Solutions Manual Structural Analysis, Principles, Methods and Modelling by Gianluca Ranzi, Raymond Ian Gilbert

solutions MANUAL FOR
structural
analysis
Principles, Methods
and Modelling



by

Gianluca Ranzi and
­Raymond Ian Gilbert

, Table of Contents

Acknowledgements i

2 Statics of structures: Equilibrium and support reactions 1

3 Internal actions of beams and frames 43

4 Statically determinate trusses 93

5 Euler-Bernoulli beam model 167

6 Slope-deflection methods 221

7 Work-energy methods 281

8 The force method 335

9 Moment distribution 431

10 Truss analysis using the stiffness 479

11 Beam analysis using the stiffness method 533

12 Frame analysis using the stiffness method 581

13 Introduction to the finite element method 653

14 Introduction to the structural stability of columns 669

15 Introduction to nonlinear analysis 679

, Chapter 2

Statics of structures:

Equilibrium and support reactions




1

, Chapter 2 – Statics of structures: Equilibrium and support reactions




2.1 Determine the components Fx and Fy parallel (c)
to the x- and y-axes for the forces F shown.
y
y y =290
y
y x x x
F = 30 kN x
 
70 20 20
35 F = 40 kN F = 50 kN
x

F = 50 kN F = 50 kN
(a) (b) (c)
Fx = F × cos() = 50 × cos (290) = 17.10 kN
_________________________________________ Fy = F × sin() = 50 × sin (290) = – 46.98 kN

y
(a) 17.1 kN
x
y
F = 30 kN

=35
x
46.98 kN
 F = 50 kN
Fx = F × cos() = 30 × cos (35 ) = 24.57 kN
Fy = F × sin() = 30 × sin (35) = 17.21 kN


y
F = 30 kN
17.21 kN

x
24.57 kN




(b)

y y
x =200
x
70
F = 40 kN F = 40 kN




Fx = F × cos() = 40 × cos (200) = – 37.59 kN
Fy = F × sin() = 40 × sin (200) = – 13.68 kN


y
37.59 kN
x
13.68 kN
F = 40 kN




3