Design Loads & Combinations

Prepared by Prof. Marcia C. Belcher, PE

Design Loads

Structural members must be designed so they are strong

enough to carry the service loads imposed on it.

We also want an “economical” structural:

Member that are not excessively large or oversized.

Design process looks like:

Determine design loads felt by each member

Model (FBD)

Determine internal forces (V, M)

Select member material/size

Check deflection limits

Types of Building Loads

Dead Loads:

Called gravity loads because they are vertical forces due to stationary

objects.

Weight of building itself

Utilities, piping, lighting, HVAC, etc.

Non-moveable objects such as carpeting, flooring, stationary service

equipment (chillers, bank vaults, etc.)

Live Loads: Vertical loads due to human occupancy, snow, rain ponding, furniture,

partition walls and moveable equipment.

Horizontal (lateral) loads due to wind, earthquake, water pressure,

blast/explosion, collision, etc.

Load Combinations

ASCE Publication 7 prescribes likely scenarios of load

combinations that a structure may feel.

ie. A structure may feel maximum human occupancy, wind and

snow at the same time.

It is not likely to feel maximum occupancy, snow, wind and

earthquake at the same time.

Load Combinations:

Strength Design Method (LRFD)

Load Combinations:

1. Ldesign = 1.4D

2. Ldesign = 1.2D +1.6 L + 0.5 (Lroof or S)

3. Ldesign = 1.2D +1.6 (Lroof or S) + (.5L or .8 W)

4. Ldesign = 1.2D + 1.6 W + 0.5 L + .5(Lroof or S)

5. Ldesign = 1.2D + Ev + Eh + 0.5 L + .2 S

6. Ldesign = 0.9D +1.6 W

7. Ldesign = .9D - Ev + Eh

Where: D = dead load

L = live load

Lr = live roof load

S = snow

W = wind

Eh = horizontal earthquake

Ev = vertical earthquake

Unit Load Calculations: Sloped Roofs

Most load computations are done in terms of plan view unit areas.

Loads for roof areas that are sloped are commonly expressed in terms of weight per unit area of horizontal projection.

Loads on Inclined Roofs

Ø

LL, W, S

Simplified Approach: Live Loads, Wind Loads, and Snow

Loads are considered to act on a

horizontal projection of an inclined

member.

SO LOAD DIAGRAM LOOKS LIKE

THIS AND CORRECTION FOR SLOPE IS

IGNORED:

LL, W, S

Length

Loads on Inclined Roof

EXCEPTION:

Full weight of dead load and vertical earthquake

load are computed to act over the entire length

and applied to the horizontal projection.

SO LOAD DIAGRAM LOOKS LIKE

THIS AND LOAD OVER FULL LENGTH IS

CALCULATED.

Ø

Projected Load: DL x 1/cosØ

Ev x 1/cosØ

Øh

cosØ=DL

h

Example: Determining Load Diagrams

Given: A simply supported roof beam inclined at a 10°

slope receives loads as follows:DL =1.2k/ft

Lr = .24 k/ft

S = 1 k/ft

Wh = 15 k

Eh = 25 k

Ev = .2 k/ft

Find: Loading diagram for the beam using strength design

method combinations.

Example: Determining Load Diagrams

Know: Load Combinations1. Ldesign = 1.4D

2. Ldesign = 1.2D +1.6 L + 0.5 (Lroof or S)

3. Ldesign = 1.2D +1.6 (Lroof or S) + (.5L or .8 W)

4. Ldesign = 1.2D + 1.6 W + 0.5 L + .5(Lroof or S)

5. Ldesign = 1.2D + Ev + Eh + 0.5 L + .2 S

6. Ldesign = 0.9D +1.6 W

7. Ldesign = .9D - Ev + Eh

Solution:a. DL adjusted for slope = 1.2 k/ft x 1/cos10° = 1.22 k/ft

b. Vertical earthquake adjusted for slope = .2 x 1/cos10° = .20 k/ft

Equations:

1. 1.4D = 1.4 (1.22 k/ft) = 1.71 k/ft

(this equation has vertical loads only)

Given Loads:DL =1.22k/ft

Lr = .24 k/ft

S = 1 k/ft

Wh = 15 k

Eh = 25 k

Ev = .2 k/ft

1.71 k/ft

L

2. 1.2D +1.6 L + 0.5 (Lroof or S) = 1.2 (1.22k/ft) + .5(1k/ft) = 1.964 k/ft

(this equation has vertical loads only)

3. This equation has vertical loads (wind) and horizontal loads

1.2D +1.6 (Lroof or S) + (.5L or .8 W) =

Example: Determining Load Diagrams DL =1.22 k/ft

Lr = .24 k/ft

S = 1 k/ft

Wh = 15 k

Eh = 25 k

Ev = .2 k/ft1.964 k/ft

L

DL (k/ft)

Lrk/ft)

S (k/ft)

Wkips

Sum

Load 1.22 ----- 1 15

Factor 1.2 ----- 1.6 .8

Factored

vertical load1.464 1.6 3.06

(k/ft)

Factored

horizontal load12 12

kips

3.06 k/ft

L

12k

4. 1.2D + 1.6 W + 0.5 L + .5(Lr or S)

5. 1.2D + Ev+ E

h +0 .5L + .2 S

Example: Determining Load Diagrams

DL(k/ft)

WKips

Lr(k/ft)

S(k/ft)

Sum

Load 1.22 15 --- 1

Factor 1.2 1.6 --- .5

Factored

vertical load1.464 --- .5 1.964

(k/ft)

Factored

horizontal load24 24 kips

1.964 k/ft

L

24k

DL(k/ft)

Ev(k/ft)

EhKips

S(k/ft)

Sum

Load 1.22 .2 25 1

Factor 1.2 1 1 .2

Factored

vertical load1.464 .2 --- .2 1.864

(k/ft)

Factored

horizontal load25 25 kips

1.864 k/ft

L

25k

DL =1.22 k/ft

Lr = .24 k/ft

S = 1 k/ft

Wh = 15 k

Eh = 25 k

Ev = .2 k/ft

6. 0.9D +1.6 W

7. 0.9D - Ev+ E

h

Example: Determining Load Diagrams

DL(k/ft)

W kips

Sum

Load 1.22 15

Factor .9 1.6

Factored

vertical load1.1 1.1

(k/ft)

Factored

horizontal load24 24

kips

1.1 k/ft

L

24k

DL(k/ft)

Ev(k/ft)

Ehkips

Sum

Load 1.22 -.2 25

Factor .9 1 1

Factored

vertical load1.1 -.2 .9

(k/ft)

Factored

horizontal load25 25

kips

.9 k/ft

L

25k

DL =1.22 k/ft

Lr = .24 k/ft

S = 1 k/ft

Wh = 15 k

Eh = 25 k

Ev = .2 k/ft

Which combination controls?

Equation 3

Equation 4

Equation 5

Example: Determining Load Diagrams

3.06 k/ft

L

12k

1.964 k/ft

L

24k

1.864 k/ft

L

25k

•The maximum effect (stress & deflection) cannot be determined by

inspection of the loads alone.

•Calculation of maximum stress and deflection for each case is required to

determine the controlling or “worse case” scenario.