 Solution: 17 80× 10 3 = [( 2σ cu + 30)× 490] + (σ cu × 210)

18 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 A B L A B B´B´ P αTL 18

19 From the above figure it is seen that B shifts to B' by an amountαTL. 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 19 Temperature stress = αTE Temperature stress = [(αTL – δ)/L] E Temperature strain = (αTL – δ) / L

20 Temperature stress in compound bars:- Material(2) Material(1) α 2 TL α 1 TL (dL) 1 P1P1 (dL) 2 P2P2 x x When a compound bar is subjected to change in temperature, both the materials will experience stresses of opposite nature. 20 σ 1 A 1 = σ 2 A 2 (there is no external load) σ 1 = ( σ 2 A 2 )/A 1 (1) Compressive force on material (1) = tensile force on material (2)

21 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) α 1 TL – (dL) 1 = α 2 TL + (dL) 2 α 1 TL – (σ 1 / E 1 ) L =α 2 TL + (σ 2 / E 2 ) L αT – (σ 1 / E 1 ) = α 2 T + σ 2 / E (2) From (1) and (2) magnitude of σ 1 and σ 2 can be found out. 21

22 (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 /˚C PROBLEMS 22

23 Free expansion αTL = 18 × ×(44-24) × 30 × 10 3 = 10.8mm i)No expansion joints provided:- Temperature stress = αTE = 18 × × 20× 200 × 10 3 = 72N/mm 2 ii)6mm gap is provided for expansion Temperature stress = [( αTL – δ) / L] E = [(10.8 – 6)/(30 × 10 3 )] ×200 × 10 3 = 32N/mm 2 When stress = 60MPa Temperature stress = [( αTL – δ) / L] E = [(10.8 – δ ) / (30 × 10 3 )] × 200 × 10 3 Solution: δ = 1.8mm 23

24 (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. E s = 200GPa, E cu = 100GPa, α s = 12 x /° C, α cu = 17x10 -6 / ° C Steel Copper α cu TL cu (dL) cu P cu (dL) s PsPs Copper 40mm 60mm 40mm 1000mm P cu x α s TL s x Solution : 24

25 Compressive force on copper bar = tensile force on steel bar 2σ cu ×A cu = σ s ×A s 2σ cu ( 40 × 5) = σ s ( 60 × 10) α cu TL cu - (dL) cu = α s TL s + (dL) s α cu TL cu - (σ cu / E cu ) L cu = α s TL s + (σ s / E s ) L s => σ cu = 1.5σ s Actual expansion in copper = Actual expansion in steel Since L cu = L s (17 × × 80 )– 1.5σ s /(100 × 10 3 ) = (12× × 80) + σ s /(200 × 10 3 ) σ s = 20N/mm 2 (T)σ cu =1.5 × 20 = 30N/mm 2 (C) Δ = Change in length = α cu ×T×L cu – (σ cu / E cu ) L cu = 17×10 -6 × 80 × 1000 – [30/(100 × 10 3 )]× 1000 Δ = 1.06mm 25

26 (3) A horizontal rigid bar weighing 200 kN is hung by three vertical rods each of 1m length and 500mm 2 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 E s = 200GPa. E cu =100GPa. α s =1.2 x /ºC. α cu =1.8x /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 P cu PsPs Copper Steel 200kN 26

27 [ (dL) T + (dL) L ] cu = [ (dL) T + (dL) L ] st [α cu TL cu +( P cu L cu / A cu E cu ) ] = [α s TL s +( P s L s / A s E s )] -----(1) For equilibrium condition F v = 0 => P s + 2P cu = 200 × 10 3 P s = 200 × 10 3 – 2P cu Substituting in (1) [1.8 × × 40 + P cu / (500 × 100 × 10 3 )] ={ 1.2 × × 40 +[ (200 × 10 3 – 2P cu ) / (500 × 200 × 10 3 )]} P cu = 44,000N P s = 112 × 10 3 N P s = 200 × 10 3 – 2 × 44,000 Elongation = α cu TL cu + (P cu ×L cu )/(A cu ×E cu ) = 1.8 ×10 -5 × 40 ×1000 +[ × 1000/(500 × 100 × 10 3 )] dL=1.6mm 27

28 Case (ii) [ (dL) T ] cu = [ (dL) L + (dL) T ] s => α cu TL cu = P s L s / A s E s +α s TL s => [1.8 × ×T] = [(200 × 10 3 ) / (500 × 200 × 10 3 ) +1.2 × × T] => T = ºC P cu PsPs (dL) T (dL) L (dL) T 28 CopperSteel (dL) T 200kN Copper

29 (4) A rigid bar AB is hinged at A and is supported by copper and steel bars as shown each having c/s area 500mm 2. If temperature is raised by 50ºC, find stresses in each bar. Assume E cu = 100 Gpa. E s = 200GPa, α s = 1.2 x /ºC α cu = 18 x /ºC C B B'B' α s TL s PsPs (dL) s (dL) cu C'C' α cu TL cu P cu C"C" B"B" RARA A 29 Copper 200mm D Steel 150mm E AC B 200mm300mm

30 Copper:- Temp.Stress = [(α cu TL cu )– (dL) cu / 200] E cu P cu = {[(α cu TL cu )– (dL) cu ]/ 200}× (E cu × A cu ) = [(18 × × 50 × 200) – (dL) cu ) /200] 100 ×10 3 × 500 P cu = [45 × ×10 3 (dL) cu ] (1) Steel:- P s = {[(dL) s – α s TL s ] / 150} E s A s P s = {[(dL) s – 12 x × 50 × 150]/150 ]} × (200× 10 3 × 500) P s = 6.66 ×10 5 (dL) s – 6 × (2) From similar Δle (dL) cu /200 = (dL) s /500 => (dL) s = 2.5(dL) cu --- (3) P cu = 2.5P s (4) 30 + ve Σ MA = 0 => -Ps × 500 +Pcu × 200 = 0

31 Substituting (1) & (2) in (4) (45 × 10 3 )– [250 ×10 3 (dL) cu ]= 2.5 {[ 6.66× 10 5 (dL) s ] – (6×10 4 )} 45 ×10 3 – 250 ×10 3 (dL) cu = 2.5 {[ 6.66 × 10 5 × 2.5(dL) cu ]– (6×10 4 )} (dL) cu = 0.044mm and (dL) s =0.11mm Substituting in (1) & (2) P cu = (45 × 10 3 ) – (250 × 10 3 × 0.044) = 34,000N P s = 34,000/2.5 = 13,600N σ s = P s / A s =13,600/500 = 27.2N/mm 2 (T) σ cu = P cu / A cu = 34,000/500 = 68N/mm 2 (C) 31

32 (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 E Al = 70GPa, E s = 200GPa, α Al = 11 x /ºC, α s = 12 x /ºC. A = 600mm 2 A = 300mm 2 40kN Aluminium 1m Steel 3m B A 32 A = 600mm 2 A = 300mm 2 40kN ALAL 1m Steel 3m RBRB RARA Solution:

33 R A + R B = 40 ×10 3 => R A = 40× 10 3 – R B [ (dL) T + (dL) L ] al + [ (dL) T + (dL) L ] s = 0 {[(α T L)– [(R A x L) /(A × E )]} al + {[(αTL)+[( R B ×L )/(A×E)] s = 0 Aluminium:- RARA RARA Steel:- RBRB RBRB 33 40kN

34 (11 × × 20 × 1000) –[{(40 × 10 3 – R B )×1000}/(600 × 70 × 10 3 )] + (12 ×10 -6 × 20 ×3000 )+ (R B × 3000)/(300 × 200 × 10 3 )= – (2.38 × R B ) ( 5 ×10 -5 R B )= × R B = R A = N ( ) R B = 31,084N ( )

35 (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, E cu =100GPa, E al = 70GPa, α s = 12 x /ºC, α cu = 18 x /ºC, α al = 24 x /ºC. A=200mm 2 A=400mm 2 A=600mm 2 150mm 200mm 150mm Steel CopperAluminium 35

36 Case(i): Supports are perfectly rigid (dL) s + (dL) cu + (dL) al = α s TL s + α cu TL cu + α al TL al = (12× ×50 × 150 ) +(18 × × 50 × 200) +(24 × × 50 ×150) = 0.45mm (σ s /E s ) L s + (σ cu /E cu ) L cu + (σ al /E al ) L al = 0.45mm [σ s /(200× 10 3 )]×150+[σ cu /(100×10 3 )]×200+[σ al /(70× 10 3 )]×150=0.45 -(1) From principle of compound bars σ s A s = σ cu A cu = σ al A al => σ s × 200 = σ cu × 400 = σ al × 600 σ s =2σ cu σ al = 0.67σ cu 36

37 Substituting in (1) [2σ cu /(200 × 10 3 )]×150+[σ cu /(100 × 10 3 )]×200+[0.67σ cu /(70×10 3 )] ×150=0.45 σ cu = 91.27N/mm 2 σ s = 2σ cu = N/mm 2, σ al = 0.67 σ cu = 61.15N/mm 2 Case (ii) Right hand support yield by 0.2mm (σ s /E s ) L s + (σ cu /E cu ) L cu + (σ al /E al ) L al = 0.45 – 0.2=0.25 [2σ cu /(200 × 10 3 )]×150 + [σ cu /(100 × 10 3 )]×200 + [0.67σ cu /(70×10 3 )] ×150 = 0.25 σ cu = 50.61N/mm 2 σ s = 2σ cu = N/mm 2 ; σ al = 0.67 σ cu = 33.91N/mm 2 37

38 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

39 39 2) Determine the stresses & deformation induced in Bronze & steel as shown in figure. Given A s =1000mm 2, A b =600mm 2, E s = 200Gpa, E b = 83Gpa ANS: ( σ b =55Mpa, σ s =93.5Mpa, dL s =dL b =0.093mm) 160kN Bronze Steel

40 40 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= mm T= C 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 20 0 C 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. E s =200GPa, E b =80GPa, α s =12×10 -6 / 0 C, α b =19×10 -6 / 0 C. ANS: σ b =8.81N/mm 2 ( C ), σ s =47.99N/mm 2 ( T )

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