Design the fillet welds parallel to the applied load to develop the full allowable tensile load of the
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- Situation 2: The lap splice shown will develop a full strength as shown in the figure. Using E70 electrodes. The width of the plate is 150 mm and the thickness is 10 mm. Use Fy = 248 MPa T4 46 mm 48 mm 1. Which of the following nearly gives the diameter of the slot weld using the LRFD method. O 45 mm T O 47 mm T Warrow_forward200 mm 300 mm 42818 N B = 60° A lifting lug is to be mounted. The lug is shown on the drawing and the force is expected on the lug. The lug has a thickness of 30 mm. Calculate the minimum size of the weld for the given load case to ensure a minimum FOS of 3. Assume an AWS number E70xx electrode is used.arrow_forwardAn angle iron made of AISI 1040 CD steel is welded to metal as shown and loaded with a dynamic load varying between +40kN and +60kN. E70XX (Sut = 482MPa, Sy = 393MPa) will be used as the welding electrode. Find the factor of safety of the weld zone. Note: that both static and dynamic loads are applied to the source.arrow_forward
- The weldment shown in the figure is subjected to a static force of 100 kN. The hot-rolled steel bar has a thickness h=6 mm and is of AISI 1015 HR steel. The vertical support is AISI 1015 HR steel. The electrode is given in the table below. Use the welding code method. (a) is the weld metal strength satisfactory? (b) Is the attachment strength satisfactory? h h=6 mm b=60 mm d=90mm electrode=E6010 |-b- h Farrow_forwardThe weldment shown in Figure 2 is subjected to an alternating force F. The hot-rolled steel bar is 10 mm thick and is of AISI 1010 steel. The vertical support is likewise of 1010 steel. The electrode is 6010. (a) Estimate the fatigue load F the bar will carry if three 6-mm fillet welds are used. 6. 6 50 F -60 Dimensions in millimetersarrow_forwardShow your full work. Don't provide the wrong solution.arrow_forward
- Solve correctarrow_forwardUsing SMAW process, E70 electrodes and an A572 steel. Determine the length of the side fillet welds for the full capacity the 125x100x10mm angle tension member as shown below. Assume that the member is subjected to repeated stress variations making any connection eccentricity undesirable. Check the block shear strength of the member. Assume that the WT chord has sufficient strength to develop the weld strengths and the thickness of its web is 12mm. Use U = 0.87 for class use and LRFD Method onlyarrow_forwardTwo plates, 25 mm thick, are welded together by means of trans verse-fillet welds. The ultimate tensile strength of a weld metal is 415 MPa. The surface finish factor of the weld surface is 0.5, and the size factor is 0.85. The reliability is 90%. Determine the length of the weld if the factor of safety is 2. The transverse force on the welds is 100 kN, which is completely reversed under fatigue loading. Take: A=10mm 25 Toe Р.arrow_forward
- 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 /Ymwarrow_forwardstrength of materialsarrow_forwardParvinbhaiarrow_forward
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