EBK PRINCIPLES OF GEOTECHNICAL ENGINEER
9th Edition
ISBN: 8220103611718
Author: SOBHAN
Publisher: CENGAGE L
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Chapter 9, Problem 9.8P
To determine
Find the factor of safety against heaving at point A.
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STRUCTURES I
Homework #1: Force Systems
Name:
TA:
PROBLEM 1
Determine the horizontal and vertical components of
the force in the cable shown.
PROBLEM 2
The horizontal component of force F is 30 lb. What is the
magnitude of force F?
6
10
4
4
F = 600lbs
F = ?
18:02
28%
50
同
こ
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3 m
35 kN
4m
84 kN
+3m
EA = constant
E = 200 GPa
A
T5
4m
Add
1,200 mm²
Find Horizontal and vertical displacement at B
4 kN
m
m-
B
Determine Vertical displacement at C
Determine the displacement or point B of the steel beam shown in
Take E200 GPa, I = 500(106) mm4.
5 m 2/
40 kN
B
12 kN/m
4 m.
12 kN/m
3 m
B
10 m-
E
constant = 70 GPa
I 554 (10) mm
Determine Displacement at C
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In a floor of an industrial building, boilers are supported symmetrically on secondary
beams A and B, which have a centre-to-centre distance of 5 m and which are in turn
supported by the main beam, which has a span of 9 m (see Fig. 10.62). Design the
main beam given the following data:
Chapter 9 Solutions
EBK PRINCIPLES OF GEOTECHNICAL ENGINEER
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- When an open-ended square tube is placed vertically into a pool of water, the water rises 4 mm up inside of the tube. A) Determine the inner length of the square tube. A solid cylindrical rod is then placed vertically down the center of the open-ended square tube and the water rises an additional 4 mm up the tube. B) Determine the diameter of the solid rod that was inserted. 0.073 @ 20N m T C s = = o .arrow_forwardPlease use the following labels in the image such as Va, Vbr, Vbl and etc. Show step by step solution for each. Thanks!arrow_forwardThe single story building shown in Fig. 2 has an applied uniform load of 300 psf (0.3ksf) including the self weight of the beams and the girders. The roof has a 16 ft x 15 ft opening as shown. 1. Determine the axial loads on Columns C1 and C2 using reactions from the beams supported on the columns. 2. Determine the axial loads on Columns C1 and C2 using the concept of tributary areas.arrow_forward
- A built-up beam section is formed by welding 2xL8x6x1 angles to the bottom flange of W36x210 as shown in Fig. 1. Determine the following section properties of the built-up cross section: a. Cross sectional area, A (in2) b. The location of the centroids CG-X and CG-Y id of the built up section from the bottom of angle ( this value is given, so you need to check that you get this value) c. Moments of inertia ICG-X (in4) and ICG-Y (in4) d. Section modulus Sx (bot) (in3) and Sx (top) (in3) e. Radius of gyration rx (in) and ry (in) f. Weight of the built up section, w (lb/ft) (use density of steel = 490 pcf) g. Surface area of the built-up section, S(ft2/ft)arrow_forward11. Design the main beam of a building supporting concrete floor slab as shown in Fig. 10.61 and with the following data: (i) Beam centres: 6 m (ii) Span (simply supported): 7.4 m (iii) Concrete slab (spanning in two directions): 240-mm thick (iv) Finished screed: 40-mm thick (v) Imposed load: 4 kN/m² (vi) Take weight of concrete slab as 24 kN/m³ and total weight of 40-mm thick screed as 1.0 kN/m² Assume Fe 410 grade steel and take initial weight of beam as 1.0 kN/m. H H- Main beam 7.4 m 6.0 m H 40 mm screed I -H. Fig. 10.61 k 240 mm slab 6.0 m Typical bay of large floor areaarrow_forwardConsider the situation in Question 2. If all the cables are made of the same material and have a maximum tensile force of 500 lb, what is the heaviest load that can be supported by the system?arrow_forward
- A flexible circular area is subjected to a uniformly distributed load of 148 (see the figure below). The diameter of the load area is 2 . Estimate the average stress increase () below the center of the loaded area between depths of 3 and 6 . Use the equations: and (Enter your answer to three significant figures.) =arrow_forwardA square flexible foundation of width B applies a uniform pressure go to the underlying ground. (a) Determine the vertical stress increase at a depth of 0.625B below the center using Aσ beneath the corner of a uniform rectangular load given by Aσ = Variation of Influence Value I qoI. Use the table below. n 0.8 1.0 m 0.2 0.4 0.5 0.6 0.2 0.01790 0.03280 0.03866 0.04348 0.05042 0.05471 0.4 0.03280 0.06024 0.07111 0.08009 0.09314 0.10129 0.5 0.03866 0.07111 0.08403 0.09473 0.11035 0.12018 0.6 0.04348 0.08009 0.09473 0.10688 0.12474 0.13605 0.8 0.05042 0.09314 0.11035 0.12474 0.14607 0.15978 1.0 0.05471 0.10129 0.12018 0.13605 0.15978 0.17522 (Enter your answer to three significant figures.) Ασ/90 = (b) Determine the vertical stress increase at a depth of 0.625B below the center using the 2 : 1 method equation below. 90 x B x L Aσ = (B+ z) (L + z) (Enter your answer to three significant figures.) Ασ/90 = (c) Determine the vertical stress increase at a depth of 0.625B below the center using…arrow_forwardPoint loads of magnitude 100, 200, and 360 act at , , and , respectively (in the figure below). Determine the increase in vertical stress at a depth of 6 below point . Use Boussinesq's equation. (Enter your answer to three significant figures.) =arrow_forward
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