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 19.7P
Select the number and arrangement of
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3. An M16 x 2 bolt with a slotted hex nut is used to clamp together two 25 mm steel plates. The joint
is assembled with a size 16 N washer underneath the bolt head. The bolt is a 5.8 class metric grade,
with a proof strength of 380 MPa. The joint is intended to be permanent. Determine a suitable length
for the bolt, rounded up to the nearest standard size in millimeters. Calculate the bolt stiffness, the
elongation of the bolt after the necessary preload is developed, and the bolt torque required to apply
the preload. Assume that coefficients of friction between the bolt, nut, and collar are all ~0.15. For
steel, use E = 207 GPa.
Standard threaded fastener lengths, dimensions of hexagonal nuts, and dimensions of metric plain
washers are provided in the textbook appendices.
Two 1/4 in × 8 in A36 steel plates are joined with a lap joint using 3/4 in A325-N bolts. If a concentric unfactored live load of 50 kips (dead load = 0) is required to be resisted by the connection and the bolt shear strength controls the design, the minimum required number of bolts is?
4. Determine the design stress for bolts in a cylinder cover where the load is fluctuating due to gas
pressure. The maximum load on the bolt is 50 kN and the minimum is 30 kN. The load is unpredict-
able and factor of safety is 3. The surface of the bolt is hot rolled and the suface finish factor is 0.9.
During a simple tension test and rotating beam test on ductile materials (40 C 8 steel annealed), the
following results were obtained :
Diameter of specimen =12.5 mm; Yield strength= 240 MPa; Ultimate strength= 450 MPa; Endurance
limit = 180 MPa.
[Ans. 65.4 MPa]
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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- A 3-in Aluminum plate and a 2-in cast-iron plate are compressed with one bolt and nut. The bolt is 7/8 in-9 UNC. (1) Determine a suitable length for the bolt, rounded up to the nearest in. (2) Determine the bolt stiffness. (3) Determine the stiffness of the members.arrow_forwardUse LRFD and design a 13-foot-long tension member and its connection for a service dead load of 8 kips and a service live load of 24 kips. No slip of the connection is permitted. The connection will be to a 3⁄8-inch-thick gusset plate, as shown in Figure . Use a single angle for the tension member. Use Group A bolts and A572 Grade 50 steel for both the tension member and the gusset plate.arrow_forwardA 5⁄8-inch-thick tension member is connected to two 1⁄4-inch splice plates, as shown in Figure The loads shown are service loads. A36 steel and 5⁄8-inchdiameter, Group A bolts will be used. If slip is permissible, how many bolts are required? Each bolt centerline shown represents a row of bolts in the direction of the width of the plates.arrow_forward
- As shown in figure below, two plates are clamped by washer-faced 2 in-20 UNF SAE grade 5 bolts each with a standard 2 N steel plain washer. The top plate is steel and the bottom pplate is gray cast iron. If a total of 8 bolts where used to close 1 feet diameter pipe which applied at maximum pressure of 150 psi. Calculate the static and fatigue safety factor of this joint. Used Goodman theory for tha fatigue safety 1/2 factor. 3/4arrow_forwardCondition #1: A structural support for a machine is subjected to a static compression load of 20 kN. The support is manufactured from a circular rod made from SAE 1040 Hot Rolled steel. Specify suitable diameter for the cross section of the rod based on the basic size. Steel data are available in Table A-10 from the textbook. Condition #2: The same structural support of the basic size determined in Condition 1 is subjected to a tensile load of 15 kN that is repeated several thousand times over the life of the machine. This load is not an addition to the 20 kN. Specify a suitable steel that is suitable to this application based on the basic size determined in Condition #1. Loading of Condition #1 does not apply here. Condition #3: The same structural support from Condition 2 is heated from room temperature of 25°C. The support is placed inside a frame on both ends. There is a total clearance of 0.2 mm between the support and its frame. Initial length of the rod is 200 mm. Specify the…arrow_forwardAgui - 4550/2 =2276 two plates each with thickness t - 16 mm are bolted together with 6-222 mndia. bolts firming alap Joint Bolt spacing are follows St- $40mm S2 - 80mm, S3=150mm effective bolt hole dia. is 25mm A36 is used. Compute the allowable tensile Strength P t = 16 mm &t 52 747₁ 0 O O 40 60 80 40 +=16mm S-400 $3-100 S2-40 7 @ 4umm Parrow_forward
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