Question

a)

Write down expressions for shear force and bending moment at a point in the interval A to B as a function of x:

100 lb 800 lb ft A. 5 ft 5 ft

The Problem.

The Support Reactions and Definition of x.

$V=100$ lb and $M=100\left(5-x\right)-1800$ lb.ft

$V=100$ lb and $M=100x-1800$ lb.ft

$V=100$ lb and $M=-100x-1800$ lb.ft

$V=-100$ lb and $M=100x-1800$ lb.ft

b)

Write down expressions for shear force and bending moment at a point in the interval B to C as a function of x:

The Problem.

The Support Reactions and the Definition of x.

$V=-100$ lb and $M=100x-1000$ lb.ft

$V=100$ lb and $M=-100x-1000$ lb.ft

$V=100$ lb and $M=100x$ lb.ft

$V=100$ lb and $M=100x-1000$ lb.ft

c)

Sketch the bending moment diagram.

The Problem.

The function for the interval from A to B:
$M=100x-1800$ lb.ft
The function for the interval from B to C:
$M=100x-1000$ lb.ft

Answer 1

MA Goollbft 400 loolle

ас Step-2 F

loo xo 6 00 asれest.ve MA1300 Ub-ft 0

terce Memanit ve ve

ouce loolkb l ooLb to aC loo xx - I800

Note÷ Mt loo (5-%)-1500 meana the lead 100 db :

600 lb-ft stthe

Nhe hendiuą memeult .s WYm a v again loon-1800 uh.ft Cχ#0ft to n:#5ft) .x 0ft, M.--1800 lb.ft

@x: 5 ft, Mt. 100x5-1000 פ50-o llo.ft. -300

EXPLANATION OF BENDING MOMENT DIAGRAM:

Initially there was only fixed end moment (resisting moment) of magnitude 1800 lb ft and it is taken to be negative as per sign convention.

But moving along the horizontal direction to the right of point A it can be clearly seen that there is a moment due to reaction at A as well.

So, the BM is a linear variation as it can be seen clearly from the equation. (R_a times x - 1800) And at point B an additional moment of 800 lb-ft is applied in the opposite direction of resisting moment.

Hence it decreases from 1300 lb-ft to 500lb-ft and continues to vary linearly with x again and finally becomes 0 at point C