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Part One

Analysis of Determinate Structures

1 Principles of statics

Notation

F force fi angle in a triangle opposite side F (^) i H horizontal force l length of member M bending moment P axial force in a member R support reaction V vertical force W (^) LL concentrated live load w (^) DL distributed dead load θ angle of inclination

1.1 Introduction

Statics consists of the study of structures that are at rest under equilibrium conditions. To ensure equilibrium, the forces acting on a structure must bal- ance, and there must be no net torque acting on the structure. The principles of statics provide the means to analyze and determine the internal and external forces acting on a structure. For planar structures, three equations of equilibrium are available for the determination of external and internal forces. A statically determinate struc- ture is one in which all the unknown member forces and external reactions may be determined by applying the equations of equilibrium. An indeterminate or redundant structure is one that possesses more unknown member forces or reactions than the available equations of equilib- rium. These additional forces or reactions are termed redundants. To determine the redundants, additional equations must be obtained from conditions of geo- metrical compatibility. The redundants may be removed from the structure, and a stable, determinate structure remains, which is known as the cut-back structure. External redundants are redundants that exist among the external reactions. Internal redundants are redundants that exist among the member forces.

Principles of statics 5

Forces acting in one plane are coplanar forces. Space structures are three- dimensional structures and, as shown in Figure 1.4, may be acted on by non- coplanar forces.

50 kips 20 kips 30 kips

1

2 3 2 3

4 1 4

(i) (ii)

Figure 1.

Figure 1.

Figure 1.

In a concurrent force system, the line of action of all forces has a common point of intersection. As shown in Figure 1.5 for equilibrium of the two-hinged arch, the two reactions and the applied load are concurrent.

6

It is often convenient to resolve a force into two concurrent components. The original force then represents the resultant of the two components. The direc- tions adopted for the resolved forces are typically the x - and y -components in a rectangular coordinate system. As shown in Figure 1.6, the applied force F on the arch is resolved into the two rectangular components:

H F V F

cos sin

θ θ

F V

H

θ

Figure 1.

F

Fl Fl Fl

F M^ ^ Fl

F F F 1 2 1 2 1 2

 

l

(i) (ii) (iii)

Forces

Moment

Figure 1.

The moment acting at a given point in a structure is given by the product of the applied force and the perpendicular distance of the line of action of the force from the given point. As shown in Figure 1.7, the force F at the free end of the cantilever produces a bending moment, which increases linearly along the length of the cantilever, reaching a maximum value at the fixed end of:

M  Fl

The force system shown at (i) may also be replaced by either of the force systems shown at (ii) and (iii). The support reactions are omitted from the fig- ures for clarity.

Structural Analysis: In Theory and Practice

8 Structural Analysis: In Theory and Practice

Example 1.

Determine the support reactions of the pin-jointed truss shown in Figure 1.9. End 1 of the truss has a hinged support, and end 2 has a roller support.

(i) Applied loads (ii) Support reactions

V 3  20 kips V 4  20 kips

V 1  15 kips

H 1  10 kips

H 3  10 kips

V 2  25 kips

3 4

1 8 ft

8 ft

8 ft

5 2

Figure 1.

Solution

To ensure equilibrium, support 1 provides a horizontal and a vertical reaction, and support 2 provides a vertical reaction. Adopting the convention that hori- zontal forces acting to the right are positive, vertical forces acting upward are positive, and counterclockwise moments are positive, applying the equilibrium equations gives, resolving horizontally:

H H H H

1 3 1 3

  kips …acting to the left

Taking moments about support 1 and assuming that V 2 is upward:

16 V 2 (^)  8 H 3 (^)  4 V 3 (^)  12 V 4  0 V 2 8 10 4 20 12 20 16 25

kips …acting upward as assumed

Resolving vertically:

V V V V V

1 2 3 4 1

 kips …acting upward

The support reactions are shown at (ii).

Principles of statics 9

1.5 Triangle of forces

When a structure is in equilibrium under the action of three concurrent forces, the forces form a triangle of forces. As indicated in Figure 1.10 (i), the three forces F 1 , F 2 , and F 3 are concurrent. As shown in Figure 1.10 (ii), if the initial point of force vector F 2 is placed at the terminal point of force vector F 1 , then the force vector F 3 drawn from the terminal point of force vector F 2 to the ini- tial point of force vector F 1 is the equilibrant of F 1 and F 2. Similarly, as shown in Figure 1.10 (iii), if the force vector F 3 is drawn from the initial point of force vector F 1 to the terminal point of force vector F 2 , this is the resultant of F 1 and F 2. The magnitude of the resultant is given algebraically by:

( F 3 ) 2  ( F 1 ) 2  ( F 2 )^2  2 F F 1 2 (^) cos f 3

F^3

Equilibrant F 3

f 2 f 3

f 1 F (^2) F 2

F 1 F 1

F 3 F 1

F 2

Resultant

(i) (ii) (iii)

Figure 1.

and:

F 3 (^)  F 1 (^) sin f 3 (^) csc f 1

or:

F f F f F f

1 1 2 2 3 3

sin sin sin

Example 1.

Determine the angle of inclination and magnitude of the support reaction at end 1 of the pin-jointed truss shown in Figure 1.11. End 1 of the truss has a hinged support, and end 2 has a roller support.

Principles of statics 11

The reaction R may also be resolved into its horizontal and vertical components:

H R

V R

1

1

cos

sin

θ

θ

kips

kips

1.6 Free body diagram

For a system in equilibrium, all component parts of the system must also be in equilibrium. This provides a convenient means for determining the internal forces in a structure using the concept of a free body diagram. Figure 1.12 (i) shows the applied loads and support reactions acting on the pin-jointed truss that was analyzed in Example 1.1. The structure is cut at section A-A, and the two parts of the truss are separated as shown at (ii) and (iii) to form two free body diagrams. The left-hand portion of the truss is in equilibrium under the actions of the support reactions of the complete structure at 1, the applied

(i) Applied loads and support reactions

(ii) Left hand free body diagram

(iii) Right hand free body diagram

V 3  20 kips

V 2  25 kips

V 2  25 kips

12.5 kips

5.59 kips

5.59 kips

12.5 kips

15 kips

H 3  10 kips H 3 ^ 10 kips 15 kips

H 1  10 kips

V 1  15 kips (^) V 1  15 kips

H 1  10 kips

cut line

V 4  20 kips

V 4  20 kips

V 3  20 kips A 3

1 5 A 2

4

Figure 1.

12 Structural Analysis: In Theory and Practice

loads at joint 3, and the internal forces acting on it from the right-hand por- tion of the structure. Similarly, the right-hand portion of the truss is in equilib- rium under the actions of the support reactions of the complete structure at 2, the applied load at joint 4, and the internal forces acting on it from the left- hand portion of the structure. The internal forces in the members consist of a compressive force in member 34 and a tensile force in members 45 and 25. By using the three equations of equilibrium on either of the free body diagrams, the internal forces in the members at the cut line may be obtained. The values of the member forces are indicated at (ii) and (iii).

Example 1.

The pin-jointed truss shown in Figure 1.13 has a hinged support at support 1 and a roller support at support 2. Determine the forces in members 15, 35, and 34 caused by the horizontal applied load of 20 kips at joint 3.

8 ft

8 ft

8 ft

5

3 4

1 2 H 1  20 kips

H 3  20 kips

V 1  10 kips V 2  10 kips

P 15

V 2

P 35

P 34

2

3

5

Cut line

A

A

(i) Loads and support reactions (ii) Free body diagram

θ

Figure 1.

Solution

The values of the support reactions were obtained in Example 1.2 and are shown at (i). The truss is cut at section A-A, and the free body diagram of the right-hand portion of the truss is shown at (ii). Resolving forces vertically gives the force in member 35 as:

P 35 (^) V 2 10 63 43 11 18

kips compression

sin sin. .

θ

…

Taking moments about node 3 gives the force in member 15 as:

P 15 (^) 12 V 28 12 10 8 15

kips …tension

14 Structural Analysis: In Theory and Practice

case are shown. As shown at (iii), the principle of superposition and the two loading cases may be applied simultaneously to the truss, producing the com- bined support reactions indicated.

Supplementary problems

S1.1 Determine the reactions at the supports of the frame shown in Figure S1.1.

48 ft

24 kips 24 kips

48 ft

100 ft 20 ft 100 ft 20 ft 100 ft

V 1 V 2

1 2 3 4

Figure S1.

10 kips

3 ft 7 ft

2

M 1 H 1

V 1

H 4

V 4

3

4 1

10 ft

20 kips

Figure S1.

S1.2 Determine the reactions at supports 1 and 2 of the bridge girder shown in Figure S1.2. In addition, determine the bending moment in the girder at support 2.

Principles of statics 15

S1.3 Determine the reactions at the supports of the frame shown in Figure S1.3. In addition, determine the bending moment in member 32.

10 kips 10 kips

H 1 H 7

V 1 V 7 10 ft 10 ft

20 ft

10 ft 10 ft

3 4 5

7

2 6

1

Figure S1.

30 ft

50 kips

20 ft 10 ft

H 1^ H 2

V 1 V 2

1

3

2

Figure S1.

S1.4 Determine the reactions at the supports of the derrick crane shown in Figure S1.4. In addition, determine the forces produced in the members of the crane.

Principles of statics 17

4 ft

20 kips

4 ft 4 ft

10 kips 3 4

H 1

V 1 V 6

2

1 6 5

Figure S1.

10 ft

1

3

2

H 1

V 1

V 3

H 3

20 ft

10 ft

1 kip/ft

Figure S1.

S1.7 Determine the reactions at the supports of the pin-jointed truss shown in Figure S1.7.

S1.8 Determine the reactions at the supports of the bent shown in Figure S1.8. The applied loading consists of the uniformly distributed load indicated.

18 Structural Analysis: In Theory and Practice

S1.9 Determine the reactions at the support of the cantilever shown in Figure S1.9. The applied loading consists of the distributed triangular force shown.

21 ft

2

(^1) H 1

M 1

1 kip/ft

Figure S1.

6 ft

4 ft

7 ft

4 kips

V 1

M 1 H 1

2

1

4

3

Figure S1.

S1.10 Determine the reactions at the support of the jib crane shown in Figure S1.10. In addition, determine the force produced in members 24 and 34.