Solutions for Statics and Strength of Materials for Construction, Engineering Technology, and Architecture, 1st Edition by Askar (All Chapters included)

Statics and Strength of Materials for
Construction, Engineering
Technology, and Architecture, 1st
Edition by Mohamed Askar




Complete Chapter Solutions Manual
are included (Ch 1 to 19)




** Immediate Download
** Swift Response
** All Chapters included

,Table of Contents are given below


1. Introduction
2. Coplanar Force Systems
3. Equilibrium of Particle and Rigid Body
4. Design Load Distribution
5. Principles of Statics of Beam and Frame Reactions
6. Principles of Statics of Truss Reactions and Axial Forces
7. Arches, Cables and Pulleys
8. Space Force Systems
9. Centroids of Area
10. Moment of Inertia of Area
11. Friction
12. Properties of Materials
13. Axial Deformation
14. Bending and Shear Stress in Beams and Frames
15. Torsion
16. Combined Streets
17. Stress Transformation
18. Deflection of Beams
19. Stress in Columns

, Solutions Manual organized in reverse order, with the last chapter displayed first, to ensure that all
chapters are included in this document. (Complete Chapters included Ch19-1)




Solution of Chapter 19
Stress in Columns

Problem 19.1

A 12 ft long steel column is fixed supported at both ends with no side sway. The cross-section of
the column is 8 in. by 6 in. Calculate the critical buckling load of the column. Steel has a modulus
of elasticity of 29,000 ksi.

Solution:

Modulus of elasticity, E = 29,000 ksi
Length of column, L = 12 ft
Effective length factor, K = 0.5 [both ends fixed with no side sway]
6 in.(8 in.)3
Moment of inertia about one axis, I x = = 256 in.4
12
8 in.(6 in.)3
Moment of inertia about another axis, I y = = 144 in.4 [this is the minimum and
12
the buckling will occur about this axis.]
I min 144 in.4
Minimum radius of gyration, rmin = = = 1.73 in.
A 8 in. 6 in.
 2E  2 (29,000 ksi)
Euler buckling stress, Fe = 2
= 2
= 165 ksi (Answer)
 KL   0.5(12  12in.) 
   
 rmin   1.73 in. 
(165 ksi may be very high to consider within the linear elastic limit)




19-1

, Problem 19.2

A 20 ft long steel column in a braced frame is fixed at one end and pinned at other end with no
side sway. The column has the outer diameter of 7 inch and inner diameter of 6 inch. Calculate the
critical buckling load of the column. Steel has a modulus of elasticity of 29,000 ksi.

Solution:

Modulus of elasticity, E = 29,000 ksi
Length of column, L = 20 ft
Effective length factor, K = 0.70 [fixed at one end and pinned at other without sway]
  d   d     7   6  
4 4 4 4

Moment of inertia about one axis, I =  o  −  i   =   −    = 54.24in 4
4  2   2   4  2   2  

I min 54.24 in.4
Minimum radius of gyration, rmin = = = 2.30in.

A
( 7 in )2 − ( 6 in )2 
4 
 E2
 (29,000 ksi)
2
Euler buckling stress, Fe = 2
= 2
= 53.6 ksi (Answer)
 KL   0.7(20  12in.) 
   
 rmin   2.30 in. 




19-2