# Roof truss -- distributed load

## Homework Statement

Snow on a roof supported by the Howe truss of the figure can be approximated as a distributed load of 20 lb/ft (measured along the roof). Treat the distributed load as you would the weight of the members; that is, replace the total load on each of the upper members as a vertical force, half applied to the joint at each end of the member. Determine the force in members BC, BG, CG, and GH.

## Homework Equations

joint equilibrium, method of joints and method of sections.

Method of joints: Sum of vertical forces = 0 and Sum of horizontal forces =0.

## The Attempt at a Solution

The only part I'm not sure about is the distributed load reduction. The book shows the figure I'm attaching. I also happen to have (supposedly) the solutions with which I contrasted my solution. The solutions manual reduces the distributed load differently than what I did: I thought that because the load was uniform in "magnitude" the load would be just 20 lb/ft multiplied by the HORIZONTAL length, not the diagonal lenght as they do in the solution (where they use Pythagoras' theorem).

Am I right or wrong? shouldn't the resultant load on members AB, BC, CD and DE be equal to 8ft * 20 lb/ft = 160 lb. And then I can proceed and put half of this onto the joints?

#### Attachments

• SNOW DISTRIBUTED LOAD.jpg
63.5 KB · Views: 307

## Answers and Replies

haruspex
Science Advisor
Homework Helper
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2020 Award
I thought that because the load was uniform in "magnitude" the load would be just 20 lb/ft multiplied by the HORIZONTAL length, not the diagonal lenght as they do
It clearly states
measured along the roof

It clearly states
Hi, yeah but looking at the direction of the load, it is vertical, therefore, I can imagine that the "area" of the distributed force would be computed by using a horizontal span. I can also imagine that I can transmit each differential element dF downward, leveling the height and getting a horizontal rectangle.
Why can't I do this?

haruspex
Science Advisor
Homework Helper
Gold Member
2020 Award
I can imagine that the "area" of the distributed force would be computed by using a horizontal span.
Imagine what you like, but it tells you, very clearly, that the given weight is 20lb per foot, measuring along the slope of the roof. As you go up the slope, the weight of snow on each foot you cover is 20lb.
If you prefer, you can use that to calculate the total weight and then calculate the weight per horizontal foot, but it will be more than 20lb.