Sambungan baut & las
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This document discusses two examples of structural steel connections: a bolted connection and a welded connection. For the bolted connection, it analyzes the connection strength using both an elastic method and a method accounting for eccentricity reduction. It determines the connection has a strength of 50.6 kips using the elastic method and 62.124 kips using the eccentricity reduction method. For the welded connection, it is given a load of 10,000 kg and the weld is on three faces. It calculates the weld lengths needed based on the material strengths and dimensions provided. It determines the weld lengths on each side should be 7 cm.
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The maximum stress in the reinforced weld of a bracket plate is calculated to be 10,408.5 psi. The plate is subjected to a load of 2,200 lbs applied 6.5 inches from the weld. The geometry and load are used to calculate the polar moment of area, torque, and radial distance to determine the torsional stress. This stress is resolved into vertical and horizontal components, and combined with the direct vertical stress from the load to find the total vertical and resultant stresses. The resultant stress is then multiplied by a concentration factor to determine the maximum stress in the weld.
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A 10m long concrete pile with a square cross section of 450x450mm driven into clay with an undrained cohesion of 35 kPa has an ultimate load capacity of 600 kN. If the cross section is reduced to 250x250mm and length increased to 20m, the ultimate load capacity will be 614.6 kN.
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1) The document discusses calculating forces and deflections in a rigid frame structure made of steel links when a 2.4 kN force is applied. It determines that the force in the upper link is 7.2 kN and in the lower link is 4.8 kN.
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The document describes a simulation with a pipe diameter of 54.9 mm and length of 3.8 m using 18 lakh elements. Simulations were also run for meshes with 5 lakh, 7 lakh, 10 lakh, 13 lakh, 15 lakh, and 18 lakh elements, but all failed to converge after 500 iterations. Mesh quality metrics such as minimum orthogonal quality and maximum aspect ratio are reported for each simulation case.
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The document contains solutions to multiple problems involving determining stresses and angles of twist in shafts with various cross sectional shapes when subjected to torques. The problems calculate shear stresses at different points in the shafts using equations relating torque, shear stress, area, and moment of inertia. Ratios of shear stresses and angles of twist between different shaft designs are also determined.
Resolucion de problema 27-4
Two equally charged particles separated by 3.20 mm are released from rest. The first particle has an initial acceleration of 7.22 m/s^2 and the second has 9.16 m/s^2. The mass of the first particle is 6.31 × 10^-7 kg. The document determines (a) the mass of the second particle and (b) the magnitude of the common charge. It is found that the mass of the second particle is 4.97 × 10^-7 kg and the common charge is 7.21 × 10^-11 C.
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Sambungan baut & las
- 1. Konstruksi Baja Sambungan Baut dan Las • GIYANA SITI MARYATUN • RIZKY KRISTANTAMA • TRI DJULIANTO
- 2. SAMBUNGAN dengan BAUT 1. Suatu sambungan terdiri dari 4 baut seperti gambar. Ru baut = 27 kip. Diminta menentukan Pu dengan : a). Cara elastis b). Cara reduksi eksentrisitas
- 3. JAWAB : 1. Cara Elastis Mu = Pu . e = Pu . 5 = 5 Pu = (32 + 32 + 32 + 32 ) + (1,52 + 1,52 + 1,52 + 1,52 ) = 45 in2 Akibat geser sentris : Pva = Pu/n = Pu/4 Akibat momen lentur : Pvb = (M . x)/ΣR2 = (5Pu . 1,5)/45 = Pu/6 Phb = (M . y)/ ΣR2 = (5Pu . 3)/45 = Pu/3 ΣPv = (1/4 + 1/6) Pu = 0,416667 Pu
- 4. = 0,534 Pu ≤ 27 Kips Kekuatan Sambungan Pu = 50,6 Kips
- 5. 2. Cara Reduksi Eksentrisitas E efektif = 5 – (1+2)/2 = 3,5 in (baut 2 baris) Mu = 3,5 Pu = (32 + 32 + 32 + 32 ) + (1,52 + 1,52 + 1,52 + 1,52 ) = 45 in2 Akibat geser sentris : Pva = Pu/n = Pu/4 Akibat momen lentur : Pvb = (M . x)/ΣR2 = (3,5Pu . 1,5)/45 = 0,11666 Pu Phb = (M . y)/ ΣR2 = (3,5Pu . 3)/45 = 0,2333 Pu ΣPv = (1/4 + 1/6) Pu = 0,416667 Pu
- 6. = 0,4346 Pu ≤ 27 Kips Kekuatan Sambungan Pu = 62,124 Kips
- 7. 2. Diket: Gaya P = 10.000 kg; Las pada tiga muka. σ =1400 kg/cm² Τ = 0,6 σ = 840 kg/cm² Hitung panjang las..? Penyelesaian: s = 10 mm = 1 cm; a = 0,707 s = 0,707 cm b = 100 mm = 10 cm; e1 = e2 = ½ b = 5 cm SAMBUNGAN dengan las
- 8. Las ujung dipasang sepanjang l3br = b = 10cm l3n = l3br – 3a = 10 cm – 3 · 0,07 cm = 7,879 cm P3 = F3gs τ = l3n · a P3 = 7,879 · 0,707 cm · 840 kg/cm² = 4680 kg e1 = e2 » P1 = P2 Σ H = 0 P1 + P2 + P3 – P = 0 P1 + P2 = P – P3 P1 = 𝑃 −𝑃3 2 P1 = 10.000 −4.680 𝑘𝑔 2 = 2.660 𝑘𝑔
- 9. • 𝜏1 = P1 Fgs1 ⇒ 𝒂𝒎𝒃𝒊𝒍 𝜏1 = 840 𝑘𝑔/𝑐𝑚² • 𝐹𝑔𝑠1 = 𝑃1 𝜏 = 2.660 𝑘𝑔 840 𝑘𝑔/𝑐𝑚² = 3,167 𝑐𝑚² • 𝐹𝑔𝑠1 = 𝑎 ⋅ 𝑙1𝑛 = 0,707 𝑐𝑚 ⋅ 𝑙1𝑛 • 3,167 𝑐𝑚2 = 0,707 𝑐𝑚 ⋅ 𝑙1𝑛 • 𝒍𝟏𝒏 = 𝟑,𝟏𝟔𝟕 𝒄𝒎² 𝟎,𝟕𝟎𝟕 𝒄𝒎 = 𝟒, 𝟒𝟕𝟗 𝒄𝒎 • 𝑙1𝑏𝑟 = 𝑙1𝑛 + 3𝑎 = 4,479 𝑐𝑚 + 3 ⋅ 0,707 𝑐𝑚 • 𝒍𝟏𝒃𝒓 = 𝟔, 𝟔 𝒄𝒎 ≈ 𝟕 𝒄𝒎 > 𝑙𝑚𝑖𝑛 = 4 𝑐𝑚 ⇒ (𝑶𝑲) Jadi, Panjang Las l1 = l2 = 7 cm Panjang Las l3 = 10 cm
- 10. TERIMA KASIH