COLLEGE OF ENGINEERING DEPARTMENT OF CIVIL & ARCHITECTURAL ENGINEERING

CVEN 220 : ANALYSIS OF STRUCTURES

WAEL I. ALNAHHAL, Ph. D., P. Eng

Spring, 2015

Ch.2: Reactions

Idealized Structure

In real sense exact analysis of a structure can never be carried out.

Support connection

Pin support and pin connection

Fixed support & fixed connection

Roller support

Support for coplanar structures

Idealized Structure Idealized Structure

Consider the crane & trolley in the following Figure

For analysis, we neglect the thickness of the 2 main member & will assume that the joint at B is fabricated to be rigid

The support at A can be modeled as a fixed support

Details of trolley can be excluded

Idealized Structure

L2 = 5m and L1 = 3m

L2 /L1 = 5/3 = 1.67 < 2

The peak intensity = (2kN / m2 )(1.5) = 3kN / m

3kN/m

1.5m

3m

1.5m

Example 2.2 Correction:

Example - Gravity Loads

A typical hotel building floor is constructed using a series of two-bay frames that are spaced 6 m apart. The frames are connected using hot-rolled beams that are spaced at 2.3 m. A lightweight 150 mm reinforced concrete one-way slab is then supported by the beams. Frame girders are continuous over interior columns, and the beams are attached to the frame girders using simple shear connections.

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Example - Gravity Loads

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Example - Gravity Loads

(Continued) Assuming the floor is to be used for guest rooms, draw the gravity load diagrams that need to be considered in the design of an interior beam and an interior frame girder. Assume that structural steel weight is about 287 N/m2 for beams and 383 N/m2 for girders, and that additional flooring weight is about 478.8 N/m2. The floor must also support 24.2 N/m2 of mechanical and plumbing load, 239.4 N/m2 for fixed partitions, and a channel suspended ceiling system with a 13 mm gypsum board (4.8 N/m2 and 9.7 N/m2 ).

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Example - Gravity Loads Solution --- Beam Loads Magnitude (N/m2) Dead Loads: 150 mm thick slab (LWC, 17287 N/m3) 2593.1 Floor system 478.8 Mechanical and plumbing 24.2 Fixed partition 239.4 Channel suspended ceiling system 4.8 13-mm gypsum board 9.7 Beam weight 287.0 -------- 3637.0 Live Loads: Hotel guest rooms 1915.2

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Example - Gravity Loads

Solution-----Beam Reactions Beam tributary width = 2.3 m Beam uniformly distributed load: Dead Load = 3637 * 2.3 = 8365.1 N/m Live Load = 1915.2 * 2.3 = 4405.0 N/m Span = 6 m Reaction due to Dead Load = 25,095 N Reaction due to Live Load = 13,215 N

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Example - Gravity Loads Solution Girder Loads

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Equation of Equilibrium

In x-y plane

= 0

= 0 F = 0

M o Fy

x

Internal loading

Determinacy & Stability Determinacy: when all the forces in structure can be

determined from equilibrium equation , the structure is referred to as statically determinate. Structure having more unknown forces than available equilibrium equations called statically indeterminate

If n is number of structure parts & r is number of unknown forces:

r = 3n, statically determinate

r > 3n, statically indeterminate

Classify determinate & indeterminate structure

Statically Determinate

r = 3

n = 1

3 = 3(1)

2nd degree Statically indeterminate

r = 5

n = 1

5 > 3(1)

Statically determinate

r = 6

n = 2

6 = 3(2)

n = 3

10 > 3(3)

1st degree

Statically indeterminate

r = 10

1st degree Statically indeterminate,

r = 7

n = 2

7 > 3(2)

r = 10

n = 2

10 > 3(2) Statically indeterminate

4th degree

Stability

Partial Constraints

= 0 will not be satisfied unstable with this loading case Fx

Improper Constraints This can occur if all the support reactions are concurrent at a point.

P d 0

This can occur also when the reactive forces are all parallel

In General

r < 3n, Then the structure is Unstable

r >= 3n, Also, Unstable if member reactions are concurrent or parallel or some of the components form a collapsible mechanism

Classify The structure Stable or Unstable

no special cases Sable

r = 3

n = 1

3 = 3(1)

Sable

r = 8

n = 2

8 > 3(2) no special cases

Unsable the three reactions are concurrent ar B

r = 3

n = 1

3 = 3(1)

Unsable

r = 7

n = 3

7 < 3(3)

Application of Equilibrium Equation

Application of Equation of Equilibrium Procedure steps

1. If not given, establish a suitable x - y coordinate system.

2. Draw a free body diagram (FBD) of the object under analysis.

3. Apply the three equations of equilibrium to solve for the unknowns.

Example 1 Determine the Reactions

M A = 0

By = 38.5k

Fy = 0 60 sin 60 + 38.5 + Ay = 0

Ay = 13.4k

60 sin 60(10) + 60 cos 60(1) + By (14) 50 = 0

Fx = 0 Ax 60 cos 60 = 0

Ax = 30k

Example 2 Determine the Reactions

M A = 600kN 60(4) 60(6) + M A = 0

Ax = 120kN

M A = 0

Ax = 0

Ax 60 60 = 0 Fx = 0 Fy = 0

Example 3 Determine the Reactions

Fx = 0 Ax 5 (1331.5) = 0

F = 0 Ay 3500 + 5 (1331.5) = 0 Ay = 2700Ib

Ax = 1070Ib

NB = 1331.5Ib

3500(3.5) + 4 N (4) + 3 N (10) = 0 5 B 5 B M A = 0

3 y

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Example 4 Determine the Reactions

M B (Right ) = 0 15Cy 6000 = 0

Ay = 7.6 k. ft

Ay + 400 8000 = 0

400(35) 6000 8000(10) + M A = 0

= 72 k.ft

= 0 Ax = 0

= 0 Fx Fy

M at point A = 0 M A = 72000Ib.ft

Cy = 400Ib

Example 5 Determine the Reactions

M B (Right ) = 0 2Cy 6(1) = 0

Ay = 9.4kN = 0 Ay + 3 6 8( 5 ) = 0

Ax = 9.87kN = 0 Ax 14.7 8( 5 ) = 0 Fx

M A = 0

Cx = 14.7kN

Cx (1.5) + 3(4) 6(3) 8(2) = 0

Cy = 3kN

Fy 4

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Example 6 Determine the Reactions

M B (Left ) = 0 6 Ay + 8(3) = 0

Ey = 5.33kN

M D (right ) = 0 8(2) + 5.33(4) M E = 0

M E = 5.33kN.m

= 0 Ex = 0

= 0 Ey 8 8 + 4 + 6.67 = 0 Fx Fy

Ay = 4kN

M D ( Left ) = 0 3Cy 4(11) + 8(8) = 0

Cy = 6.67kN

Example 7 Determine the Reactions

M B (Right ) = 0 Cx (6) + 240(3) + 60(4.5) + 84.9( ) = 0

Ay = 120kN Ay + 240 254.6 sin 45 + 84.9 sin 45 = 0

Ax = 285kN Ax 195 +180 + 60 + 254.6 cos 45 + 84.9 cos 45 = 0

Cx = 195kN

Fx = 0

254.6 sin 45(1.5) 84.9 cos 45(4.5) + 84.9 cos 45(4.5) = 0

Cy = 240kN

Cy (6) 180(1.5) 60(1.5) 254.6 cos 45(4.5) M A = 0

Fy = 0

2 3 2

Problem 1 Determine the Reactions

Problem 2 Determine the Reactions

Problem 3 Determine the Reactions

Slide Number 1Idealized StructureSlide Number 3Slide Number 4Slide Number 5Slide Number 6Slide Number 7Slide Number 8Slide Number 9Slide Number 10Slide Number 11Idealized StructureIdealized StructureSlide Number 14Slide Number 15Slide Number 16Slide Number 17Slide Number 18Slide Number 19Slide Number 20Slide Number 21Slide Number 22Slide Number 23Slide Number 24Slide Number 25Example - Gravity LoadsExample - Gravity LoadsExample - Gravity LoadsExample - Gravity LoadsExample - Gravity LoadsExample - Gravity LoadsEquation of EquilibriumDeterminacy & StabilitySlide Number 34Slide Number 35Slide Number 36Slide Number 37Slide Number 38Slide Number 39Slide Number 40Slide Number 41Slide Number 42Classify The structure Stable or UnstableSlide Number 44Slide Number 45Slide Number 46Slide Number 47Slide Number 48Application of Equation of EquilibriumSlide Number 50Slide Number 51Slide Number 52Slide Number 53Slide Number 54Slide Number 55Slide Number 56Slide Number 57Example 5Slide Number 59Slide Number 60Slide Number 61Slide Number 62Slide Number 63Slide Number 64Slide Number 65Slide Number 66Slide Number 67Slide Number 68