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Technical Roof Deck Example

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Technical Information

General - Presented in the load tables are maximum uniformly distributed specified loads.

LIMIT STATES DESIGN (LSD)

Strength - Limit States Design principles were used in the development of the load tables in accordance with CSA-S136-01, Cold Formed Steel Structural Members and the National Building Code of Canada. The factored resistance under consideration, oR, must be equal to or greater than the eect of the factored loads, i.e.,

ΦR > EFFECT OF FACTORED LOADS.

Hence, a short calculation must be carried out to compute the specied live load. See example.

Serviceability - Maximum specified deflection loads given in the tables must be compated with their respective specified live loads.

STEEL

Specification - Conforms to ASTM A653/A653M steel. Grade 33/230; Yield strength 33 ksi/230 MPa and tensile strength of 45 ksi/310 MPa. Grade 80/550; Yield strength 80 ksi/550 MPa and tensile strength 82 ksi/570 MPa.

Finishes - A25/ZF075 or G90/Z275. For heavier galvanizing, refer to ASTM A653/A653M.

DESIGN CONSIDERATIONS

Strength - The maxiumum uniformly distributed speciedload obtained from the load table must be equal to orgreather than (Specied live load + 0.833 times thespecied dead load).
Where 0.833 = 1.25/1.5.

Conservative Strength Approach - The maximum uniformly distributed specied load obtained from theload table must be equal to or greater than (Specied liveload + specied dead load).

Serviceability (Deflection) - The effective movement of inertia for deflection determination has been calculated at an assumed specified live load stress of 0.6Fy.).

EXAMPLE (Use of Load Table)

WF Roof Deck A & B

Given:

~ Double span continuous, L = 7 ft each span
~ Deck thickness, t = 0.030 in., 33 ksi
~ L/240 deflection limit
~ Bearing length, n = 3 in.
~ Specified loads

  1. Dead loads (DL)
    a) Deck                 1 psf
    b) Superimposed    8 psf
                        DL = 9 psf
  2. Live load (LL)
                      LL = 40 psf

Solution:
Strength

  1. Specified load
    [LL +0.833DL]
    [40 + 0.833(9)] = 47.5 psf
  2. Maximum specified load (from table under "S")
    Is 51 psf
    Since 51 > 47.5 ∴OK
  3. Check web crippling (N = 3 in.)
    a) At end
    End reaction
    0.375(47.5)7 = 125 lb/ft
    (from section property table)
    Pe = Pe1 + Pe2 [(N/t)]½
    Pe = 178 + 44.6[3/0.030]½ = 624 lb/ft
    Since 624 > 125 ∴OK
    b)At interior
    Interior reaction
    1.25(47.5)7 = 416 lb/ft
    (from section property table)
    Pi = Pi1 + Pi2[(N/t)]½
    Pi = 433 + 73.6[3/0.030]½ = 1169 lb/ft
    Since 1169 > 416 ∴OK
Deflection
From table under "D" (L/180) = 103 psf
For L/240, multiply 103 by 180/240 = 77.3 psf
Since 77.3 psf is > 40 psf ∴OK