Structural Steel Design (6th Edition)
6th Edition
ISBN: 9780134589657
Author: Jack C. McCormac, Stephen F. Csernak
Publisher: PEARSON
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Question
Chapter 2, Problem 2.6PFS
To determine
The maximum factored uniform load per foot given loading.
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A simply supported timber beam 20 cm wide and 40 cm deep carries a point load W at the middle point. The permissible stresses in flexure (bending) and shear is 100 N/cm2 and 150 N/cm2 respectively. Calculate the span length below which the shear stress will govern the safe load and above which the bending stress will govern the safe load.
6. The timber beam has a cross-sectional area of 2000 mm? and its modulus of elasticity is 12 GPa. Axial
loads are applied at points B, C, and D as shown. Calculate the total change in length of the beam. Answer:
Stotal = 2.33 mm
40 kN
35 kN
20 kN
D
A
1.60 m
3.20 m
3.20 m
An eight Storey city library is proposed to be constructed
in Baguio.
The structure shall be a reinforced concrete building (no
shear walls).
The weight per floor (including 1st floor) is estimated to
be 1800 KN.
The roof deck weighs 1700 KN. Solve for the Base Shear,
V, for the
condition above using the following formulas:
1. V =ZICKSW (SEAOC CODE 1974 - 4 TH Revision)
Adopted by National Structural Code of the Philippines,
3rd edition
(NSCP 1987)
2. V = ZICW/Rw National Structural Code of the
Philippines, 4th
edition (NSCP 1992), Volume I
3. Most critical formula for V as indicated in Sec.
208.5.2 of the
NSCP Vol.1 5th edition (2001).
4. Most critical formula for V as indicated in the NSCP
Vol.1 6th
edition (2015).
1700 KN
100 KN
32
1800 KN
La Tide
Toeang
Warg
32
1800 KN
32
259
32
1800 KN
32
1800 KN
City Cam Cena, Nagun C Bengt
Nearet Active failt Trace t
Fae tunao Fa
Segnet Nane Tut Fat
Vear Maper 2012
Ma t et 10 00
32
1800 KN
3.5
20
Floor height: 1st floor 3.5m and the rest of the…
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- Determine the ultimate load distribution acting on all beams and girders in the given floor framing below. Assume that the supports are simply supported. Compute for the reactions and loads in each beam. Assume unit weight of concrete to be 23.6 kN/m³. Always simplify your load distributions (distributed loads acting on the same points/locations should be added up). Monolithic Floor All concrete slabs, t = 100 mm All concrete beams, b = 250 mm; h= 500 mm All concrete columns, 300 mm x 300mm Superimposed Loads Floor Finish: Ceiling: Ceramic tile ½ inch mortar Frame Partition: Exterior Walls Wood furring Gypsum board 14 inch thick Mechanical Duct Allowance (MDA) Utilities: twice the weight of MDA Live Load Wood, studs with gypsum board h=2.8m 5m A H 3m Normal Unit Weight Concrete Masonry/ Concrete Hollow Blocks, CHB 4 inch thick with plastering on both faces, full grout height=3m (no interior walls) Residential NOTE: REFER TO NSCP 2015 CHAPTER 2 FOR THE MINIMUM DESIGN LOADS. USE LRFD…arrow_forwardThe concrete girders shown in the photo of the passenger car parking garage span 30 ft and are 15 ft on center. If the floor slab is 5 in. thick and made of reinforced stone concrete, and the specified live load is 50lb/ft2 , determine the distributed load the floor system transmits to each interior girder.arrow_forward1. The flooring of a warehouse is made up of double-tee joists (DT) as shown. The joists are simply supported on a span of 7.5 m. and pre tensioned with one tendon in each stem with an initial force of 745 kN each, located at 75 mm above the bottom fiber, loss of stress at service load is 18%. Load imposed on the joists are: Dead load = 2.3 kPa Live load = 6 kPa Properties of DT: A = 200,000 mm² yt = 88 mm a = 2.4 m I = 1880 x 106 mm4 yb = 267 mm a=2.4 m N.A. 75 mm 192 mm 88 mm 267 mm a. Compute the stress at the bottom fibers of the DT at mid-span due to the initial pre stressing force alone. b. Compute the stress at the bottom fibers of the DT at mid-span due to the service loads and pre stress force. c. What additional super imposed load can the DT carry such that the resulting stress at the bottom fibers at mid-span is zero.arrow_forward
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