struc lec. no. 1

Theory of Structures(1)

Lecture No. 1

Loads

Reactions

Supports

Conditional Equations

Stability and Determinancy

Chapter 1 : Loads and Reactions

Chapter 2 : Statically Determinate Beams Types of Beams

Normal Force, Shearing Force, and Bending Moment

Relationship between Loads, Shearing Force

and Bending Moment

Standard Cases of S.F. and B.M.D(s) for Beams

Principal Of Superposition

Inclined Beams

A Beam: is a structural member subjected

to some external forces.

Chapter 3 : Statically Determinate Rigid FramesInternal Stability and Determinacy

Method of Frames Analysis

A Frame: is a structure composed of a number of members

connected together by joints all or some are rigid.

Chapter 4 : Statically Determinate Arches

Reactions , Thrust, Shearing Force and Bending Moment using Analytical Method Reactions Thrust, Shearing Force, and Bending Moment at any Point in the Arch using Graphical Method

An Arch :is a curved beam or two curved beams connected together by intermediate hinge.

Chapter 5 : Statically Determinate Trusses

Stability and DeterminacyAnalysis for Trusses: Methods of: joints, Sections, and Stress Diagram

A truss: Consists of a number of straight members pin -

connected together and subjected to concentrated

loads at hinges.

Chapter 1

Loads and Reactions

Loads classified according to:

Cause of Loading : dead ,live, and lateral Shape of Load : concen. and distrib. Rate of application : static and dynamic

Reactions: resistance by supports to counteract the action of loadsSupports : Roller, Hinge, Fixed, and Link

Roller Hinged

Fixed Link

Determination of the Reactions:

”External loads acting on the structure together

with the reactions at the supports must constitute a

system of non concurrent forces in equilibrium”

∑x = 0 (sum of horizontal forces) = 0

∑y = 0 (sum of vertical forces) = 0

∑M = 0 (sum of moments ) = 0

YB

XA

YA

Solution of Example 1

1- ∑ X = 0 XA – 5 = 0 XA = 5t

2- ∑MB = 0 YA × 6 – 2 × 6 × 3 + (4 × 3)/2 × 1 = 0 YA= 5t

3- ∑ Y = 0 YA + YB = 2 × 6+ (4 × 3)/2=18 YB = 13t

Solution of Example 2

1-     ∑ X = 0 XA = 3t

2-     ∑ MB = 0 YA× 4 + 8 + 4 × 1 – 3 × 1= 0 YA= -2.25t ↑

YA=2.25t ↓

3- ∑ y = 0 YA + 2× 2 = YB

2.25+ 4 = YB YB = 6.25 t ↑

YA

XA

YB

Solution of Example 3

1- ∑ X = 0 XB – 4 = 0 XB = 4t

2 -∑MA = 0

10YB – 4 × 12 – 2 × 9 – 2 × 5× 2.5 – 4 × 3= 0

YB = 10.3 t

3- ∑ Y = 0 YA + YB = 4 + 2 + 10 = 16 YA = 5.7 t

YBYA

XB

Solution of Example 4

1- ∑ X = 0 XA – 4 = 0 XA = 4t

2- ∑ Y = 0 YA – 1 × 6 – 5 – 5 – 2.5 = 0 YA = 18.5 t

3- ∑ M/A= 0 MA – 5 × 2 – 5 × 4 – 2.5 ×6 - 1×6×3 = 0

MA = 63 t.m

Solution of Example 5

R1 = 1/2 × 4 × 8 = 16 t

R2 = 0.5 × 4 = 2 t

1- ∑ X = 0 XA – R2 = 0 XA = 2t

2- ∑ MA= 0 11 YB + 2 × 13 – R2 × 2 – 2 × 1 – R1 × 7= 0

YB = 8.36 t

3- ∑ Y = 0 YA+ YB+ 2 – 2 – 16 = 0 YA= 7.64t

Solution of Example 6

1- ∑ X = 0 XB – 20 = 0 XB = 20t

2- ∑ MA = 0

6 YB + 20 × 3 – 10 × 6 – 20 × 6 = 0 YB = 20t

3- ∑Y = 0

YA + YB = 20 + 30 + 10 = 60 YA = 40t

Solution of Example 7

∑MO = 0

3 × 0 + 8 × 1 + 1 × 3 – Rd × 5 = 0

Rd = 2.2 t ↑

∑Mp = 0

1 × 2 + 8 × 4 + 3 × 5 – Rb × 5 = 0

Rb = 6.93t

∑Mq = 0

1 × 2 + 8 × 4 + 3 × 5 – Rc × 5 = 0

Rc = 6.93t

2

2

8t

Conditional Equations Some structures are

divided to several parts connected together by hinges, links or rollers.

Forces are transmitted through these connections

from one part to the others .

00 McC

BrightMcandMc

C

AleftCM

Solution of Example 8

Method I1- Part FD

1) ∑ MD = 0 8 × 2 = 4 YF YF = 4t

2) ∑Y = 0 YD = 8 – 4 YD = 4t2- Part ECF 1) ∑ ME = 0 10×2.5+4×5= 4YC YC= 11.25t

2) ∑Y = 0 YE=10+4-11.25 YE=2.75t3- Part ABE 1) ∑MA=0 12×3+2.75×6=5YB YB=10.5t

2) ∑Y = 0 YA= 2.75 + 12-10.5 YA= 4.25t

Solution of Example 8 “ Continued ”

Method II

1)∑MF(right)= 0 8 × 2 – 4 YD = 0 YD = 4t 2) ∑ME(right)= 0 18×4.5 – 4×9 – 4YC= 0 YC=11.25t

3) ∑MA = 0 30×7.5 – 4×15 – 11.25× 10 - 5YB = 0 YB=10.5t

4) ∑Y = 0 30 – 4 – 11.25 – 10.5 – YA=0 YA=4.25t

c

c

d

M = 0 = Xd × 8 – 8 × 2

Xd = 2 t

b

dbM = 8 ×12 + 2.5 × 12 × 6

Xd

- 2 × 8 - 10 ×Yd=0

Yd = 26 t

∑Y = 0 = Ya + 26 – 2.5 × 12 – 8

Ya = 30 + 8 – 26 = 12 t.

∑X = 0 = Xa – 0.5 × 8 – 2

Xa = 4 + 2 = 6 t. ←

Solution of Example 9

= 0 = Ma + 0.5 × 8 × 4 – 6 × 8 Ma = 48 – 16 = 32 m.t (anticlockwise)

Check : ∑Mc for part ac Xa Ya

32 + 0.5 × 8 × 4 + 2.5 × 12 × 4 – 6 × 8 – 12 × 10= 32 + 16 + 120 – 48 – 120 = 0

b

abM

c

aCM

Solution of Example 9 “Continued”

1) ∑MF(right) = 0 RC cosα × 4.5 – 4 × 4.5 × 2.25 – 10 × 10.5 = 0 RC = 40.4 t

2) ∑MA = 0 YB×9 + XB × 2 + RC cosα × 18 + RC sinα × 8 –10×24

- 20×4.5 – 5 × 5 – 4 × 9 × 13.5 + 5 × 2 = 0

4.5 YB = 27.57 – XB (1)

3) ∑ME (right) = 0 4.5 YB - 6XB + RC cosα × 13.5 – 4 × 9 × 9 - 10 × 19.5 = 0

4.5 YB = 6XB + 82. 545 (2)

Solution of Example 10

By solving equations (1) & (2) we get

YB = 7.87 t. XB= - 7.85t

4) ∑Y = 0 YA = 36 + 10 + 20 + 5 – 32.33 – 7.87

YA = 30.8 t.

5) ∑ME(left) = 0 4.5 YA – 8 XA – 5 × 3 – 5 × 6.5 = 0

XA = 11.4 t.

Solution of Example 10 “Continued”

Stability and Determinancy

A. External concerned with the reactions.

B. Internal concerned with the internal forces and moments.

A Stable Structure is one that support: a system of loads. These loads together with the support reactions have to be in equilibrium and satisfy the three equilibrium equations in addition to conditional equations (if any).

Stability of Structures

Statically Unstable Structure is one in which: the number of the unknown reactions is less than the sum of the available equilibrium and conditional equations (if any).

Externally Statically Unstable Structures

Geometrical Unstable Structures

Statically Determinate Structure is one in which the reactions can be determined by application of the equations of

equilibrium in addition to conditional equations(if any).

Determinancy of Structures

Statically Indeterminate Structure is one in whichthe number of unknown reactions is more than the number of availableequations

Stable-Once Statically Indeterminate

Stable-Three Times Statically Indeterminate

Check the stability and determinacy for the following structures. Then, show how to modify it for stability and determinancy

(a) (b)

(c)

(d)

Solution of Example 11

Structure (a) :Is unstable because there are only two

unknown reactions.

The modification : Change either of the two roller supports to

a hinge.

Structure (b) : Is unstable because the beam is rested on

four link members connected at a point.

The modification : Separate the link members

Modification for structure (b)

Structure (C) : Is stable and three times statically indeterminate because there are six unknown support reactions.The modification : Add three intermediate hinges

Modification for structure (C)

Structure (D) : Is stable and five times statically indeterminate because there are

8 unknown support reactions

The modification : One of fixed supports is changed to a roller In addition to three intermediate hinges

arranged as given in Struc. (c).