II. Calculation Questions 1. The fillet welds are used in the joint. The design value of the weld strength is f =160N / mm² .The leg size of the welds is h=12mm (tabs are used). The design values of the static loads are N=550KN and P=500KN, respectively. Check whether the welded connection satisfactory? (unit in Fig: mm) N 09 092
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- A tension plate shown below is used to support suspended load "T". Gusset Plate F, = 248 MPa Fu = 400 MPa 200 mm. Determine the allowable tensile capacity of the plate if L= 240 mm. (Assume weld strength is satisfactory)A U-groove weld is used to butt weld 2 pieces of 7.0-mm-thick titanium plate. The U-groove is prepared using a milling cutter so the radius of the groove is 3.0 mm. During welding, the penetration of the weld causes an additional 1.5 mm of material to be melted. The final cross-sectional area of the weld can be approximated by a semicircle with a radius of 4.5 mm. The length of the weld is 200 mm. The melting factor of the setup is 0.57 and the heat transfer factor is 0.86. (a) What is the quantity of heat (in Joules) required to melt the volume of metal in this weld (filler metal plus base metal)? Assume the resulting top surface of the weld bead is flush with the top surface of the plates. (b) What is the required heat generated at the welding source?Dynamic forces F1 = 4000N and F2 = 6000N are applied to the welded part shown in the figure as shown below. For dynamic forces, both dynamics are synchronized, and when the force in the y-axis direction is downward, it is on the right in the x-axis direction. When the force is up, the other force is on the left. The welded part is made of 1018 HR steel, 10 mm thick. a) Safety factor for welding? b) Calculate the safety factor for the base metal. (Electrode type for question is E9010.)
- a) list two advantages and two disadvantages of welded joints b) Name the type of bonding forces applied in welded joints A plate 50mm wide by 11mm thick is joined with another plate 120mm wide by 11mm thick by parallel welds of 40mm long.if the permissible tensile and shear stresses are 110MPa and 80MPa respectively calculate; i) safe load F carried by the joint ii) Efficiency of the jointA bracket is as shown in figure 2. Determine the uniform weld size for the arrangement. The permissible shear stress of weld material is 80 MPa. Consider the loading condition as shown in figure.Question No. 11: (a) Illustrate flatness, straightness and circularity with symbol. (b) A circular shaft, 80 mm in diameter is welded to support by means of a circumferential fillet weld. It is subjected to a torsion moment of 3000 N-m. Determine the size of weld, if the maximum shear stress in the weld is not to exceed 70 N/mm2 . (c) Analyze different stresses during design of screw and nut
- Question: a) A plate 100 mm wide and 10 mm thick is to be welded to another plate by means of double parallel fillets. The plates are subjected to a static load of 80 kN. Find the length of weld if the permissible shear stress in the weld does not exceed 55MPa. Determine the increase in load carrying capacity if a combination of single transverse double parallel fillet is used the permissible tensile strength of material is 80MPa. b) Draw a neat sketch of double riveted double cover but joint. If 8mm plates with 20 mm diameter rivets having a pitch of 50 mm are joint using double riveted lap joint. The permissible stresses are permissible tensile stress = 120 MPa;Permissible crushing stress= 100 MPa; Permissble shear stess= 150 MPa. Find the strengths and efficiency of joint, taking the strength of the rivet in double shear acoording to the indian boiler standards.I need right solution with clear calculations. Two steel plates having yield point stress of 350 N/mm2 are joined together by double fillet welds of 10 mm thick as shown in figure. The weldment is from the same parent plate material. If the weld has to be replaced by 3 bolts of the same material, one each at A and B and one at mid-way between C and D, design the bolts. Take factor of safety as 2 for the boltsAn RSW operation is used to join two pieces of sheet steel having a unit melting energy of 8369.85 J/cm3. The sheet steel has a thickness of 0.3125 cm. The weld duration will be set at 0.25 sec with a current of 11,000 amp. Based on the electrode diameter, the weld nugget will have a diameter of 0.75 cm. Experience has shown that 40% of the supplied heat melts the nugget and the rest is dissipated by the metal. If the electrical resistance between the surfaces is 130 micro-ohms, what is the thickness of the weld nugget assuming it has a uniform thickness?
- A GTAW operation is performed on low carbon steel, whose unit melting energy is 10.3 J/mm3. The welding voltage is 22 volts and the current is 135 amps. The heat transfer factor is 0.7 and the melting factor is 0.65. If filler metal wire of 3.5 mm diameter is added to the operation, the final weld bead is composed of 60% volume of filler and 40% volume base metal. If the travel speed in the operation is 5 mm/sec, determine (a) cross-sectional area of the weld bead, and (b) the feed rate (mm/sec) at which the filler wire must be supplied.9. determine the capacity of the details shown below.A992: Fy=50ksi, Fu=65ksiA36: Fy=36ksi, Fu=58ksieffective bolt hole diameter = bolt hole + 1/16"beam properties: tw=0.38 in., bf=7.07 in., tf=0.63 in.Fexx=70ksi i) what is the weld capacity of the shear connection in kips? (2 decimal places)Te - Effective throat thickness well in mm = 6 sin45° Consider the following drawing in which plate 1 is welded onto plate 2 as shown for tensile and shear loading. The plates are fillet welded with a weld thickness of 6 mm as shown for a length of 150 mm. The plates have an ultimate strength of 275 MPa and the welding was done with F70 welding electrodes, which possesses an ultimate strength of 483 MPa. Consider a partial factor of safety of 1.7 for the welded joint and determine the design strength of the weld joint, both in tension (Tdw) and in shear (Vdw) due to the forces acting on it. a) Determine the design strength of the following weld in tension ( ) due to vertical forces. b) Determine the design strength of the weld in shear ( ) due to horizontal forces. Attached pic is shear equation for calculation Tension equation for calculation is: Tdw = FyLwte /Ymw