Problem 6.1:The figure shows a horizontal steel bar of thickness h loaded in steady tension and welded to avertical support. Find the load F that will cause an allowable shear stress, Tallow, in the throats ofthe welds.bdhTallow30 mm30 mm5 mm140 MPadhFFor the weldments of Problem 6-1, the electrode is specified in the table. For the electrode metalindicated, what is the allowable load on the weldment?Reference ProblemElectrode7-1E6010The materials for the members being joined in Problem 6–1 is specified below. What load on theweldment is allowable because member metal is incorporated in the welds?Reference ProblemBarVertical Support7-11035 HR1035 CD

Answered: Image /qna-images/answer/f54cd842-1201-4f65-ab7e-3ec1b33bb162.jpg

Answered: Problem 6.1: The figure shows a…

Elements Of Electromagnetics

7th Edition

ISBN:9780190698614

Author:Sadiku, Matthew N. O.

Publisher:Sadiku, Matthew N. O.

ChapterMA: Math Assessment

Section: Chapter Questions

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The 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 millimeters
8. Design the length of weld (L) in the connection below to develop the full tensile capacity of the plate. Use E70 electrodes and 1/4" side fillet welds. The material is ASTM A36 Steel and the allowable axial tensile stress is 22,000 psi. (show all your work) "L" PL 1/2" PL 3/8" x 6" P
Ex Welding.pdf → The Figure shows a 0.75-in-thick AISI 1020 HR steel bar subjected to a load of (F = 1.4 kip) and welded to a vertical support using a 7010 electrode. Estimate the shear stress for the weld. For the AISI 1020 HR steel bar, S, = 30 kpsi, Su= 55 kpsi (from the table) From the pattern of weld shown below, b= d = 2 in (welding table) bd2 * = b/2 ý = d/2 A = 1.414hd Then, A = 1.414hd = 1.414(5/16)(2) = 0.884 in? I= bd/2 = 2(4)/2 = 4 in 1= 0.707hl, = 0.707(5/16)(4) = 0.884 in Primary shear: F t%3D A 1.4 = 1.584 kpsi 0.884 Secondary shear: Mr " = 1.4(7)(1) = 11.086 kpsi 0.884 Then, the shear stress is 1=Vr2 + t"2 = V2.509 + 122.9 = 11.2 kpsi . Ans II
An 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.
None
Нome work 1. A plate( 100 mm) wide and (10 mm) thick is to be welded with another plate by means of double transverse welds at the ends. If the plates are subjected to a load of (70 kN ), find the size of weld for static as well as fatigue load. The permissible tensile stress should not exceed (70 MPa) [Ans. 83.2 mm; 118.5 mm]
Note: All dimensions are in mm E - 210 000 MPa (N/mm') F-13 IN 1. A bracket is shown in the figure. Calculate the stresses on the welding. The bracket welded by E7018 clectrode. Based on stresses calculations, calculate the design factor of safety ELECTRODE TENSILE NUMBER YIELD STRENGTH STRENGTH _(MPa) (345) (393) (462) (531) (MPa) E60xx (427) (482) (551) (620) A = 0.707h(b + 2d). E70xx i = b/2 ESOxx b+ 2d E90xx 2d3 (600) (737) - 2d?y + (b+2d)3² 3 E100xx (689) (827) %3D E120xx
Problem 3.1: Determine the design strength of a 1 mm length of a 6.35 mm fillet weld formed by the shielded metal arc process (SMAW) and E70 electrodes with a minimum tensile strength Fexx = 483 MPa. Assume that the load is to be applied parallel to the weld length. Problem 3.2: Repeat problem 3.1 if the weld is 500 mm long. Problem 3.3: Repeat problem 3.1 if the weld is 750 mm long. Problem 3.4: Repeat problem 3.2 if the application of the load is at an angle of 35 degrees from the longitudinal axis of the weld. Problem 3.5: Repeat problem 3.3 if the application of the load is at an angle of 35 degrees from the longitudinal axis of the weld.
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
Show your full work. Don't provide the wrong solution.
Q-2 The permissible shear stress for the weldment illustrated is 138 MPa. Estimate the load, F, that will cause this stress in the weldment throat. 200 mm Explain your solution and your steps. 75 mm 6 mm
Solve it pls

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