EBK ENGINEERING FUNDAMENTALS: AN INTROD
5th Edition
ISBN: 9780100543409
Author: MOAVENI
Publisher: YUZU
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Chapter 1.4, Problem 4BYG
To determine
Explain the acts of nuclear engineers.
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A W16×67 of A992 steel has two holes in each flange for 7/8-inch-diameter bolts.
For A992 steel: Fy
= 50 ksi, F₁ = 65 ksi.
For a W16×67: bƒ
=
10.2 in., tf
= 0.665 in., Zx =130 in.3 and S
= 117 in.3
a. Assuming continuous lateral support, verify that the holes must be accounted for and determine the nominal flexural strength.
(Express your answer to three significant figures.)
Mn
=
ft-kips
b. What is the percent reduction in strength?
(Express your answer to three significant figures.)
%
Reduction =
A gravity retaining wall is shown in the figure below. Calculate the factor of safety with respect to overturning and sliding, given the following data:
Wall dimensions: H = 6 m, x1 = 0.6 m, x2 = 2 m, x3 = 2m,
x4 = 0.5 m, x5 = 0.75 m, x6 = 0.8 m, D= 1.5 m
Soil properties: 1 = 17.5 kN/m³, ø₁ = 32°, 12 = 18 kN/m³,
=22°, 40 kN/m²
Y₁
H
D
x2
x3
x5
X6
Use the Rankine active earth pressure in your calculation. Use Yconcrete = 23.58 kN/m³. Also, use k₁ = k₂ = 2/3 and P₁ = 0 in the equation
FS
S(sliding)
| tan(k102) + Bk₂c½ + Pp
(Σν).
Pa cos a
(Enter your answers to three significant figures.)
FS (overturning)
FS (sliding)
Q4 Use
b member
Castigliano's second theorem to determine
the structure shown
Forces
In figure below
longer than
+
In
IF
required. For all members
member EC
IS 10mm
E = 200 KN/mm²
200KN YE
FV 100 KN
A = 1800 mm² and
2
2m
3m
B
D
3m
8M
*
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- For the cantilever retaining wall shown in the figure below, let the following data be given: Wall dimensions: H = 8 m, x1 = 0.4 m, x2 = 0.6 m, x3 = 1.5 m, x4 3.5 m, x5 = 0.96 m, D= 1.75 m, a = 10° Soil properties: ₁ = 17.3 kN/m³, 1₁ = 32°, Y2 = 17.6 kN/m³, 2=28°, c₂ = 30 kN/m² The value of Ka is 0.3210. For 2=28°: N = 25.80; N₁ = 14.72; N₁ = 16.72. 3. Also, use k₁ = k₂ = 2/3 and Pp = 0 in the equation Calculate the factor of safety with respect to overturning, sliding, and bearing capacity. Use concrete = 24.58 kN/m³. A FS (sliding) (V) tan(k₁₂) + Bk₂c½₂ + Pp Pa cos a (Enter your answers to three significant figures.) FS (overturning) FS(sliding) FS (bearing)arrow_forwardUse the Force Method to analyse the structure. After the analysis, draw Bending Moment and Shear Force diagrams. Redundants need to be put at 'j' and 'k' as moments (EI is constant all across the structure)I am so confused as to how to approach the problem. I would really appreciate an answer with a little explanation, or helpful working out!arrow_forwardI got 5.97 mm please show your work clearly. thank youarrow_forward
- A 7K SK-> VE 3 F T A=52 E=29000 ksi diagonal members 6' A=30.25.72 E=1800 ksi for horizontal & vertical member ↓ B Oc AD 8 Primary Structures remove roller @C make D a roller For Primary and Cut BF For redundant Ik ↑ ec Ik = @D Ik @BFarrow_forwardConsider the geometric and traffic characteristics shown below. Approach (Width) North South East West (56 ft) (56 ft) (68 ft) (68 ft) Peak hour Approach Volumes: Left Turn 165 105 200 166 Through Movement 447 400 590 543 Right Turn 162 157 191 200 Conflicting Pedestrian Volumes 900 1,200 1,200 900 PHF 0.95 0.95 0.95 0.95 For the following saturation flows: Through lanes: 1,600 veh/h/In Through-right lanes: 1,400 veh/h/In Left lanes: 1,000 veh/h/In Left-through lanes: 1,200 veh/h/In Left-through-right lanes: 1,100 veh/h/In The total cycle length was 283 s. Now assume the saturation flow rates are 10% higher, that is, assume the following saturation flow rates: Through lanes: 1,760 veh/h/In Through-right lanes: 1,540 veh/h/In Left lanes: 1,100 veh/h/In Left-through lanes: 1,320 veh/h/In 1,210 veh/h/In Left-through-right lanes: Determine a suitable signal phasing system and phase lengths (in s) for the intersection using the Webster method. (Enter the sum of green and yellow times for…arrow_forwardThe given beam has continuous lateral support. If the live load is twice the dead load, what is the maximum total service load, in kips / ft, that can be supported? A992 steel is used: Fy = 50 ksi, Fu=65 ksi. Take L = 30 ft. bf For W40 x 149: 2tf = 7.11, = = 54.3, Z 598 in.³ tw W W40 X 149 L (Express your answers to three significant figures.) a. Use LRFD. Wtotal = kips/ft b. Use ASD. Wtotal kips/ftarrow_forward
- The beam shown in the figure below is a W16 × 31 of A992 steel and has continuous lateral support. The two concentrated loads are service live loads. Neglect the weight of the beam and determine whether the beam is adequate. Suppose that P = 52 k. For W16 × 31: d = 15.9 in., tw = 0.275 in., h/tw = 51.6, and M = M₁ = 203 ft-kip, Mn/₁ = Mp/α = 135 ft-kip. P Р W16 x 31 a. Use LRFD. Calculate the required moment strength, the allowable shear strength, and the maximum shear. (Express your answers to three significant figures.) Mu = OvVn = ft-kip kips kips Vu = Beam is -Select- b. Use ASD. Calculate the required moment strength, the allowable shear strength, and the maximum shear. (Express your answers to three significant figures.) Ma = Vn/b - Va = Beam is -Select- ft-kip kips kipsarrow_forwardDetermine the smallest value of yield stress Fy, for which a W-, M-, or S-shape from the list below will become slender. bf/2tfh/tw Shape W12 × 72 8.99 22.6 W12 × 26 8.54 47.2 M4 × 6 11.9 22.0 M12 x 11.8 6.81 62.5 M6 × 4.4 5.39 47.0 S24 × 80 4.02 41.4 S10 × 35 5.03 13.4 (Express your answer to three significant figures.) Fy = ksi To which shape does this value apply? -Select- ✓arrow_forwardCompute the nominal shear strength of an M12 × 11.8 of A572 Grade 60 steel (Fy = 60 ksi). For M12 x 11.8: d = 12 in., tw = 0.177 in., h/tw = 62.5. Vn = kipsarrow_forward
- A flexural member is fabricated from two flange plates 1/2 × 71/2 and a web plate 3/8 × 19. The yield stress of the steel is 50 ksi. a. Compute the plastic section modulus Z and the plastic moment Mp with respect to the major principal axis. (Express your answers to three significant figures.) Z = Mp = in. 3 ft-kips b. Compute the elastic section modulus S and the yield moment My with respect to the major principal axis. (Express your answers to three significant figures.) S = My = in.3 ft-kipsarrow_forward= 65 ksi. A W16×36 of A992 steel has two holes in each flange for 7/8-inch-diameter bolts. For A992 steel: Fy = 50 ksi, Fu For a W16×36: bƒ = 6.99 in., tƒ = 0.430 in., Z = 64.0 in.³ and Sx = 56.5 in.³ a. Assuming continuous lateral support, verify that the holes must be accounted for and determine the nominal flexural strength. (Express your answer to three significant figures.) Mn = ft-kips b. What is the percent reduction in strength? (Express your answer to three significant figures.) Reduction = %arrow_forwardFind the reinforcements for the mid span and supports for an interior 9 in. thick slab (S-2) in thefloor from Problem 1. Ignore the beams and assume that the slab is supported by columns only (i.e.a flat plate). Sketch the slab and show the reinforcements including the shrinkage andtemperature reinforcement steel. Use f c’ = 4,000 psi and f y = 60,000 psi.NOTE: Problem 3 requires additional column placements at locations such as C and D. The stripof slab between these two columns will behave as a beam support to the one-way slab (with 10 ft.span). Problem 1. The figures below shows the framing plan and section of a reinforced concrete floor system.Floor beams are shown as dotted lines. The weight of the ceiling and floor finishing is 6 psf,that of the mechanical and electrical systems is 7 psf, and the weight of the partitions is 180psf. The floor live load is 105 psf. The 7 in. thick slab exterior bay (S-1) is reinforced with #5rebars @ 10 in. o.c. as the main positive…arrow_forward
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