Introduction to Bracing¶

In order to provide stability by resisting lateral loads we need to provide braces. Braces are used in wall, column, and slab forms. They are also needed in custom wooden scaffolding.

What types of loads do you expect bracing to be designed for?

Wallform Brace¶

If you design a brace in compression you will need to account for buckling; which means that you will have the reduction factor $C_P$ applied to the strength of the brace, i.e. a larger brace would be needed. For this reason we always design braces to be in tension, (we use $F_t$).

However, for argument's sake let's design a brace in compression. You also manage to get some 16 ft long rough sawn lumber for your bracing.

For now we will ignore the tension brace.

The wallform is 10' high and is expected to be exposed to a wind force of 20 psf (about a 110 mph wind gust event that would occur once every 25 years). We want the braces to be made from No.2 Douglas Fir-Larch (DFL) and placed at 4 feet on center.

Compression Brace Load¶

The line load at the top of the wall $(w_{hor})$ is...

$$w_{hor} = p \times h \times \left(\frac{h/2}{h} \right)$$$$w_{hor} = 20 psf \times 10 ft \left(\frac{5 ft}{10 ft} \right)$$$$w_{hor} = 100 plf$$

To convert $w_{hor}$ to $P_{brace}$ we need to to remember a little vector math and that boils down to...

$$w_{hor}\times W_{trib} = P_{brace} \times \frac{Hor}{L_{brace}}$$$$P_{brace}=w_{hor} \times \frac{L_{brace}}{Hor} \times W_{trib} $$$$Hor = \sqrt{L_{brace}^2 - h^2}$$$$Hor = \sqrt{(16 ft)^2 - (10 ft)^2}$$$$Hor = 12.5 ft$$$$P_{brace} = 100 plf \times \frac{16 ft}{12.5 ft} \times 4 ft$$$$P_{brace} = 512.4 lb$$

Sizing the Compression Brace¶

To get an estimate of the size of the brace we'll need, let's use the fact that the slenderness ratio of the brace needs to less than or equal to 50.

$$S.R. = \frac{L_e}{d_{min}}$$$$d_{min} = \frac{L_e}{S.R.}$$$$d_{min} = \frac{16 ft}{50} \times \frac{12 in}{1 ft}$$$$d_{min} = 3.84"$$

So we'll try a rough sawn 4x4 which has the actual dimensions of $3.875" \times 3.875"$.

Section Properties¶

Since we are using rough sawn lumber we will calculate the properties we would normally look up in Table 1B

$d = 3.875"$

$b = 3.875"$

$A = b \times d $

$A = 3.875" \times 3.875" = 15.02 in^2$

Column Stability Factor¶

Since the this loading is due to wind pressure alone the load duration factor $C_D = 1.6$. This is one of those few occasions in this course where $C_D \ne 1.25$.

Using the stress values and adjustment factors found in Table 4A we can calculate $C_P$.

(You may have to scroll down to see the whole calculation.)

In [11]:

brace.C_P.work

Timber/ColumnBuckling.py:-1: UserWarning: Note: Using default value for 'C_P_l_e.c', (c = 0.8), for a sawn lumber column.

Out[11]:

$$\frac{l_{e,x}}{d} = 49.55$$ $${}$$ $$\frac{l_{e,y}}{b} = 49.55\text{ }\Leftarrow \text{ Controls} \therefore SR = 49.55$$ $${}$$ $$F_c^* = F_c \times C_D \times C_M \times C_t \times C_F \times C_i $$ $${}$$ $$F_c^* = 1350 psi \times 1.6 \times 1 \times 1 \times 1.15 \times 1 = 2484 psi$$ $${}$$ $$F_{cE} = \frac{0.822 \times E_{min}'}{(SR)^2} = \frac{0.822 \times 580000 psi}{(49.55)^2} = 194.2 psi$$ $${}$$ $$\frac{F_{cE}}{F_c^*} = 0.07818$$ $${}$$ $$\textbf{Note:}\text{ Using default value for }\texttt{'C_P_l_e.c'}\text{, }(c=0.8)\text{, for a sawn lumber column.}$$ $${}$$ $$c = 0.8 \text{ (from section 3.7.1.5 Timber NDS)}$$ $${}$$ $$\frac{1+F_{cE}/F_c^*}{2 \times c}=\frac{1+0.07818}{2 \times 0.8}=0.6739$$ $${}$$ $$\frac{F_{cE}/F_c^*}{c}=\frac{0.07818}{0.8}=0.09772$$ $${}$$ $$C_P = \frac{1+F_{cE}/F_c^*}{2 \times c} -\sqrt{\left(\frac{1+F_{cE}/F_c^*}{2 \times c}\right)^2-\frac{F_{cE}/F_c^*}{c}}$$ $${}$$ $$C_P = 0.6739 -\sqrt{\left(0.6739\right)^2-0.09772} =0.07690$$

Brace Capacity¶

The capacity of the brace can now be found.

In [12]:

brace.column_capacity.work

Out[12]:

$$F_c' = F_c \times C_D \times C_M \times C_t \times C_F \times C_i \times C_P$$ $${}$$ $$F_c' = 1350 psi \times 1.6 \times 1 \times 1 \times 1.15 \times 1 \times 0.07690$$ $${}$$ $$F_c' = 191.0 psi$$ $${}$$ $$P_{allowed} = F_c' \times A$$ $${}$$ $$P_{allowed} = 191.0 psi \times 15.02 in^2$$ $${}$$ $$P_{allowed} = 2868 lbs$$

Comparing Allowable vs Actual Stress¶

The actual stress is,

In [13]:

brace.f_c_latex

Out[13]:

$$f_c = \frac{P}{A}$$ $${}$$ $$f_c = \frac{512.4 lb}{15.02 in^2}$$ $${}$$ $$f_c = 34.13 psi$$

and doing the comparison with the allowable stress we find,

In [14]:

brace.column_adequacy_check_latex

Out[14]:

$$ F_c' = 191.0 psi \geq f_c = 34.13 psi$$ $$\textbf{This column is adequate}$$

To be Continued¶

The next presentation will look at the same problem except we will assume the brace it only in tension.

Go ahead and start thinking about what that might mean for the design.

References:¶

Class website (Use this link to if you are taking the course on e-learning.)

Github.io version of course website (Do not use this link if you are taking this course in Summer A or B.)

IPython.org (IPython is the opensource software used in the development of much of this course.)

Complete Software List