Week 10 Challenge

Solution.1)

Aim: To calculate the total number of purlins and girts required for the building with following inputs:

Rafter span = 50 m

Bay spacing = 8 m

Roof slope = 1 in 10

Height of column = 12 m

Brick wall height = 2 m

Purlin spacing = 1.2 m

Girt spacing = 1.5 m

Procedure:

• Determination of Number of Purin:

Given: Rafter Span = 50 m.

Spacing of Purlin = 1.2 m.

Therefore, Half of Span of Rafter = 25 m.

Height of the Truss = 2.5 m (Calculated from the given value of slope)

Hence, Inclined length of Rafter = `sqrt(25^2 +2.5^2)` = 25.124 m

Now, since we know that Number of Purlin = (Inclined Length of Rafter/ Spacing of Purlin) + 1

= (25.124/1.2) + 1 = 22 ( round up nearest integer).

Hence, Total Number of Purlins on entirre span of truss = 22 x 2 = 44.

• Calculation of number of Girts:

Given: ColumnHeight = 12 m.

Girt Spacing = 1.5 m.

Brick Wall Height = 2 m. [No Girts are required]

So, Height of column for which Girt is required = 12-2 = 10 m.

Therefore, Number of Girts on one side = (10/1.5)+1 = 8

Hence, Total Number of Girts on both sides = 8 x 2 = 16

Result: Total Number of Purlins required = 44 and Total number of Girts required on both sides = 16.

Solution.2)

Aim: For given loading we need to taper the beam for the BMD envelope and check utility ratio within limit.

Given untappered frame:

Procedure:

• We will run the analysis to view.
• Now, in Post processing mode, we will see the bending moment diagram and utility ratio for the current designed frame.
• We will select Load as Envelelope as we wish to view the bending moment diagram due combination of loads.

• In the below figure we can see the Utilisation ratio and Bending Moment diagram, now we will try to taper the section such that where bending moment is not high, we will provide section will lesser moment of inertia, as for such part, lesser Moment of inertia is enough to resist the Load.
• After giving the appropriate tappered section, final utilisation ratio for the different members will be as shown in below figure.

• Dimensions of the Different Beams are shown below.

For R1:

For R2:

For R4:

For R5:

For R6:

For R9:

For R10:

For R11:

For R13:

Result: For the Given Frame, we have tappered all the section to reduce the steel use and tried to keep the utilisation ration within 0.65 to 0.95, for optimum result.

Solution.3)

Aim: We need to find the wind pressure on the wall of the building.

Building Details: An Industrial building of plan dimension 10m x 5m to be constructed (as in Figure below). Building is situated in Chennai. Assume wall opening as 15 %. Bay spacing = 2.5m.

Procedure:

Given: Location : Chennai

Plan and elevation view dimension as given above.

Wall Opening as 15 %

Bay Spacing = 2.5 m.

For Calculation of Wind Speed we will refer IS 875 Part-3.

As per IS 875 Part-3, Clause 5.2

Basic Wind Speed`V_b` = 50m/s

Since, Design Wind Speed `V_z` = `V_b.K_1.K_2.K_3.K_4`

Values of `K_1,K_2,K_3,K_4` as per IS 875 Part-3 is 1, 0.91, 1.0, 1.15 respectively.

Hence, Design Wind Speed = 52.33 m/s.

Since, Wind Pressure `P_z` = 0.6x`(V_z)^2` where `P_z`= Wind Pressure in N/`(m)^2` = 1.64 kN/`m^2`.

Therefore, `P_z`= 1643.06 N/`m^2` = 1.64 kN/`m^2`.

Now, Design Wind Pressure `P_d`= `K_d.K_a.K_c.P_z`

As per IS 875 Part-3, values of `K_d, K_a,K_c` are 1.0, 0.9 and 0.8 respectively.

Therefore, `P_d`= 1.181 kN/`m^2`

Now, Wind Load acting in a direction normal to the individual structural element or cladding unit F = {`C_(pe) - C_(p i)`} X `A` X `P_d` where `C_(p i)` = internal pressure coefficient = 0.5 for opening of 15% of wall area i.e. medium permeability (as per clause 6.2.2 of IS :875, part 3)

External pressure coefficient:

l= 10 m ; b = 5 m ; h = 6 m

`h/b=6/5=1.2`

`l/b=10/5=2`

Thus as per table 5 of clause 6.2.3.1 of IS:875, Part 3

`1/2lth/wle 3/2` and `3/2lel/blt4`

Wind angle`theta`	`C_(pe)`	`C_(pe)`	`C_(pe)`	`C_(pe)`
	Wall A	Wall B	Wall C	Wall DF
`0^0`	+0.7	-0.3	-0.7	-0.7
`90^0`	-0.5	-0.5	+0.7	-0.1

Case-I `theta=0^0`

`C_(pn et)` for wall A

`C_(pn et)= C_(pe)-C_(p i)`

(+0.7)-(0.5)= +0.2 ( Pressure)

(+0.7)-(-0.5)= +1.2 ( Pressure)

Design wind force for wall A

F= `C_(pn et)xxAxxP_d`

`F_A`= +0.2x 6x2.5x1.181 = 3.54 KN ( Pressure)

`F_A`= +1.2x 6x2.5x1.181 = 21.29 KN KN (Pressure)

`C_(pn et)` for wall B

`C_(pn et)= C_(pe)-C_(p i)`

(-0.3)-(0.5)= -0.8 ( Suction)

(-0.3)-(-0.5)= -0.2 ( Pressure)

Design wind force for wall B

F= `C_(pn et)xxAxxP_d`

`F_B`= -0.8x 6x2.5x1.181 = 14.17 KN ( Suction)

`F_B`= -0.2x 6x2.5x1.181 = 3.54 KN KN (Suction`)

`C_(pn et)` for wall C & D

`C_(pn et)= C_(pe)-C_(p i)`

`C_(pn et)`for wall C

`C_(n et) = C_(pe-C_(p i)`

= (-0.7)-(0.5)= -1.2( Suction)

=(-0.7)-(-0.5)= -0.2 ( Suction)

Design wind force for wall C & D

`F_C= -1.2xx30xx1.181 = 42.516 KN`(Suction)

`F_D=-0.2xx30xx1.181 = -7.09 KN`( Suction)

Case-II-`theta = 90^0`

`C_(pn et)`for wall A & B

`C_(pn et)= C_(pe)-C_(p i)`

(-0.5)-(+0.5)= -1.00 ( Suction)

(-0.5)-(-0.5)= 0.00

Design wind force for wall A & B

`F= C_(pn et)xxAxxP_d`

`F_A=`1x15x1.181 = 17.715 KN

`F_B=0`

`C_(pn et)` for wall C

`F= C_(pn et)xxAxxP_d`

= =(+0.7) - (+0.5) = +0.2 ( Pressure)

= (+0.7)-(-0.5) = 1.2 (Pressure)

Design force on wall C

`F= C_(pn et)xxAxxP_d`

= +0.2x 30x1.118 = + 7.09 KN(Pressure)

= +1.2x 30x 1.181 = 42.52 KN (Pressure)

`C_(pn et)`for wall D

`F= C_(pn et)xxAxxP_d`

`C_(pn et)`= (-0.1)-(+0.5) = -0.6 ( Suction)

= (-0.1)- (-0.5) = +0.4 ( Pressure)

`F_D`= -0.6x30x1.181 = -21.26 KN( Suction)

`F_d =` +( 0.4)x 30x 1.181 = + 14.17 KN ( Pressure)

Maximum Pressure for design= 42.52 KN

`C_(pe)` for pitched roof for rectangular clad building as per table-6 . clause 6.2.3.2 of IS: 875 Part-3

A= surface area of structural element or cladding unit

Windward side- `theta= 0^0`

Design wind Pressure `=C_(pe)-C_(p i)xxP_d`

= (-0.9+ 0.5)x 1181 = -472N/Sqm

= (-0.9-0.5)x 1181 = -1653.5 N/Sqm

Leeward side `F=(C_(pe)-C_(p i))xx P_d`

= (-0.6+ 0.5)x 1181 = 118.1 N/Sqm

= (-0.6-0.5)x 1181 = -1299.1N/Sqm

Windward side- `theta=90^0`

Design wind Pressure = `(C_(pe)-C_(p i))xx P_d`

= (-0.9+ 0.5)x 1181 = -472N/Sqm

= (-0.9-0.5)x 1181 = -1653.5 N/Sqm

Leeward side

`(C_(pe)-C_(p i))xxP_d`

= (-0.6+ 0.5)x 1181 = 118.1 N/Sqm

= (-0.6-0.5)x 1181 = -1299.1N/Sqm

Maximum uplift pressure = - 1653.5 N/Sqm

Result: The maximum pressure on Wall is found to be 42.52 kN.

Solution.4)

Aim: To find support reaction for given load on the Mono sloped structure (20 degree) of width 10 m & length 30 m with bay spacing of 6 m having no wall sheeting (fully open all around) has following loads.

Self Weight factor – 1.1

Roof Sheet loading = 0.5 kN/m²

Purlin load = 6 kN/m²

Live Load on Roof = 1.5 kN/m2

Wind Load on rafter (Pressure) = 2 kN/m²

Procedure:

• Open the new Staad File and create the basic model in Staad Pro, as per given data.

• Now, we will provide fixed support to the columns.

• Now, we will apply the loads on the Model.

NOTE: Apply the factor of -1.1 for SelfWeight as per given condition.

• Apply all the loads as per given loading data.

• Now, we will run the analysis.

• Go to Post Processing, and view the suppport reactions.

• Now, go to output file, under Utilities Tab and see the net reaction on structure for all the loading case.
• Net Support Reaction for the loading cases are given below.

Result: Model is prepared as per the given dimension and data, loading is applied on the modelled structure and Support reaction has been found out successfully.