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7. STATICALLY INDETERMINATE MEMBERS & THERMAL STRESSES

7. STATICALLY INDETERMINATE MEMBERS & THERMAL STRESSES. STATICALLY INDETERMINATE MEMBERS.

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7. STATICALLY INDETERMINATE MEMBERS & THERMAL STRESSES

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  1. 7. STATICALLY INDETERMINATE MEMBERS & THERMAL STRESSES

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

  3. Material(2) COMPOUND BAR L1 L2 Material(1) W • 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 • Change in length in all the materials are same. • (ii) Applied load is equal to sum of the loads carried by eachbar.

  4. (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

  5. Q.7.1 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. Illustrative Problems

  6. Deformation is same (dL)s = (dL)c (σs / Es )× Ls = (σc / Ec)× Lc σs / 15Ec= σc/Ec Steel Case(i) As = 4800mm2 Ac = (250× 250) – 4800 = 57,700 mm2 250mm 250mm 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 (1) σs = 15σc

  7. 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

  8. Q.7.2A 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

  9. Steel 5mm Since deformation are same (dL)s = (dL)cu (σs / Es)×Ls =( σcu / Ecu )× Lcu σs / (200 × 103 )= σcu / (100 × 103) σs = 2 σcu W = σs As+ σcu Acu Copper 25mm 600mm 33mm 50KN 50 × 103=2σcu ×( π/4) (5)2 + σcu × π/4 [(33)2 – (25)2] σcu = 123.86N/mm2 σs = 247.72 N/mm2 (dL)s = (σs / Es )× Ls (dL)s =[247.72/(200 ×103)] ×600 =0.74mm = (dL)cu

  10. Q.7.3 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: Bar AB :- L=500mm, A=100mm2, E=200GPa Bar CD:- L=900mm, A=300mm2, E=100GPa 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: C F A 600mm 900mm 500mm D B E 20kN

  11. 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

  12. Q.7.4 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) 120mm Steel (40mm×40mm) 80mm

  13. Deformations are same (dL)s = (dL)cu (σs / Es) × Ls = (σcu / Ecu )× Lcu (σs / 2Ecu )× 200 =( σcu / Ecu )× 120 Solution: σs = 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

  14. Q.7.5 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. Steel Brass 5m 3m 9m A B x 3000N

  15. W= σsAs + σbAb 3000= (1.2 σb× 445) + (σb× 1000) Deformations are same (dL)b = (dL)s (σb / Eb )× Lb = (σs / Es )× Ls (σb / 105) ×(3× 103) = [σs / (200 × 103)] × [5 × 103] σs = 1.2 σb σb = 1.95N/mm2 σs = 2.34N/mm2 +veΣMA= 0 -(3000) (x) + (2.34 × 445) × 9000 = 0 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.

  16. Q.7.6 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. 80kN 0.15mm Steel bar Copper tube 1000mm

  17. (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 Solution: σs = 2 σcu + 30 W = σs×As + σcu×Acu 80× 103 = [( 2σcu + 30)× 490] + (σcu× 210) σcu = 54.87N/mm2 σs = (2×54.87) + 30 = 139.84N/mm2

  18. Temperature Stress B A L B B´ A P L αTL A B L 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.

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

  20. Temperature stress in compound bars:- x α1TL P1 Material(1) Material(2) (dL)2 (dL)1 P2 α2TL x ∆ When 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)

  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) 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.

  22. IllustrativeProblems Q.7.7 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

  23. Free expansion αTL = 18 × 10-6 ×(44-24) × 30 × 103 = 10.8mm • No expansion joints provided:- • Temperature stress = αTE • = 18 × 10-6 × 20× 200 × 103 • = 72N/mm2 • 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 δ = 1.8mm

  24. Q.7.8 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 x (dL)cu αcuTLcu Solution: Copper 40mm Pcu αsTLs Steel (dL)s 60mm Ps Pcu 40mm Copper 1000mm ∆ x

  25. Compressive force on copper bar = tensile force on steel bar 2σcu ×Acu = σs ×As 2σcu ( 40 × 5) = σs ( 60 × 10) σcu = 1.5σs Actual expansion in copper = Actual expansion in steel αcuTLcu - (dL)cu = αsTLs + (dL)s αcuTLcu - (σcu / Ecu) Lcu =αsTLs + (σs / Es) Ls 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

  26. Q.7.9 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 Steel Copper Pcu Pcu Ps (dL)T (dL)T (dL)T (dL)L (dL)L 200kN (dL)L

  27. [ (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 = 200 × 103 – 2 × 44,000 Ps = 112 × 103N Elongation = αcuTLcu + (Pcu ×Lcu )/(Acu ×Ecu) = 1.8 ×10-5 × 40 ×1000 +[ 44000 × 1000/(500 × 100 × 103)] dL=1.6mm

  28. Case (ii) Steel Copper Pcu Pcu Ps Note: When entire load is carried by steel, copper will have extension due to temp. only (dL)T (dL)T 200kN (dL)T (dL)L [ (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

  29. Q.7.10 X 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 D Copper E 200mm Steel 150mm A C B RA C B A (dL)cu αsTLs C' B" (dL)s αcuTLcu Ps C" B' Pcu

  30. 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) + ve Σ MA = 0 -Ps× 500 +Pcu× 200 = 0 Pcu = 2.5Ps (4)

  31. 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 (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)

  32. Q.7.11 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 = 600mm2 A = 300mm2 A B Aluminium Steel 40kN 3m 1m

  33. A = 600mm2 A = 300mm2 RA 40kN RB Steel AL 3m 1m Steel:- Aluminium:- RB RB RA RA +ve ΣFX = 0 RA – 40 ×103 +RB = 0 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

  34. (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 RA = 8915.99 ≈ 8916N ( ) RB = 31,084N ( )

  35. Q.7.12 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 • Supports are perfectly rigid • 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=600mm2 A=400mm2 A=200mm2 Copper Aluminium Steel 150mm 200mm 150mm

  36. 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 σal = 0.67σcu σs=2σcu

  37. 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

  38. Exercise problems Q1.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. Q2.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) Steel Bronze Bronze 160kN

  40. Q3.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

  41. Q4.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 )

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