Resistência dos Materiais

1ª LISTA DE EXERCÍCIOS DE RM PROBLEM 977 Using the parallelaxis theorem determine the product of inertia of the area shown with respect to the centroidal x and y axes Ixy 425 in4 PROBLEM 9185 Determine by direct integration the moments of inertia of the shaded area with respect to the x and y axes Ix 128 ab3 Iy 120 a3 b PROBLEM 9189 Determine the polar moment of inertia of the area shown with respect to a Point O b the centroid of the area Jc 241x106 mm4 PROBLEM 9192 For the L5 x 3 x 12 in angle cross section shown use Mohrs circle to determine a the moments of inertia and the product of inertia with respect to new centroidal axes obtained by rotating the x and y axes 30 clockwise b the orientation of the principal axes through the centroid and the corresponding values of the moments of inertia Imax 1043 in4 Imin 1547 in4 A25 Determinar os momentos principais de inércia e a orientação dos eixos principais de inércia da área da seção transversal os quais têm origem no centróide C Usar as equações desenvolvidas na Seção A4 A26 Resolver o Problema A25 usando o círculo de Mohr A25 Imax 509 pol4 Imin 0428 pol4 θp1 224 θp2 676 A26 Imax 509 pol4 Imin 0428 pol4 θp1 224 42 Basic elasticity Note that P could have been found directly in this particular case from the axial strain Thus from the first of Eqs 152 σx Eεa 70 000 1000 106 70 Nmm2 as before References 1 Timoshenko S and Goodier J N Theory of Elasticity 2nd edition McGrawHill Book Company New York 1951 2 Wang C T Applied Elasticity McGrawHill Book Company New York 1953 3 Megson T H G Structural and Stress Analysis 2nd edition Elsevier 2005 Problems P11 A structural member supports loads which produce at a particular point a direct tensile stress of 80 Nmm2 and a shear stress of 45 Nmm2 on the same plane Calculate the values and directions of the principal stresses at the point and also the maximum shear stress stating on which planes this will act Ans σI 1002 Nmm2 θ 2411 σII 202 Nmm2 θ 11411 τmax 602 Nmm2 at 45 to principal planes P12 At a point in an elastic material there are two mutually perpendicular planes one of which carries a direct tensile stress at 50 Nmm2 and a shear stress of 40 Nmm2 while the other plane is subjected to a direct compressive stress of 35 Nmm2 and a complementary shear stress of 40 Nmm2 Determine the principal stresses at the point the position of the planes on which they act and the position of the planes on which there is no normal stress Ans σI 659 Nmm2 θ 2138 σII 509 Nmm2 θ 11138 No normal stress on planes at 7021 and 275 to vertical P13 Listed below are varying combinations of stresses acting at a point and referred to axes x and y in an elastic material Using Mohrs circle of stress determine the principal stresses at the point and their directions for each combination σx Nmm2 σy Nmm2 τxy Nmm2 i 54 30 5 ii 30 54 5 iii 60 36 5 iv 30 50 30 Ans i σI 55 Nmm² σII 29 Nmm² σI at 115 to x axis ii σI 55 Nmm² σII 29 Nmm² σII at 115 to x axis iii σI 345 Nmm² σII 61 Nmm² σI at 795 to x axis iv σI 40 Nmm² σII 60 Nmm² σI at 185 to x axis P14 The state of stress at a point is caused by three separate actions each of which produces a pure unidirectional tension of 10 Nmm² individually but in three different directions as shown in Fig P14 By transforming the individual stresses to a common set of axes xy determine the principal stresses at the point and their directions Ans σI σII 15 Nmm² All directions are principal directions P15 A shear stress τxy acts in a twodimensional field in which the maximum allowable shear stress is denoted by τmax and the major principal stress by σI Derive using the geometry of Mohrs circle of stress expressions for the maximum values of direct stress which may be applied to the x and y planes in terms of the three parameters given above Ans σx σI τmax sqrtτmax² τxy² σy σI τmax sqrtτmax² τxy² P16 A solid shaft of circular crosssection supports a torque of 50 kNm and a bending moment of 25 kNm If the diameter of the shaft is 150 mm calculate the values of the principal stresses and their directions at a point on the surface of the shaft Ans σI 1214 Nmm² θ 3143 σII 464 Nmm² θ 12143 P17 An element of an elastic body is subjected to a threedimensional stress system σx σy and σz Show that if the direct strains in the directions x y and z are εx εy and εz then σx λe 2Gεx σy λe 2Gεy σz λe 2Gεz where λ νE 1 v1 2v and e εx εy εz the volumetric strain P18 Show that the compatibility equation for the case of plane strain viz ²γxy x y ²εy x² ²εx y² may be expressed in terms of direct stresses σx and σy in the form ² x² ² y²σx σy 0 P19 A bar of mild steel has a diameter of 75 mm and is placed inside a hollow aluminium cylinder of internal diameter 75 mm and external diameter 100 mm both bar and cylinder are the same length The resulting composite bar is subjected to an axial compressive load of 1000 kN If the bar and cylinder contract by the same amount calculate the stress in each The temperature of the compressed composite bar is then reduced by 150C but no change in length is permitted Calculate the final stress in the bar and in the cylinder if E steel 200000 Nmm² E aluminium 80000 Nmm² α steel 0000012C and α aluminium 0000005C Ans Due to load σ steel 1726 Nmm² compression σ aluminium 691 Nmm² compression Final stress σ steel 1874 Nmm² tension σ aluminium 91 Nmm² compression P110 In Fig P110 the direct strains in the directions a b c are 0002 0002 and 0002 respectively If I and II denote principal directions find εI εII and θ Ans εI 000283 εII 000283 θ 225 or 675 Rm1 Lota 2 σx 80 kNmm² τxy 45 Nmm² σx σy 2 80 0 2 40 Nmm² media dos tensões σx σy 2 80 0 2 40 Nmm² Diferença dos tensões Ro13 sqrt40² 45² sqrt1600 2025 sqrt3625 6021 Tensões principos σ1 40 6021 10021 Nmm² σ2 40 6021 2021 Nmm² τmáx sqrtσx σy 2² τxy² τmáx sqrt80 0 2² 45² 6021 Nmm² planos principais θp 12 tg¹ 2τxy σx σy θp 12 tg¹ 2 45 80 0 12 tg¹ 1125 θp 2411 Continuação 11 θ₂ θ₁ 90 θ₂ 2411 90 11411 12 σₓ 50 Nmm² σᵧ 35 Nmm² τₓᵧ 40 Nmm² Tensão σₓ σᵧ 50 35 15 75 2 2 2 σₓ σᵧ 50 35 85 425 2 2 2 Ro13 425² 40² 180625 1600 5837 Tensão principal σ₁ 75 5837 6587 Nmm² σ₂ 75 5837 5087 Nmm² θₚ ½ tg 2 10 ½ tg 8085 θₚ 2164 θ₂ θₚ 90 θ₂ 2164 90 11164 Continuação 12 90 7021 1939 em relação ao eixo x 90 275 1175 P 13 Caso 1 σₓ 54 σᵧ 30 τₓᵧ 5 54 302 42 54 302 12 R 12² 5² 144 25 169 13 Tensões principais σ₁ 42 13 55 σ₂ 42 13 29 Angulo θₚ ½ tg¹ 25 54 30 ½ tg¹ 10 24 ½ 2262 1131 Caso 2 σₓ 30 σᵧ 54 τₓᵧ 5 30 542 42 30 542 12 R 12² 5² 144 25 13 Tensões principais σ₁ 42 13 55 σ₂ 42 13 29 Angulo θₚ ½ tg¹ 2 5 30 54 ½ tg¹ 10 24 θₚ 1131 Continuação 13 Caso 3 σₓ 60 σᵧ 36 τₓᵧ 5 60 362 48 60 362 12 R 12² 5² 144 25 13 Tensões principais σ₁ 48 13 35 σ₂ 48 13 61 Angulo θₚ ½ tg¹ 2 5 60 36 ½ tg¹ 10 24 θₚ 1131 Caso 4 σₓ 30 σᵧ 50 τₓᵧ 30 30 502 40 30 502 10 R 10² 30² 1000 3162 Tensões principais σ₁ 40 3162 838 σ₂ 40 3162 7162 Angulo θₚ ½ tg¹ 2 30 30 50 ½ tg¹ 3 3679 15 Tau máx σ1 σ2 2 σ2 σ1 2 Tau máx Tensão normal σmed σ1 σ2 2 σ1 σT 2 Tau máx 2 σ1 Tau máx Ralo R σ1 σ2 22 τxy2 Tau máx2 τxy2 Tensão σx σmed R σ1 Tau máx Tau máx2 τxy2 σy σmed R σ1 Tau máx Tau máx2 τxy2 σx σ1 Tau máx Tau máx2 τxy2 σy σ1 Tau máx Tau máx2 τxy2 P17 εx 1ε σx νσy σz εy 1ε σy νσx σz εz 1ε σz νσx σy Forma invertida σx ε 1 ν1 2ν 1 νεx νεy εz σx εν 1 ν1 2ν εx εy εz ε1 2ν 1 ν1 2ν εx ou λ νE 1 ν1 2ν G E 21 ν e εx εy εz Tensão σx λe 2Gεx σy λe 2Gεy σz λe 2Gεz RM Liotal 52 A1 20 x 60 Área 1200 mm² A2 30 x 24 Área 720 A3 30 x 36 Triagulo 12 30 36 540 ΣAixi 1200 10 720 35 540 30 53400 ΣAiyi 1200 30 720 12 540 36 64080 ΣAi 1200 720 540 2460 x 53400 2460 1621 mm ȳ 64080 2460 319 mm 521 d L 2 m no problema 4D 2 sen 55 4C 075 sen 55 4D 4C 2 sen 55 4C 4C 2 sen 55 075 sen 55 tg 55 d 2 sen 55 Continuación s 27 Resolviendo tg 55 d 2 sen 55 075 sen 55 tg 55 d 2 075 sen 55 125 sen 55 luego d 125 sen 55 tg 55 125 sen 55 tg 55 d 125 cos 55 d 125 05736 0739 m S 138