M5-6 CORRECTION DUE TO SAG. A 50-m steel tape weighs 0.04 kg/m and is supported at its end points and at the 8-m and 25-m marks. If a pull of 6 kg is applied, determine the following: a) Correction due to sag between the 0-m and 8-m marks, 8-m and 25-m marks, and the 25-m and 50-m marks. b) Correction due to sag for one tape length. c) Correct distance between the ends of the tape.

M5-6 CORRECTION DUE TO SAG. A 50-m steel tape weighs 0.04 kg/m and is supported at its end points and at the 8-m and 25-m marks. If a pull of 6 kg is applied, determine the following: a) Correction due to sag between the 0-m and 8-m marks, 8-m and 25-m marks, and the 25-m and 50-m marks. b) Correction due to sag for one tape length. c) Correct distance between the ends of the tape.

Structural Analysis

6th Edition

ISBN:9781337630931

Author:KASSIMALI, Aslam.

Publisher:KASSIMALI, Aslam.

Chapter2: Loads On Structures

Section: Chapter Questions

Problem 1P

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A 50m steel tape weighs 0.05 kg/m and is supported at its end points and at the 10m and 30m marks. If a pull of5 kg is applied, determine the following. a. Correction due to sag between the 0m and 10m marks, 10m and 30m marks, and the 30m and 50m marks. b. Correction due to sag for one tape length c. Correct distance between the ends of the tape
2. A 50- m steel tape weighs 0.04 kg/m and is supported at its end points andat the 9-m and 30- m marks. If a pull of 6kg is applied, determine thefollowing:a) correction due to sag between the 0-m and 9- m marks, 9-m and 30-mmarks, and the 30- m and 50- m marks.b) correction due to sag for one tape lengthc) correct distance between the ends of the tape.
A 30 m steel tape was standardized at a temperature of 20°C and under a pull 5 kg. The tape was used in catenary at a temperature 25°C and under a pull of P kg. The cross-sectional area of the tape is 0.02 cm2, its weights per unit length is 22 g/m. Young’s modulus = 2 x 10-6 kg /cm2, α = 11 x 10-6 per °C. Find the correct horizontal distance, if P is equal to (i) 5 kg, and (ii) 11 kg.
Draw a shear diagram Cx = 0 Cy=30 Mc = -210 V {0<=x<8} = 0 V { 8< x <=14} = -5 (x-8)
Situation 11. An overhang beam is loaded as shown below. The beam cross-section was built by attaching two (2) channels to a 9mm thick plate using 16mm rivets The property of the channel is given below: Depth, D=225 mm Flange Width, Bf=112.5 mm. Flange Thickness, tf=9mm Web Thickness, tw=9 mm The allowable flexural stress on the beam is 180 MPa. Rivets has a capacity of τ= 100 MPa on shear, for bearing, σb=200 MPa on single sheer, σb= 260MPa on double shear. 3. Determine the maximum allowable moment, M(all) in kn-m, base on the beam's cross-section. 4. Determine the location of the maximum moment on the beam in meters.
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A T-beam for a floor system has a slab thickness of 75mm and a web thickness of 375mm. It has an effective depth of 600mm. The beam carries a dead load moment of 270Kn.M and a liveload moment of 460Kn.m. It has a span of 5.4m and a spacing center to center of 1.8m. f’c=20.7MPa, fy=344.7MPa. a) Determine the effective width of the flange b) Determine the depth of compression block c) Determine the steel area required
Situation 11 An overhang beam is loaded as shown below. The beam cross-section was built by attaching two (2) channels to a 9mm thick plate using 16mm rivets The property of the channel is given: Depth, D=225 mm Flange Width, B=112.5 mm. Flange Thickness, tf=9mm Web Thickness, tw9 mm The allowable flexural stress on the beam is 180 MPa. Rivets has a capacity of t 100 MPa on shear, for bearing, o.-200 MPa on single sheer, a-260MPa on double shear Determine the location of centroid, y, from top of the beam in mm Determine the moment of inertia, I, of the section in mm^4. Determine the maximum allowable moment, M(all) in kn-m, base on the beam's cross-section. Determine the location of the maximum moment on the beam in meters. Determine the maximum moment, in kN-m, on the beam Determine the maximum flexural stress on the beam in MPa.
A square, tied column 410 mm wide is reinforced with 4 25-mm-diameter vertical bars with fy = 420 MPa enclosed in 10-mm-dia. ties. Using f'c = 23 MPa, calculate the nominal axial strength at balanced condition in kN with bending about the x axis. Write your answer with 2 decimal places only. ANSWER : 1392.69KN
1. Two A36 16mm Thick Steel Plates Are Connected By Four Rivets With Fv=152 MPa As Shown. Given: Rivet DIa 20 mm x=100 mm a. Calculate "P" in KN considering tension failure on gross area b. Calculate "P" in KN considering tension failure on net area c. Calculate "P" in KN considering failure on bearing on projected area. d. Calculate "P" in KN considering failure on shear on rivets e. Calculate "P" in KN considering failure on bearing on block shear f. Calculate the maximum safe load "P" in KN
Situation 11 An overhang beam is loaded as shown below. The beam cross-section was built by attaching two (2) channels to a 9mm thick plate using 16mm rivets The property of the channel is given below: Depth, D=225 mm Flange Width, B=112.5 mm. Flange Thickness, tf=9mm Web Thickness, tw9 mm The allowable flexural stress on the beam is 180 MPa. Rivets has a capacity of t 100 MPa on shear, for bearing, o.-200 MPa on single sheer, a-260MPa on double shear. 2. Determine the moment of inertia, I, of the section in mm^4.