Universiteit van Pretoria

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Universiteit van Pretoria Departement Siviele Ingenieurswese Department of Civil Engineering Struktuuranalise SIN 311 Structural analysis SIN 311 Eerste Semestertoets 2010 Volpunte = 60 First

1 Universiteit van Pretoria Departement Siviele Ingenieurswese Department of Civil Engineering Struktuuranalise SIN 311 Structural analysis SIN 311 Eerste Semestertoets 2010 Volpunte = 60 First Semester Test 2010 Full marks = 60 Tyd = 1.5 uur Time = 1.5 hours Interne eksaminator/internal examiner: Prof CP Roth Eksterne eksaminator/external examiner: Prof BWJ van Rensburg Toeboek Closed book et Op: Toon, waar van toepassing, die formules wat gebruik word en die vervanging van waardes in die formules. Toon die eenhede van waardes wat bereken word. Die toets bestaan uit 4 vrae. Note: Show, where applicable, the formulas that are used and the substitution of values into the formulas. Indicate the units of values calculated. The test consists of 4 questions. Vraag 1 (14 punte): Teken die buigmomentdiagram vir die volgende staties-onbepaalbare balk as gevolg van die kragte getoon saam met vertikale afsakking van 0.025m afwaarts by ondersteuningspunte B en C. Die balk is ingeklem by A en D en op rollers by B en C. Die buigstyfheid EI is constant. Gebruik E = 200GPa, I = 100x10 6 mm 4. Gebruik die helling-verplasing metode en simmetrie en/of antisimmetrie. Question 1 (14 marks): Draw the bending moment diagram for the following statically indeterminate continuous beam due to the loads indicated along with support settlements at B and C of 0.025m downwards. The beam is fixed at A and D and on rollers at B and C. The bending stiffness EI is constant. Take E = 200GPa, I = 100x10 6 mm 4. Use the slope-deflection method and symmetry and/or anti-symmetry. 10kN/m A B C D 0.025m 0.025m 5m 5m 5m SIN223 Page 1 of 8 5 arch 2013

2 Vraag 2 (26 punte): n Simmetriese spitsdak portaalraam CDE word getoon. Die raam is ingeklem by A en E. EI is konstant. Verontagsaam verplasing as gevolg van aksiale krag en skuif. Drie puntlaste word aangewend soos getoon. Gebruik die metode van helling-verplasing en teken die buigmomentdiagram van die struktuur. Wenk: gebruik simmetrie en/of antisimmetrie. Question 2 (26 marks): A symmetrical pitched portal frame CDE is shown. The frame is fixed at A and E. EI is constant. Neglect deformation due to axial forces and shear. Three point loads are applied as shown. Use the method of slope-deflection and draw the bending moment diagram of the structure. Hint: use symmetry and/or anti-symmetry. 10kN C 20kN B 20kN D 2m A E 2m 4m 4m Vraag 3 (8 punte): Teken die buigmomentdiagram vir die raam deur middel van die metode van superposisie. A is ingeklem en D rus op n roller. Die temperature in BC styg met 10 C en die temperatuur in CD styg met 20 C. Kies die reaksie by D as die oortollige krag. Verontagsaam deformasie as gevolg van aksiale kragte en skuif. Gebruik E = 200GPa, I = 300x10-6 m 4, α = 12x10-6 ( C) -1. Question 3 (8 marks): Draw the bending moment diagram for the frame using the method of superposition. A is fixed and D is a roller. The temperature in BC rises by 10 C and the temperature in CD rises by 20 C. Select the reaction at D as the redundant force. Neglect deformation due to axial forces and shear. Take E = 200GPa, I = 300x10-6 m 4, α = 12x10-6 ( C) -1. B C A D 4m SIN223 Page 2 of 8 5 arch 2013

3 Vraag 4 (18 punte): n Simmetriese raam CDEF word getoon. Die raam is ingeklem by E en F, en A en D rus op rollers. EI is konstant. Verontagsaam verplasing as gevolg van aksiale krag en skuif. Twee puntlaste word aangewend soos getoon. Gebruik die metode van helling-verplasing en teken die buigmomentdiagram van die struktuur. Wenk: gebruik simmetrie en/of antisimmetrie Question 4 (18 marks): A symmetrical frame CDEF is shown. The frame is fixed at E and F, and on rollers at A and D. EI is constant. Neglect deformation due to axial forces and shear. Two point loads are applied as shown. Use the method of slope-deflection and draw the bending moment diagram of the structure. Hint: use symmetry and/or anti-symmetry. 4kN 4kN A B C D 4m E F 12m Formules / Formulas: F 1 = Fv + v dx + AE EI m + r = 2 j 3 m + r = 3 j + e c = α T Virtual Work and Superposition Equations: Superposition / Superposisie F v F k P, Q, R etc = redundancies with a particular direction and sense. In the further notation one must understand "deflection of P" is the deflection of the primary structure at the point of application of P in the direction of P. The deflection may also be a rotation. P = The total deflection as a result of the cause P0 = Deflection of P with P = Q = R = 0 (zero) PT = Deflection of P as a result of temperature changes PS = Deflection of P as a result of support displacement PE = Deflection of P as a result of lack-of-fit f PP = Flexibility coefficient (deflection of P as a result of P = 1) f PQ = Flexibility coefficient (deflection of P as a result of Q = 1) f PR = Flexibility coefficient (deflection of P as a result of R = 1) SIN223 Page 3 of 8 5 arch 2013

4 The superposition equations for a number of redundancies can be written as follows: P = P0 + PT + PS + PE + P f PP + Q f PQ + R f PR Q = Q0 + QT + QS + QE + P f QP + Q f QQ + R f QR R = R0 + RT + RS + RE + P f RP + Q f RQ + R f RR Slope-deflection / Helling-defleksie Standard form (both ends fixed) / Standaard vorm (albei ente vas) 2 EI = ( 2 θ + θ 3 ψ ) + FE 2 EI = ( θ + 2 θ 3 ψ ) + FE A B BA A B BA The odified Slope-Deflection Equation with a Hinge at A/ Aangepaste hellingdefleksie vergelyking met skarnier by A. 3 EI 1 = ( θ ψ ) + FE FE 2 BA B BA SIN223 Page 4 of 8 5 arch 2013

5 Symmetrical Structures Slope-Deflection Equations Symmetry: 2EI = ( 2θ + θ 3ψ ) + FE A B 2EI = ( 2θ θ 0) + FE A A 2EI = ( θ A ) + FE Note that with symmetry the fixed end moment at B is the negative of the fixed end moment at A. 2EI = ( θ + 2θ 3ψ ) + FE BA A B BA 2EI = ( θ 2θ 0) FE BA A A 2EI BA = ( θ A ) FE Note that in this case the moments at A and B have the opposite sign. Anti-symmetry: With anti-symmetry the rotation at B is equal in sign and size to the rotation at A. 2EI = ( 2θ + θ 3ψ ) + FE A B 2EI = ( 2θ + θ 0) + FE A A 6EI = ( θ A ) + FE The fixed end moment at B is equal in size and sign in the anti-symmetrical loading case. SIN223 Page 5 of 8 5 arch 2013

6 2EI = ( θ + 2θ 3ψ ) + FE BA A B BA 2EI = ( θ + 2θ 0) + FE BA A A BA 6EI BA = ( θ A ) + FEBA Note that in this case the moments at A and B have the same sign. SIN223 Page 6 of 8 5 arch 2013

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DEPARTEMENT SIVIELE EN BIOSISTEEM-INGENIEURSWESE DEPARTMENT OF CIVIL AND BIOSYSTEMS ENGINEERING MEGANIKA SWK 122 EKSAMEN MECHANICS SWK 122 EXAMINATION

UNIVERSITEIT VAN PRETORIA UNIVERSITY OF PRETORIA Kopiereg voorbehou / Copyright reserved DEPARTEMENT SIVIELE EN BIOSISTEEM-INGENIEURSWESE DEPARTMENT OF CIVIL AND BIOSYSTEMS ENGINEERING MEGANIKA SWK 122