Construction Materials, Methods and Techniques (MindTap Course List)
4th Edition
ISBN: 9781305086272
Author: William P. Spence, Eva Kultermann
Publisher: Cengage Learning
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Chapter 8, Problem 11RQ
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The styles of reinforcing bars used
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Question 2
(a) A simple circular hollow section (CHS) tubular K-joint in a steel structure, subjected to balanced
axial loading, is illustrated in Figure 2a. Determine the maximum hot spot stress at the joint
intersection of the chord and the loaded brace B.
(b) The steel structure is installed in the seawater with cathodic protection. Determine the number of
stress cycles to failure based on the maximum hot stress range obtained in part (a). Use the
NORSOK standard. (Refer to S-N curves for tubular joints in air environment and seawater with
cathodic protection).
(c) Estimate the number of load repetitions required to induce fatigue failure in the tubular joint, based
on the load history provided in Figure 2b. The nominal yield and ultimate tensile strength are 355
N/mm² and 510 N/mm², respectively. Assume a damage limit of 1.0. Use the Modified Goodman
formulation to determine the equivalent completely reversed stress.
(d) Describe briefly the procedure to determine the hot…
The steel member is a fillet welded built-up section that comprises two flange plates (100mm x 20mm) and a
web plate (250mm x 10mm) as depicted in Section A-A. The leg size of the weld is 8 mm. Use an
appropriate consequence class. Based on the damage tolerant method and the modified Goodman equation.
Determine an equivalent completely reversed stress. Ignore the vibration and dynamic amplification. Use
Euro-code 1993-1-9.
(a) Calculate the maximum and minimum stresses at steel member section A-A.
(b) Check the fatigue resistance of the steel member at Section A-A using the fatigue limit.
(c) Discuss the possible failure mode of the steel member due to fatigue loading.
State your design assumptions, if any.
Steel plate (Flange)
100mm x 20mm
10.0 m
Fillet weld (manual)
(Typical)
Steel plate (Web)
250mm x 10 mm
Steel plate (Flange)
100mm x 20mm
Section A-A
Fixed end
Welded built-up
steel section
5.0 m
A
2.5m
3.0 m
Fatigue load range 5 kN
A
Total weight of steel section
Total weight of…
Chapter 8 Solutions
Construction Materials, Methods and Techniques (MindTap Course List)
Ch. 8 - How long can ready-mix concrete remain in the...Ch. 8 - What devices are used on site to move concrete to...Ch. 8 - How is concrete consolidated?Ch. 8 - What may happen if the temperature of a concrete...Ch. 8 - Why are freshly poured concrete slabs bullfloated?Ch. 8 - Prob. 6RQCh. 8 - What is meant by hydration?Ch. 8 - Prob. 8RQCh. 8 - Prob. 9RQCh. 8 - Prob. 10RQ
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- 30 20 10 Stress N/mm² 0 -10 -20 -30 Time Question 1 A Grade S355 steel member, which forms part of the structural framework supporting a storage tank in a warehouse, is subjected to various loads, as shown in Figure 1. The yield and tensile strength of the steel member are 355 N/mm² and 510 N/mm², respectively. The steel member is subjected to axial tension due to its self-weight and appurtenances of 40.0kN. The 10.0kN storage tank is positioned 1.0 m from the centreline of the steel member, and it experiences a fatigue load range of 5.0kN. The steel member is a fillet welded built-up section that comprises two flange plates (100mm x 20mm) and a web plate (250mm x 10mm) as depicted in Section A-A. The leg size of the weld is 8 mm. Use an appropriate consequence class. Based on the damage tolerant method and the modified Goodman equation. Determine an equivalent completely reversed stress. Ignore the vibration and dynamic amplification. Use Euro-code 1993-1-9. (a) Calculate the maximum…arrow_forwardPlease do not use design aid - R. Show step by step and every formular usedarrow_forwardFollowing is the variation of the field standard penetration number (№60) in a sand deposit: Depth (m) N60 1.5 6 3 8 4.5 9 6 8 7.5 9 13 14 The groundwater table is located at a depth of 6 m. Given: the dry unit weight of sand from 0 to a depth of 6 m is 16 kN/m³, and the saturated unit weight of sand for depth 6 to 12 m is 18.2 kN/m². Use the relationship given in the equation CN = 1 σo/Pa 0.5 to calculate the corrected penetration numbers. (Round your answers to the nearest whole number.) Depth (m) Neo (N1)00 1.5 3 6 8 4.5 9 6 7.5 9 14 8 13arrow_forward
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