Applied Statics and Strength of Materials (6th Edition)
6th Edition
ISBN: 9780133840544
Author: George F. Limbrunner, Craig D'Allaird, Leonard Spiegel
Publisher: PEARSON
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Chapter 19, Problem 19.12P
Design an end connection using longitudinal welds and an end transverse weld to develop the full tensile capacity of the angle shown. Use
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A mild steel ammonia receiver is a cylinder 2 m in diameter with a pressure load of 1250 kPa and a design stress of 65 MPa. The lap welding tensile strength is 450 MPa and the weld -joint efficiency is assumed to be 100%. Compute the bursting pressure of the receiver in kN/m².
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.
PROBLEM 02: Design a SMAW fillet welds to resist the load applied on the connection shown. Use E70 electrodes and A572 gr. 50 steel for both members and assume the gusset plate is adequate for the load applied. Determine the L1 and L2.
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Chapter 19 Solutions
Applied Statics and Strength of Materials (6th Edition)
Ch. 19 - Prob. 19.1PCh. 19 - Rework Problem 19.1 assuming a bearing-type...Ch. 19 - Rework Problem 19.1 assuming a bearing-type...Ch. 19 - Compute the allowable tensile load for the...Ch. 19 - Rework Problem 19.4 assuming a bearing-type...Ch. 19 - Rework Problem 19.4 assuming that the bolts are 34...Ch. 19 - Select the number and arrangement of 34 in....Ch. 19 - Calculate the allowable tensile load for the...Ch. 19 - In the connection shown, 14 in. side and end...Ch. 19 - Design the fillet welds parallel to the applied...
Ch. 19 - A fillet weld between two steel plates...Ch. 19 - Design an end connection using longitudinal welds...Ch. 19 - Calculate the allowable tensile load for the butt...Ch. 19 - Calculate the allowable tensile load for the lap...Ch. 19 - Calculate the allowable tensile load for the butt...Ch. 19 - Rework Problem 19.10 assuming that both plates are...Ch. 19 - Rework Problem 19.12 assuming that the angle is an...Ch. 19 - Two ASTM A36 steel plates, each 12 in. by 12 in. ,...Ch. 19 - Rework Problem 19.20 changing the fasteners to 34...Ch. 19 - Calculate the minimum main plate thickness for the...Ch. 19 - A roof truss tension member is made up of 2L6412...Ch. 19 - Rework Problem 19.23 changing the fasteners to six...Ch. 19 - Determine the allowable tensile load that can be...Ch. 19 - The welded connection shown is subjected to an...Ch. 19 - In Problem 19.26, use a 38 in. fillet weld, change...
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- 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_forwardDesign the above joint using welded connection. The w=15mm. The required length of weld is: Answer: _____________________.arrow_forwardQuestion 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 nutarrow_forward
- 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 jointarrow_forwardA cylindrical air-compressor tank with an inside diameter of 50 mm is to be designed to withstand safely a working pressure of 0.86 MPa. Boiler plate with an allowable tensile stress of 76 MPa. All seams will be butt welds with an expected efficiency of 100 percent. a.) Determine the force per lineal mm of longitudinal seam, b.) Determine the required thickness “t” of wall, c.) Determine the unit stress in the circumferential seam.arrow_forwardQuestion 41. Two 10-mm-thick by 60-mm wide hot rolled SAE 1050 steel plates are to be butt-welded together with a CJP weld. Choose an appropriate electrode for tensile loading of the joint. 2. Find the torque that a 2-surface dry disk clutch can transmit if the outside and inside lining diameters are 120 mm and 70 mm, respectively, and the applied force is 10 kN. Assume uniform wear and μ = 0.4.arrow_forward
- Design a single riveted lap joint to connect two mild steel plates, 14.5 mm thick, the joint being designed for maximum efficiency. The allowable stresses are σt = 55 N/mm2 , = 32.5 N/mm2 and σc = 63.5 N/mm2 . Also calculate the minimum force per pitch which will rupture the jointarrow_forwardSketch a V-grooved butt joint, and label all of the joint's dimensions.arrow_forwardCan a tee weld be strong if the welds on both sides do not have deep penetration? Why or why not?arrow_forward
- Use LRFD and design a welded connection to resist the available strength of the tension member in Figure. All steel is A36. Show your results on a sketch, complete with dimensions.arrow_forwardA 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?arrow_forwardThe unit melting energy for a certain sheet metal is 10J/mm^3 .The thickness of each of the two sheets to be spot-welded is = mm. To achive required strength, it is desired to form a weldnugget that is =7 mm in diameter and 3.5 mm thick. The weld duration will be set at 44 sec .If the electrical resistance between the two surfaces is 144 micro-ohms and only one half the electricity generated will be used to form the weld nugget Determine the minium current level required in this operationarrow_forward
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