Wall Form Design Part II: Temporary Structures and Calculations - Prof. Saeed Daniali, Study notes of Introduction to Business Management

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Wall Form Design - Part II

Professor Kamran M. Nemati

CM 420CM 420 Temporary Structures

Temporary Structures

CM 420 CM 420CM 420CM 420

Wall Form Design

Part II

Temporary

Structures

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Wall Form Design Example (Continued)

ƒ Design forms for 14 ft high wall to be concreted

at the rate of 3 ft per hour, internally vibrated.

Assume the mix is made with Type I cement,

with no pozzolans or admixtures, and that the

temperature of concrete at placing is 60°F.

Slump is 4 in. The forms will be used only

once, so short-term loading stresses will apply.

ƒ Form grade plywood sheathing ¾ in. thick is

available in 4 x 8-ft sheets, and 4500-lb coil ties

are on hand. Framing lumber of No. 2 Douglas

Fir-Larch is to be purchased as required.

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Last session we did:

1) Determined the pressure on the form

2) Used 4 x 8 sheets of 3/4” thick plywood

for sheathing in strong way.

3) designed for stud spacing of 1-ft O.C.

3.1) Checked for Bending

3.2) Checked for Deflection

3.3) Checked for Rolling Shear

4) Stud spacing of 12-inch O.C. was

verified.

Wall Form Design - Part II

Professor Kamran M. Nemati

CM 420CM 420 Temporary Structures

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Wall Form Design Example

ƒ STEP 3: STUD SIZE and SPACING OF WALES

(Wales support the studs)

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Wall Form Design Example

ƒ Design for 2x4 S4S studs. Find the

maximum span that can support a lateral

pressure of 600 psf.

ƒ Equivalent uniform load, w , is the max.

lateral pressure times the stud spacing.

Hence:

Studs can be considered as continuous beams

subjected to uniform loading. Like the previous set

of calculations, check for allowable span for

bending, deflection, and shear.

( ) 600 lb/lf

12 in./ft.

600 psf 12 in.

stud =

×

w =

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Wall Form Design Example

ƒ CHECK BENDING

ƒ Assume using No. 2 Douglas Fir-Larch

studs. From Table 4-2, the extreme fiber

bending stress, F b

, is 875 psi. However,

this value should be adjusted.

Wall Form Design - Part II

Professor Kamran M. Nemati

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Wall Form Design Example

ƒ The first adjustment factor is theshort-term

loading factor of 1.25. The second adjustment

factor is thesize factor obtained from Table 4-

2B, which is 1.5. Therefore:

′= 875 psi×1.25×1.5= 1640 psi

b

F

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Wall Form Design Example

ƒ The values of section modulus, S , for 2x

S4S No. 2 Douglas Fir-Larch can be

obtained from Table 4-1B as 3.06 in.

3 .

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Wall Form Design Example

ƒ The allowable stud span as a continuous

beam is:

w

FS

l

b ′

= 10. 95

31. 7 in.

10. 95 =

×

l =

Wall Form Design - Part II

Professor Kamran M. Nemati

CM 420CM 420 Temporary Structures

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Wall Form Design Example

ƒ CHECK DEFLECTION

ƒ The allowable deflection is less than l /360 of the

span and 1/8 in., whichever is less.

ƒ Using Table 4-2, the values for modulus of

elasticity for 2x4 S4S No. 2 Douglas Fir-Larch is E

= 1,600,000 psi, and in Table 4-1B the value for

the moment of inertia for is: I = 5.36 in.

4

ƒ For Δ = 1/8”:

1. 6924. 3 41. 0 in. 600

1. 693 1. 693 = × =

×

= =

w

EI

l

3. 8410. 93 42. 0 in. 600

3. 844 3. 844 = × =

×

= =

w

EI

l

ƒ For Δ = l /360:

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Wall Form Design Example

ƒ CHECK SHEAR

ƒ From Table 4-2, allowable F v (rolling shear

stress) can be found to be F

s

= 95 psi, which

should be multiplied by 1.25 for short-term

loading, as well as by a factor of 2.0 for

horizontal shear adjustment. Therefore, the

allowable shear stress is:

FV ′= 95 × 1. 25 × 2 = 238 psi

ƒ A 2x4 S4S has an actual b = 1 ½ in. and d = 3

½ in., which is obtained from Table 4-1B. Use

the equation for maximum shear for a

continuous beam and solve for l :

27. 8 7 34.8 in. 2

1 23 600

2

1 3 2

1 2381

13. 33 2 13. 33 +× = +=

× ×

• =

′ = d w

Fbd l

V

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Wall Form Design Example

ƒ Horizontal Shear Adjustment

ƒ Theshear stress factor can be applied to the base design value to increase the allowable horizontal shear stress when the length of

splits or size of shakes and checks is known, as shown in Table 6-3.

ƒ Designers may estimate an appropriate adjustment factor when

they have general knowledge of the lumber quality available. Conservative practice would suggest use of the factor 1.

whenever there is absolutely no information on splits, checks and shakes.

Wall Form Design - Part II

Professor Kamran M. Nemati

CM 420CM 420 Temporary Structures

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Wall Form Design Example

ƒ SPACING OF THE WALES

ƒ From the stud spans calculated above, the

shortest span is based on bending which is

31.7 inches.

ƒ This means the wales, which are the stud

supports CANNOT be spaced more than 31.

inches apart (this span can be increased near the

top, since in the top 4 ft., the pressure decreases

linearly from 600 psf to 0).

ƒ The top and bottom wales are often set

about 1 ft from top and bottom of wall forms.

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Wall Form Design Example

ƒ Place wales 12 inches from both top and

bottom of the wall form.

ƒ Then, 14' − 1' − 1' = 12 ft. or 144 inches

remains for spacing the other wales, which

can be no more than 31.7 inches apart.

ƒ Set them at 30 in., except one span at 24

in. (smaller spans at the bottom).

ƒ We place the smaller span near the bottom

of the form where theoretically a higher

pressure could occur.

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Wall Form Design Example

30” = 2’ − 6”

Wall Form Design - Part II

Professor Kamran M. Nemati

CM 420CM 420 Temporary Structures

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Wall Form Design Example

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Wall Form Design Example

STEPS 4 & 5: TIE DESIGN, WALE SIZE

and TIE SPACING

ƒ From the pressure diagram, the

equivalent uniform load per lineal foot of

wale is determined to be 1500 lb/lf.

ƒ The problem statement indicates that

4500-lb coil ties are available and will be

used.

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Wall Form Design Example

STEPS 4 & 5: TIE DESIGN, WALE SIZE

and TIE SPACING (Cont’d)

ƒ With the maximum load per lineal foot of

wale being 1500 lbs, then the maximum

tie spacing is:

3 ft.

1500 lb/ft

4500 lb

Waleload

Tiecapacity Tie Spacing= = =

Wall Form Design - Part II

Professor Kamran M. Nemati

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Wall Form Design Example

ƒ CHECK BENDING (Cont’d)

ƒ Maximum bending moment for a uniformly

loaded continuous beam (more than 3

supports) is:

in.-lb.

2

wl

M Max =

ƒ The maximum moment of the member

being designed is:

M FS Max b

= ′

ƒ Therefore:

2

wl

Fb ′ S =

ƒ Or:

Fb

wl S ′

=

120

2

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Wall Form Design Example

ƒ CHECK BENDING

ƒ F'b is the allowable stress in the extreme

fiber and was calculated to be 1640 psi

(refer to top of page 16 of the handout).

The span, l , is 3 ft. or 36 inches, and w =

1500 lb/lf.

ƒ Therefor the required section modulus, S ,

can be calculated using the above

equation:

3

2 2

9. 88 in.

=

×

×

=

′

=

F b

wl

S

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Wall Form Design Example

ƒ In order to avoid drilling of timbers, they

commonly use double-member wale. So the

required section modulus of 9.88 in.

3

is for two

members.

ƒ Referring to Table 4-1B, double 2x4s will yield a

section modulus of 2x3.06 or 6.12 in.

3

, which is

less than 9.88 in.

3

, and therefore not

acceptable.

ƒ Checking the next larger size, 3x4, will result in:

S = 2x 5.10 = 10.20 in.

3

> 9.88 in.

3

, which

satisfies the section modulus requirements. ⇒

Use double 3x4 wales.

Wall Form Design - Part II

Professor Kamran M. Nemati

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Wall Form Design Example

ƒ CHECK SHEAR

ƒ To check the horizontal shear for the double 3x

wales, use the horizontal shear stress formula for a

uniformly loaded continuous beam.

( l d )

bd

w f (^) V 2

13. 33

= −

77. 14 2. 42 186 psi 238 psi O.K. 12

⎟= × ≅ 〈 ⇒

⎠

⎞

⎜

⎝

⎛ ×

× −

×

×

fV =

ƒ From Table 4-1B, the value of bd for a 3x4 member

can be obtained as: 8.75 in.^2 ( or simply: 2.5” x 3.5”=8.75 in.^2 )

ƒ Therefore the stress in the double 3x4 members

meets the requirements. (The value of the adjusted

allowable shear stress of 238 psi was calculated before - refer to page

20 of the handout).

⎟

⎠

⎞

⎜

⎝

⎛

= −

1. 9 2 d L bd

w

or f^ V

(They are exactly the same formulas)

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Wall Form Design Example

ƒ STEP 6: BEARING CHECK

ƒ Check:

ƒ 1) bearing of the studs on wales and

ƒ 2) bearing between the tie washer or tie

holders and wales.

ƒ From Table 4-2, the value of compression

Perpendicular to grain, F c ⊥ , for No. 2 2x

Douglas Fir-Larch is 625 psi.

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Wall Form Design Example

Length of bearing, in

1/2 1 1 1/2 2 3 4

6 0r more

Factor…. 1.79 1.37 1.25 1.19 1.13 1.09 1.

From page 13 of the handout, the multiplying factors for

indicated lengths of bearing on small area plates and

washers are shown below: 3.5”

(1.13+1.09)/2 = 1.

Wall Form Design - Part II

Professor Kamran M. Nemati

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Wall Form Design Example

ƒ TIES: Assume a 3½ in.-square tie washer.

ƒ Then the bearing area is:

(3½″)

2 − ¾″×3½″= 12.25 − 2.63 = 9.63 in.

2

ƒ Since this is a short

bearing length, F c ⊥ should

be multiplied by a factor

of 1.11 (refer to page 13

for 3½ in. bearing

length):

ƒ Adjusted F c ⊥

= 625 (1.11)

= 694 psi

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Wall Form Design Example

ƒ The actual bearing stress is then:

467 psi

9.63in.

4500 lb

bearingarea

Maximum tieload

2

= = < 694 psi ⇒ O.K.

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Wall Form Design Example

ƒ STUDS ON WALES:

ƒ The bearing area between

2x4 studs and double 3x

wales can be calculated as:

2 x (1½” x 2½”) = 2 x 3.75 = 7.5 in. 2

ƒ Load transfer to the wale = ½ the stud

span above and below the wale x the lateral

pressure x the stud spacing

600 psf 1 ft 1500 lb

× × =

+

200 psi

7.5in.

1500 lb

bearing stress

2

= = < 625 psi ⇒ O.K.

Wall Form Design - Part II

Professor Kamran M. Nemati

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4x8-ft 3/4” thick plywood sheathing

2x4 S4S No. 2 Douglas Fir- Larch studs 12-in. O.C.

Double 3x4 wales

3 1/2 in.-square tie washer

4500-lb coil ties with 3-ft spacing

Wall Form Design Example

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Wall Form Design Example

Bracing for Lateral Loads

ƒ Consider the necessary bracing for a wall

form 14 ft. high, above grade, in an area

where the local building code specifies a

minimum 20 psf wind loading.

ƒ Table 5-7 (page 6 of the handout) indicated

that 140 lb per lineal foot should be used

for design of bracing, since the wind force

prescribed by local code gives a value

larger than the 100 lb/ft minimum

established by ACI Committee 347.

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Wall Form Design Example

Bracing for Lateral Loads

Wall Form Design - Part II

Professor Kamran M. Nemati

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Wall Form Design Example

Bracing for Lateral Loads

ƒ Strut Bracing

ƒ If wooden strut bracing is provided, strong

enough to take either a tension or

compression load, then single side bracing

may be used.

ƒ Nailed connections at either end must be

strong enough to transmit the tension load,

and wales or other form members must be

strong enough to transmit accumulated

horizontal forces to the strut bracing.

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Wall Form Design Example

Bracing for Lateral Loads

ƒ Strut Bracing

ƒ If wooden bracing is attached 1 or 2 feet

below the top of the wall, the bracing must

carry more than the 140 lb per ft load applied

at the top.

14’

12’

H

H’

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Wall Form Design Example

Bracing for Lateral Loads

ƒ Strut Bracing

ƒ H’ the horizontal bracing force 2 feet from

the top of the wall would have to be

(14’/12’)(140 lb/ft) = 163 lb per ft

in order to balance the 140 lb/lf design load

applied at the top of the wall.

ƒ If end of the brace is put 8 feet from the

wall, use the relationship between sides of

the right triangle to find the the length of

brace and load it must carry.

Wall Form Design - Part II

Professor Kamran M. Nemati

CM 420CM 420 Temporary Structures

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Wall Form Design Example

Bracing for Lateral Loads

ƒ Strut Bracing

h= 12’

X= 8’

t

H’=163 lb./ft. 2 2 t = h + x

( 12 ) ( ) 8 208 14. 15 ft

2 2

t = + = =

( ) 290 lbperftofwall

tension compressioninstrut= 163 × =

8'

14.15'

(tension) compressioninstrut=H′×

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Wall Form Design Example

Bracing for Lateral Loads

ƒ If struts are spaced every 8 feet along

the wall, then 8x290 = 2320 lb must be

carried by each brace.

ƒ Many wood members strong enough to

carry this load in compression will also

be adequate in tension. However, the

strength of connections (nails, etc.) must

be made adequate for the tension load.