CIV2503-Projeto Estrutural-Analise de Vigas e Deflexao

1 University of Southern Queensland Faculty of Health Engineering and Sciences Assignment 1 CIV2503 STRUCTURAL DESIGN I Due 22 August 2022 Marks 200 Note to students This assignment uses the last three digits denoted x y z of your student ID number to individualise your design data Student must use these three digits to obtain your own design data For example a student with ID number of 0060123456 will have x 4 y 5 z 6 Use 4 significant figures for midstep values and 3 significant figures for the value of a final answer Question 1 60 marks A simply supported beam has a span of 8 m and is subject to a half distributed load w 3x 0z kNm and a concentrated load P 10y 0z kN as shown in Figure 1 The beam is made of OneSteel 300PLUS universal beam section 410UB 537 with E200x103 MPa and loaded in its strong axis direction For simplicity you can ignore the beam selfweight Use the method of superposition and carry out the following tasks a Analyse and draw neat labelled Shear Force and Bending Moment Diagrams for the beam ie for each of the given individual loads and then superpose for the combined loading Find critical values of Shear Force and Bending Moment for strength design 30 marks b Determine the total maximum deflection of the beam using approximate method not the exactrigorous methods from CIV3505 Then evaluate it against the deflection limit span250 and give comments 20 marks c State all possible solutions on how to reduce the deflection of beam assuming you are not subject to any design constraints 10 marks Figure 1 Simply Supported Beam Example A student with xyz 456 will have w 34 06 4 kNm P 105 06 111 kN Question 2 70 marks a An indeterminate beam is subject to a two distributed loads w1 6x 0z kNm and w2 5y 0z kNm as shown in Figure 2 Analyse and indicate where you would add conceptual pins to convert this to a corresponding determinate beam 15 marks b Using the approximate method with conceptual pins analyse and draw a neat and fully labelled Bending Moment Diagram showing all critical values and other necessary details for the beam Include all the working for part marks 45 marks 2 m 2 m 4 m 8 m 2 c Summarise the differences in design load effects ie shears moments and deflections between using a continuous beam of 4 equal spans and using 4 identical simply supported beams to cover the same length Some extra beam formulae are included in the final page to complement those already provided in the learning materials 10 marks Figure 2 Indeterminate Beam Example A student with xyz 456 will have w1 64 06 7 kNm w2 55 06 61 kNm Question 3 70 marks Plan view of a steel portal frame building with a 3 m cantilever roof on one side is shown in Figure 3 Roof layers roof sheeting insulation and roof bracing are supported by purlins which are in turn supported by universal beams B1 to B5 There is a mechanical equipment hung at bottom of beam B3 causing a dead load PG 6x kN and a live load PQ 4y kN both unfactored a Develop line load diagrams using the basic design load combination 12G 15Q for the following members i A typical purlin 15 marks ii Beam B1 20 marks iii Beam B3 25 marks b Calculate the basic design load combination 12G 15Q axial applied at the top of column Y3 10 marks Roof sheeting selfweight allowance g1 005z kPa selfweight of insulation and roof bracing g2 6z kgm2 For selfweights of beams 360UB567 and purlins Z15015 see OneSteel and Lysaght catalogues Use Table 32 of ASNZS 11701 to determine imposed action on roof using only distributed load is sufficient 3 10000 3000 X Y Z Figure 3 Plan view of portal frame building Example A student with xyz 456 will have PG 64 kN PQ 45 kN g1 0056 kPa g2 66 kgm2 Final note to students See overpage for some extra beam formulae for question 2 3000 1 5000 2 5000 purlin at 1250 mm centres mechanical equipment hanged at this point 3 5000 4 5000 5 B1 B3 B5 B4 B2 R1 V1 R4 V4 0400wL R2 R3 110wL V21 V32 0500wL V22 V31 0600wL M1 M5 at 0400L from R1 or R4 0080wl2 M2 M4 at R2 or R3 0100wl2 M3 at mid centre span 0025wl2 Maximum deflection Δmax at 0446L from R1 or R4 00069wL4 EI R1 V1 R5 V5 0393wL R2 R4 1143wL R3 0928wL V21 V42 0607wL V22 V41 0536wL V31 V32 0464wL M1 at 0393L from R1 M7 at 0393L from R5 00772wl2 M2 at R2 01071wl2 M3 at 0536L from R2 M5 at 0536L from R4 00364wl2 M4 at R3 00714wl2 Maximum deflection Δmax at 0440L from R1 and R5 00065wL4 EI 4 L 1 0 P 2m 2m 4m 1149925 E 200 103 MPa X9 Y2 Z5 P10205 107 Kw W3905 44 KNm Fx0 Ax0KN MA0 107 2 44 4 6 By 8 0 By15875KN Fy0 107 44 4 BY Ay0 Ay12425KN CORTANTE 12425 1725 15875 As12425 2 2485 A21725 2 345 A3 158752 2 44 286381 A3 17252 2 44 0338139 Digitalizado com CamScanner 5 MOMENTO 1725 44 X1 X1 0392045 m Vmáx 12425 KN 0 x 2 m Mmáx 2863814 KNm X 4392045 m Vmín 15875 KN X 8 m Mmín 0 KNm X0 m X 8 m b σ M y I σ KN m m m4 σ KN m2 410VB 537 I 188 106 mm4 y 403 2 mm x 178 2 mm M 2863814 KNm Digitalizado com CamScanner Iy 103 106 mm4 σmax 2863814 103 2015 103 106 σmáx 056025 Mla O perfil está sob tensão abaixo da tabela do perfil ou seja suporta a carga aplicada VAC 12425 VED 1725 VDB 44X C VDB 44X 19325 MAC 12425X MCD 1725X C MCD 1725X 214 MDB 44x2 2 19325X C MDB 22X2 19325X 138 X 8 V 15875 448 C 15875 C 19325 X 2 M 2485 1725 2 C 2485 C 214 X 8 M 0 44 82 2 19325 8 C 0 C 138 D DEFLEXÃO MÁXIMA 1 1 2 12 2 KN m V triangular areas 13 248512 1656 trapezoids 16 1 22485 283 2 16 2 2 2485 2 283 803 semi circle and triangle 16 2 283 2 2295 4 9893 1656 803 9893 EI 1958 EI DEFLEXÃO MÁXIMA D 1958 EI E 200 103 Mla I 103 106 mm4 D 1958 200 103 D 950485 mm D 95 102 m A deformação pode ser controlada com a mudança de material ao mesmo a modificação do perfil aumentando a inércia GPT 3 21 113 3 ADICIONADAS Modificação pontos nos pontos B C D sistemas 3 roturas a Sistema dobernindade M 07 8 8 W2 578 333 KNm 456 kNm M1 746 333 KNm W2 52 05 57 kNm W1 69 05 74 KNm 333 O momento nos pinos rótulas será sempre nulo Elimina os momentos existentes nos apoios A diferença entre uma viga continua e uma viga que apresenta o intervalo em cada apoio é como demonstrado no item anterior os apoios deixam de receber carga de momento o fato de ter momento nulo no apoio elimina os esforços de momento nos pilares deste apoio M1 0072269 2 2 1923 KNm L RESULTADO EM ANEXO DO ENUNCIADO M1 333 KN O momento máximo será maior na viga rotulada nos apoios uma vez que todo o momento estará inserido no vão e não mais no vão e apoio