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    2 Load SafetyFactor

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    brayan
    Here are the steps to solve this example problem: 1) Determine the factored loads using the load factors from ACI 318: Pu = 1.2D + 1.6L + 0.5Lr + 1.0Wt + 1.0Ec = 1.2(150) + 1.6(300) + 0.5(60) + 1.0(70) + 1.0(50) = 630 kips 2) Determine the required strength using the strength reduction factor: Pr = Pu/φ = 630/0.65 = 969 kips 3) The nominal strength Pn must be equal or greater than the required strength: Pn ≥

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    2 Load SafetyFactor

    Uploaded by

    brayan
    42 views17 pages
    Here are the steps to solve this example problem: 1) Determine the factored loads using the load factors from ACI 318: Pu = 1.2D + 1.6L + 0.5Lr + 1.0Wt + 1.0Ec = 1.2(150) + 1.6(300) + 0.5(60) + 1.0(70) + 1.0(50) = 630 kips 2) Determine the required strength using the strength reduction factor: Pr = Pu/φ = 630/0.65 = 969 kips 3) The nominal strength Pn must be equal or greater than the required strength: Pn ≥

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    Concrete Lecture 2

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    Here are the steps to solve this example problem: 1) Determine the factored loads using the load factors from ACI 318: Pu = 1.2D + 1.6L + 0.5Lr + 1.0Wt + 1.0Ec = 1.2(150) + 1.6(300) + 0.5(60) + 1.0(70) + 1.0(50) = 630 kips 2) Determine the required strength using the strength reduction factor: Pr = Pu/φ = 630/0.65 = 969 kips 3) The nominal strength Pn must be equal or greater than the required strength: Pn ≥

    Copyright:

    © All Rights Reserved
    Download as PDF, TXT or read online from Scribd
    Download as pdf or txt
    42 views17 pages

    2 Load SafetyFactor

    Uploaded by

    brayan
    Here are the steps to solve this example problem: 1) Determine the factored loads using the load factors from ACI 318: Pu = 1.2D + 1.6L + 0.5Lr + 1.0Wt + 1.0Ec = 1.2(150) + 1.6(300) + 0.5(60) + 1.0(70) + 1.0(50) = 630 kips 2) Determine the required strength using the strength reduction factor: Pr = Pu/φ = 630/0.65 = 969 kips 3) The nominal strength Pn must be equal or greater than the required strength: Pn ≥

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    © All Rights Reserved
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    You are on page 1of 17

    Load types and safety factors

    and the safety provisions of ACI Code
    Learning objectives
    • Why do we use safety factors when designing 
    structural elements?
    • What are the different types of safety factors?
    • How do we use safety factors in design?
    Classification of loads
    Loads may be described by their variability with respect to time 
    and location:
    • Permanent load remains constant once the structure is 
    completed. E.g. self weight of structure and soil pressure 
    against foundation
    • Variable load
    – Sustained loads of long duration. E.g. weight of filing cabinets in an 
    office
    – Loads of short duration. E.g. weight of people in the same office

    • Accidental loads include vehicular collisions and explosions
    Classification of loads – Cont.
    Loads may also be described by whether they cause acceleration 
    or vibration in the structure:
    • Static loads do not cause any appreciable acceleration or 
    vibration of structure
    • Dynamic loads cause acceleration or vibration 
    Small accelerations are taken into account by increasing the 
    specified static loads
    Large accelerations can be accounted for by multiplying the 
    effect of live load by an impact factor
    Classification of loads in the US codes

    • Dead loads: constant in magnitude and fixed in 
    location throughout the life of the structure
    – structural elements: beams, columns, slabs, etc.
    – some nonstructural elements: walls, floor fills, plaster, 
    permanent mechanical equipment, etc.

    • Live loads: occupancy load. Their magnitude and 
    location at any time is uncertain
    • Environmental loads:
    – Snow loads and wind pressure
    – Earthquake loads
    Weight of some common building 
    materials for estimating dead loads

    Why do we use tables like this?
    Minimum Live Loads Specified in ASCE/SEI 7‐10
    Minimum Live Loads Specified in ASCE/SEI 7‐02
    Snow loads

    SEI/ASCE 7‐02: Minimal Design Loads for Buildings and Other Structures

    How about seismic loads?
    Safety factors
    Objective: Design a column that can withstand the 
    compressive stress P P
    Design criteria: Pn  P
    • Pn :   nominal compressive strength
    • The word “nominal” implies that the strength is 
    computed value based on the specified concrete, 
    steel strength, and dimensions

    Pn  P
    But, in real life, is                  a safe design criteria?
    Safety factors – Cont.
    Pn  P

    Pn : Design stress
    Pn  P
    P  PU : Factored load
     : Strength‐reduction (resistant)factor,    1
     : Load factor,  1
    Final form of the  P Pn  D PD  L PL  ...
    design criterion:
    Strength method
    Load factors and required strength (ACI 318‐11)

    The service or working loads are the loads that are assumed to actually occur when 


    a structure is in use or service. Strength level loads are already factored. 
    Strength Reduction Factor

    From chapter 9 of ACI Code
    Limit States
    When a structure/structural element becomes 
    unfit for its intended use, it has reached a limit 
    state
    • Ultimate limit states: involves structural 
    collapse
    • Serviceability limit states
    – Excessive deflection
    – Excessive crack width
    – Undesirable vibrations
    Limit‐state design of structural elements

    Step 1: Design a structural element for the 
    ultimate limit state

    P Pn  D PD  L PL  ...

    Step 2: Check whether the structure will exceed 
    any of the serviceability limit states
    structural design steps (ultimate limit state 
    method)
    • structural analysis is performed separately for each type of 
    load 
    • based on each analysis the values of existing moment 
    (flexure), shear, and axial loads on each elements is calculated
    • Based on ACI equations 9‐1 to 9‐7 calculate the required 
    strengths (flexural strength Mu, shear strength Vu, and axial 
    strength Pu) of each element.
    • Using strength reduction factor calculate
    Mn  Mu / 
    the values of nominal strengths (Mn, Vn and Pn):
    Vn  Vu / 
    • Design the element so that it has the nominal  Pn  Pu / 
    strengths
    Example
    The axial loads for a building column (non‐spiral reinforced) have been 
    estimated with the following results: 

    • D = 150 kips
    • live load from roof, Lr = 60 kips
    • live loads from floors, L = 300 kips 
    • compression wind, Wt = 70 kips service level
    • tensile wind, Wt = 60 kips “
    • seismic compression load, Ec = 50 kips “
    • seismic tensile load, Et = 40 kips “

    Determine the required and nominal design loads

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