Types of Structures and Loads · Types of Structures and Loads ... Ex.1-1 The floor beam in Fig....

March 2016

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Types of Structures

and Loads

Week 1

من يحط اإلحسان

يتلقونه من قدر

نوت استراكشر

. تتكون النوت من عشرة أسابيع

على شرح كل نوتيحتوي

من وتمارينألمثلة وحلول

سابقة.هوموركات وامتحانات

(hs.net-engتابع )

النوت يتم تنقيحها وتحديثها

ي، وإضافة الجديد إليها بشكل دور

داوم على التواصل معنا.

March 2016




Types of Structures and Loads

Introduction

A structure refers to a system of connected parts used to support a

load.

Concrete beams generally have a rectangular

cross section, since it’s easy to construct this

form directly in the field.

Because concrete is rather weak in resisting

tension, steel ''reinforcing rods'' are cast into

the beam within regions of the cross section

subjected to tension.

Columns

Members that are generally vertical and resist axial

compressive loads are referred to as columns,

Tubes and wide-flange cross sections are often

used for metal columns, where circular and square

cross sections with reinforcing rods are used for

those made of concrete.

Occasionally, columns are subjected to both an

axial load and a bending moment as shown in the

figure. These members are referred to be as beam

columns.

يحبك أن أردت إذا: الكالم خالصة

.أيديهم بين فيما ازهد الناس،

March 2016




Beams

Beams are usually straight horizontal members

used primarily to carry vertical loads.

Beams are primarily designed to resist bending

moment; however, if they are short and carry

large loads, the internal shear force may become

quite large and this force may govern their

design.

When the material used for a beam is a metal

such as steel or aluminum, the cross section is most efficient when

it’s shaped as shown in the Fig. 1-3.

Here the forces developed in the top and bottom flanges of the

beam form the necessary couple used to resist the applied moment

M, whereas the web is effective in resisting the applied shear V.

This cross section is commonly referred to as a "wide flange".

الهـدف، إلى وصولـك عدم في الحياة مأساة تكمن ال

.إليه الـوصول تحاول هدف لك يكون أال في لكن

March 2016




The design loading for a structure is often specified in codes. In

general, the structural engineer works with two types of codes:

general building codes and design codes.

General building codes specify the requirements of governmental

bodies for minimum design loads on structures and minimum

standards for construction.

Design codes provide detailed technical standards and are used to

establish the requirements for the actual structural design.

Table1-1 lists some of the important codes used in practice. It

should be realized, however, that codes provide only a general

guide for design.

The ultimate responsibility for the design lies with the structural

engineer.

Table 1-1

General Building Codes

Minimum Design Loads for Buildings and Other Structures,

SEI/ASCE 7-05, American Society of civil Engineers

International building Code

Design Codes

Building Code Requirements for Reinforced Concrete, Am. Conc. Inst.

(ACI)

Manual of Steel Construction, American Institute of Steel Construction

(AISC)

Standard Specifications for Highway Bridges, American Association of

State Highway and Transportation Officials (AASHTO)

National Design Specifications for Wood Construction, American Forest

and Paper Association (AFPA)

Manual for Railway Engineering, American Railway Engineering

Association (AREA)

لالقلي إعطاء من تستح ال

.منه أقل الحرمان فإن

Loads

March 2016




Dead Loads

Dead loads consist of the weights of the various structural

members and the weights of any objects that are permanently

attached to the structure.

In some cases, a structural dead load can be estimated satisfactorily

from simple formulas based on the weights and sizes of similar

structures.

Through experience one can also derive a "feeling" for the

magnitude of these loadings. For example, the average weight for

timber buildings is 1.9-2.4 kN/m2, for steel framed building it’s

2.9-3.6 kN/m2, and for reinforced concrete it’s 5.3-6.2 kN/m2.

Table 1-2

Minimum Densities for Design Loads from Materials

kN / m3

Aluminum

Concrete, plain cinder

Concrete, plain stone

Concrete, reinforced cinder

Concrete, reinforced stone

Clay, dry

Clay, damp

Sand and gravel, dry, loose

Sand and gravel, wet

Masonry, light weight solid concrete

Masonry, normal weight

Plywood

Steel, cold-drawn

Wood, Douglas Fir

Wood, Southern Pine

Wood, spruce

26.7

17.0

22.6

17.4

23.6

9.9

17.3

15.7

18.9

16.5

21.2

5.7

77.3

5.3

5.8

4.5

الحياة على تزد لم إذا

.عليها زائد فأنت شيئا

March 2016




Table 1-3

Minimum Design Dead Loads

Walls kN / m2

100 mm clay brick

200 mm clay brick

300 mm clay brick

1.87

3.78

5.51

Frame Partitions and Walls

Exterior stud walls with brick veneer

Window, glass, frame and sash

Wood studs 50 x 100 mm unplastered

Wood studs 50 x 100 mm plastered one side

Wood studs 50 x 100 mm plastered two sides

2.30

0.38

0.19

0.57

0.96

Floor fill

Cinder concrete, per mm

Light weight concrete, plain, per mm

Stone concrete, per mm

0.017

0.015

0.023

Ceilings

Acoustical fiberboard

Plaster on tile or concrete

Suspended metal lath and gypsum plaster

Asphalt shingles

Fireboard, 13 mm

0.05

0.24

0.48

0.10

0.04

أو حياتنــا نحيــا أن نختــار إمــا

.عنا نيابة يحياها غيرنا نترك

March 2016




Ex.1-1 The floor beam in Fig. 1-8 is used to

support the 1.83-m width of a lightweight

plain concrete slab having a thickness of

102 mm. The slab serves as a portion of the

ceiling for the floor below, and therefore its

bottom is coated with plaster. Furthermore,

a 2.44-m-high, 305-mm-thick lightweight

solid concrete block wall is directly over the

top flange of the beam. Determine the

loading on the beam measured per foot of

the length of the beam.

Solution:

Using the data in Tables 1-2 and 1-3, we have:

Concrete slab: [0.015 kN / (m2 . mm)] (102 mm) (1.83 m)

= 2.80 kN / m

Plaster ceiling: (0.24 kN / m2) (1.83 m) = 0.44 kN / m

Block wall: (16.5 kN / m3) (2.44 m) (0.305 m) = 12.28 kN / m

_____________

Total Load: 15.52 kN / m

أكثر هو المرء غاية نسيان

.انتشارا الغباء أشكال

March 2016




Live Loads

The minimum live loads specified in codes are determined from

studying the history of their effects on existing structures.

Table 1-4

Minimum Live Loads

Occupancy or Use Live Load

kN / m2

Assembly areas and theaters

Fixed seats

Movable seats

Dance halls and ballrooms

Garages (passenger cars only)

Office buildings

Lobbies

Offices

Storage warehouse

Light

Heavy

Residential

Dwelling (one – and two – family)

Hotels and multifamily houses

Private rooms and corridors

Public rooms and corridors

Schools

Classroom

Corridors above first floor

2.87

4.79

4.79

2.40

4.79

2.40

6.00

11.97

1.92

1.92

4.79

1.92

3.83

موعدها في األعمال إنجاز يتم

.فقط المفكرة في الصحيح

March 2016




For some types of buildings having very large floor areas, many

codes will allow a reduction in the uniform live load for a floor,

since it’s unlikely that the prescribed live load will occur

simultaneously throughout the entire structure at any one time. For

example, ASCE 7-05 allows a reduction of live load on a member

having an influence area (KLL AT) of 37.2 m2 or more. This

reduced live load is calculated using the following equation:

L = Lo (0.25 + 4.57

√KLL AT

) 𝐸𝑞. 1 − 1

Where,

L = reduced design live load per square meter of area supported

by the member.

Lo = unreduced design live load per square meter of area

supported by the member.

KLL = live load element factor, for interior columns KLL = 4.

AT = tributary area in square meters.

The reduced live load defined by Eq. 1-1 is limited to not less than

50% of (Lo) for members supporting one floor, or not less than

40% of (Lo) for members supporting more than one floor. No

reduction is allowed for loads exceeding 4.97 kN/m2, or for

structures used for public assembly, garages, or roofs.

الخطوة يهم، ال المسافة بعد

.بةصعو األكثر هي فقط األولى

March 2016




(Example 1-2) A two-story office building

has interior columns that are spaced 6.71 m

apart in two perpendicular directions. If the

(flat) roof loading is 0.96 kN/m2, determine

the reduced live load supported by a typical

interior column located at the ground level.

Solution:

Each interior column has a tributary area or effective loaded area of:

𝐴𝑡 = (6.71 𝑚) (6.71 𝑚) = 45.0 𝑚2

A ground–floor column therefore supports a roof live load of:

𝐹𝑅 = (0.96 𝑘𝑁/𝑚2) (45.0 𝑚2) = 43.2 𝑘𝑁

This load can’t be reduced, since it’s not a floor load. For the second

floor, the live load is taken from Table 1-4:

Lo = 2.4 kN / m2. Since KLL = 4, Then: KLL AT = 4 (45.0 m2)

= 180 m2 > 37.2 m2, the live load can be reduced:

L = Lo (0.25 + 4.57

√180 ) = (1.42) 𝑘𝑁/𝑚2

The reduced load here is (1.42/2.4) ∗ 100 = 59.2% > 50% OK.

𝐹𝐹 = (1.42 𝑘𝑁/𝑚2) (45.0 𝑚2) = 63.9 𝑘𝑁

𝐹𝑡𝑜𝑡𝑎𝑙 = 𝐹𝑅 + 𝐹𝐹 = 43.1 𝑘𝑁 + 63.9 𝑘𝑁 = 107.0 𝑘𝑁

المثالية الظروف ينتظر البعض

.انتظارهم يطول وغالبا يبدأ، كي

March 2016




1-2 The building wall consists of

200-mm clay brick. In the interior,

the wall is made from 50 mm * 100

mm wood studs, plastered on one

side. If the wall is 3 m high,

determine the load in kN per meter

of length of wall that the wall exerts

on the floor.

Solution:

𝑊𝑐𝑙𝑎𝑦 = 3.78 𝑘𝑁/𝑚2 ∗ 3 𝑚

= 11.34 𝑘𝑁/𝑚

𝑊𝑠𝑡𝑢𝑑𝑠 = 0.57 𝑘𝑁/𝑚2 ∗ 3 𝑚

= 1.71 𝑘𝑁/𝑚

𝑊𝑇𝑜𝑡𝑎𝑙 = 11.34 + 1.71

= 13.05 𝑘𝑁/𝑚

نقطة هي الملحة الحاجة

.اإلنجازات كل بداية

March 2016




Solution:

𝐴𝑟𝑒𝑎 = 6 ∗ 8

= 48 𝑚2

𝑊𝑑𝑒𝑎𝑑 = 0.015 ∗ 125 ∗ 48

= 90 𝑘𝑁

𝑊𝑙𝑖𝑣𝑒 = 1.92 ∗ 48

= 92.2 𝑘𝑁

𝑊𝑇𝑜𝑡𝑎𝑙 = 90 + 92.2

= 182.2 𝑘𝑁

ينتج ال الواهنة الرغبة

.عظيمة إنجازات عنها

1-5. The floor of a classroom is made of 125-mm thick lightweight plain

concrete. If the floor is a slab having a length of 8 m and a width of 6 m,

determine the resultant force caused by the dead load and the live load.

March 2016




1-6. The pre-cast T-beam has the

cross-section shown. Determine its

weight per meter of length if it’s

made from reinforced stone concrete

and eight 20-mm cold-formed steel

reinforcing rods.

Solution:

قطع تتكون من مستطيلين وشبه منحرفمساحة الم*

Area concrete = (1.2 ∗ 0.2) + (0.45 + 0.15

2 * 0. 2) + (0.15 ∗ 0.5)

= 0.375 𝑚2

𝑊𝑐𝑜𝑛𝑐𝑟𝑒𝑡𝑒 = 0.375 ∗ 23.6

= 8.85 𝑘𝑁/𝑚

𝑊𝑠𝑡𝑒𝑒𝑙 = 8 ∗ (𝜋 ∗ 0.012) ∗ 77.3

= 0.194 𝑘𝑁/𝑚

𝑊𝑇𝑜𝑡𝑎𝑙 = 8.85 + 0.194

= 9.044 𝑘𝑁/𝑚 نادرا القمة إلى الطريق

.مزدحما يكون ما

March 2016




Solution:

125 mm concrete slab = (125)(0.023) = 2.875 kN/m2

100 mm cinder fill = (100)(0.017) = 1.7 kN/m2

metal lath & plaster = 0.48 kN/m2

ρtotal = 5.06 kN/m2

فة،مختل أشياء يفعلون ال الناجحون

.تلفةمخ بطريقة األشياء يفعلون بل

1-7 The second floor of a light manufacturing

building is constructed from a 125-mm-thick

stone concrete slab with an added 100-mm

cinder concrete fill as shown. If the suspended

ceiling of the first floor consists of metal lath

and gypsum plaster, determine the dead load

for design in kN per square meter of floor

area.

March 2016




Solution:

Weight per square m

= (7.86) (38

1,000) = 0.299 kN m2⁄

From table 1-3

Shingles = 0.10 kN/m2

𝜌 = 0.299 + 0.10 = 0.399 kN/m2

𝜌𝑥 = (0.399) sin 30° = 0.2 kN m2⁄

𝜌𝑦 = (0.399) cos 30° = 0.35 kN m2⁄

أخطاء يرتكب لم إنسان

.أبدا يحاول لم باألغلب

1-9 The beam supports the roof

made from asphalt shingles and

wood sheathing boards. If the

board have a thickness of 38 mm

and a specific weight of 7.86

kN/m2, and the roof's angle of

slope is 30°, determine the dead

load of the roofing – per square

meter – that is supported in the x

and y direction by the purlins.

March 2016




𝐴𝑇 = 4.5 ∗ 4.5 = 20.25 𝑚2

𝑊𝑟𝑜𝑜𝑓 = 20.25 ∗ 1 = 20.25 𝑘𝑁

(This load cannot be reduced since it’s not a floor load)

𝐾𝐿𝐿 = 4 (𝐼𝑛𝑡𝑒𝑟𝑖𝑜𝑟 𝑐𝑜𝑙𝑢𝑚𝑛)

𝐾𝐿𝐿 𝐴𝑇 = 4 ∗ 20.25

= 81 > 37.2 (𝐿𝑖𝑣𝑒 𝑙𝑜𝑎𝑑 𝑐𝑎𝑛 𝑏𝑒 𝑟𝑒𝑑𝑢𝑐𝑒𝑑 𝑓𝑜𝑟 𝑓𝑙𝑜𝑜𝑟 𝑙𝑜𝑎𝑑)

L = Lo (0.25 + 4.57

√KLL AT

)

= 1.92 (0.25 + 4.57

√81 )

= 1.45 𝑘𝑁 / 𝑚2

𝑊𝑓𝑙𝑜𝑜𝑟 = 1.45 𝑥 20.25 = 29.36 𝑘𝑁

(𝑎) 𝑊𝑔𝑟𝑜𝑢𝑛𝑑 = 𝑊𝑟𝑜𝑜𝑓 + 𝑊𝑓𝑙𝑜𝑜𝑟

= 20.25 + 29.36

= 49.6 𝑘𝑁

(𝑏) 𝑊𝑠𝑒𝑐𝑜𝑛𝑑 𝑓𝑙𝑜𝑜𝑟 = 𝑊𝑟𝑜𝑜𝑓

= 20.25 𝑘𝑁 أن المتعة من نوع إنه

.المستحيل تفعل

1-10. A two-story school has interior columns that are spaced 4.5 m apart in

two perpendicular directions. If the loading on the flat roof is estimated to be

1kN/m2, determine the reduced live load supported by a typical interior

column at: (a) the ground-floor level, (b) the second-floor level.

March 2016




Wind loads

According to ASCE 7-05 standard, the wind pressure can be calculated asing this

equation.

𝑞𝑧 = 0.613 𝑘𝑧𝑡 𝑘𝑑 𝑉2 𝐼

Where,

𝑞𝑧 − Wind pressure (𝑁/𝑚2)

𝑉 − Wind velocity in m/s. value are obtained for the wind map.

𝐼 − Importance factor.

𝑘𝑧 − Velocity pressure Exposure Coefficient (from Table 1-5)

Table 1-5

z (m) 0-4.6 6.1 7.6 9.1 12.2 15.2

𝑘𝑧 0.85 0.90 0.94 0.98 1.04 1.09

𝑘𝑧𝑡 − Constant that depends apron the __________ .

𝑘𝑧𝑡 = 1 (for far ground).

𝑘𝑑 − is used when the structure is subjected to combinations of loads.

𝑘𝑑 = 1 (for wind acting alone).

Design wind pressure for signs:

𝐹 = 𝑞𝑍 𝐺 𝐶𝐹 𝐴𝐹

Where,

𝐹 − Resultant force acting on the face of the sign . (N)

𝐺 − wind gust coefficient factor, which depends upon the exposure.

𝐺 = 0.85 (for rigid structures).

𝐶𝐹 − force coefficient that depends wpon the ratio of sign dimensions

(M/N) – (from table 1-6).

𝑀 − large dimension.

𝑁 − small dimension. .نفسك إلى اآلخرين أفكار تنسب ال

March 2016




Table (1-6)

M/N L 6 10 20 40 60

𝑘𝑧 0.85 0.90 0.94 0.98 1.04

𝐴𝐹 − Area of the face of the sign projected into the wind.

Snow Loads

𝑃𝑓 = 0.7 𝐶𝑒 𝐶𝑡 𝐼 𝑃𝑔

Where,

𝑃𝑓 − Design snow load on the roof (kn/m2)

𝐶𝑒 − Exposure factor which depends upon the terraive.

𝐶𝑒 = 0.8 (for roofs in unobstructed areas).

𝐶𝑒 = 1.3 (for sheltered roofs in large cities).

𝐶𝑡 − Thermal factor.

𝐶𝑡 = 1.2 (for unheated structures, kept below freezing).

𝐶𝑡 = 1.0 (for normally heated structures).

𝐼 − Importance factor.

𝐼 = 0.8 (for agriculture − − − − − storage facilities).

𝐼 = 1.2 (for hospitals)

𝑃𝑔 − Ground snow loading (kn/m2).

If 𝑃𝑔 ≤ 0.96 𝑘𝑁/𝑚2, use the largest value for 𝑃𝑓 from the equation or from

𝑃𝑓 = 𝐼 𝑃𝑔.

If 𝑃𝑔 > 0.96 𝑘𝑁/𝑚2, use the largest value compared to

𝑃𝑔 = 𝐼 (0.96 𝑘𝑁/𝑚2).

التضحية يتم أال اإلنسانية من

.غاية سبيل في بإنسان

March 2016




From the wind map V = 38 m / s

𝑘𝑧 = 0.85

𝑘𝑧𝑡 = 1

𝑘𝑑 = 1

𝑞𝑧 = 0.613 𝑘𝑧 𝑘𝑧𝑡 𝑘𝑑 𝑉2 𝐼

= 0.613 (0.85)(1)(1)(382)(0.87)

= 654.58 𝑁/m2

𝐹 = 𝑞𝑧 𝐺 𝐶𝐹 𝐴𝐹

𝐺 = 0.85

𝑀 𝑁⁄ = 6 3 =⁄ 2 < 6 , 𝐶𝐹 = 1.2

𝐴𝐹 = 3(6) = 18 𝑚2

𝐹 = 654.58 (0.85)(1.2)(18)

= 12.0 𝑘𝑁

قدونيعت ألنهم يستطيعون إنهم

.يستطيعون أنهم

1-13 Determine the resultant force acting on

the face of the truss-supported sign if it is

located near Los Angeles, California on

open flat terrain. The sign has width of 6 m

and a height of 3 m as indicated. Use an

importance factor of I = 0.87.

March 2016




𝐶𝑒 = 0.8

𝐶𝑡 = 1.0

𝐼 = 1.2

𝑃𝑔 = 1.2 𝑘𝑁 𝑚2⁄

𝑃𝑓 = 0.7 𝐶𝑒 𝐶𝑡 𝐼 𝑃𝑔

= 0.7 × 0.8 × 1.0 × 1.2 × 1.2

= 0.81 𝑘𝑁 𝑚2⁄

OR

𝑃𝑓 = 𝐼 (0.96 𝑘𝑁/𝑚2)

= 1.2 (0.96)

= 1.15 (𝑘𝑁/𝑚2)

( نختار𝑷𝒇.األكبر )

∴ 𝑃𝑓 = 1.15 𝑘𝑁/𝑚

ز،الفو في الرغبة لديهم الناس أغلب

.الثمن ببذل يرغب من قليل لكن

1-19 A hospital located in Chicago, Illinois, has a flat roof, where the ground

snow load is 1.2 kN/m2. Determine the design snow load on the roof of the

hospital.

Types of Structures and Loads · Types of Structures and Loads ... Ex.1-1 The floor beam in Fig....

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