2019-2101ENG Tutorial 8 solution PDF

Title 2019-2101ENG Tutorial 8 solution
Course Mechanics Of Material I
Institution Griffith University
Pages 4
File Size 368.2 KB
File Type PDF
Total Downloads 109
Total Views 182

Summary

Download 2019-2101ENG Tutorial 8 solution PDF


Description

2101ENG Mechanics of Materials 1

Tutorial 8

Week 10-11

Problems taken from Ferdinand P. Beer E. Russell Johnston, Jr & John T. DeWolf, “Mechanics of Materials”,7th Edition, McGraw Hill.

________________________________________________________________________ Problem 6.1 Three full-size 50 x 100-mm boards are nailed together to form a beam that is subjected to a vertical shear of 1500 N. Knowing that the allowable shearing force in each nail is 400 N, determine the largest longitudinal spacing s that can be used between each pair of nails.

Problem 6.3

1

2101ENG Mechanics of Materials 1

Tutorial 8

Week 10-11

Problems taken from Ferdinand P. Beer E. Russell Johnston, Jr & John T. DeWolf, “Mechanics of Materials”,7th Edition, McGraw Hill.

________________________________________________________________________ Three boards, each 50 mm thick, are nailed together to form a beam that is subjected to a vertical shear. Knowing that the allowable shearing force in each nail is 600 N, determine the allowable shear if the spacing s between the nails is 75 mm.

2

2101ENG Mechanics of Materials 1

Tutorial 8

Week 10-11

Problems taken from Ferdinand P. Beer E. Russell Johnston, Jr & John T. DeWolf, “Mechanics of Materials”,7th Edition, McGraw Hill.

________________________________________________________________________ Problem 6.6

A column is fabricated by connecting the rolled-steel members by bolts of 18 mm diameter spaced longitudinally every 125 mm. Determine the average shearing stress in the bolts caused by a shearing force of 12 kN parallel to the y-axis.

y'o bf

tf

tw O’

d

x 'o

x C250×37

Figure 2

Material property of C250×37 Designation A C250×37

d

bf

tf

tw

Ix (x10 6mm 4) 4740 254 73.4 11.1 13.4 37.9

Iy x 6 4 (x10 mm ) 1.39 15.7

Solution I = 197.83 × 106 mm 4 254 10 + ) = 462000 mm3 2 2 3 VQ 12× 10 × 462000 = = 28.024 N / mm q= I 197.83 ×106 qs 28.024 × 125 Fbolt = = = 1751N 2 2 F 1751 τ= = = 6.88MPa A π9 2 Q = (350 ×10) ⋅(

3

2101ENG Mechanics of Materials 1

Tutorial 8

Week 10-11

Problems taken from Ferdinand P. Beer E. Russell Johnston, Jr & John T. DeWolf, “Mechanics of Materials”,7th Edition, McGraw Hill.

________________________________________________________________________ Problem 6.10 For the beam and loading shown, consider section n-n and determine (a) the largest shearing stress in that section, (b) the shearing stress at point a.

4...


Similar Free PDFs