Static Equilibrium Balance of forces Simply Supported Beams

Static Equilibrium, Balance of forces & Simply Supported Beams What you need to know to perform basic static analysis on simply supported beams with point loads and evenly distributed loads.

Static Equilibrium and balance of forces Static equilibrium in a mechanical system is the condition that exists when there is NO acceleration • • F=ma must equal 0. More simply put when considering the analysis of simple beams There is no motion – everything is static and not moving Since mass and acceleration are not zero for simple beams that must mean that: Ø the sum of the forces acting on the mechanical system must equal zero AND Ø With no rotation (static) the sum of the moments must also equal zero. Formulas we will be using For analysis of simple beams we will be using each of the formulas above

How to read the mathematical formula with a ∑ in it. This formula is read as The sum of the moments ∑ M about a point equals zero P An example of taking the sum of would be finding your bowling average: Scores 150 125 155 110 = bowling average

Moment – What is it and what are Distance - d its units. It is a rotational force – a force applied at a distance from a pivot point Point of rotation Formula for Moment Force - F M=F*d M – moment or a torque F – force in US customary units is pounds d – distance from the point of rotation is in inches or feet What are the units for a moment? M=F*d = lbs * ft ft*lbs or read as foot pounds Yes you can have a force times a distance and the result is foot pounds ft*lbs UNITS - ft*lbs OR in*lbs Moment caused by A M=F*d F = 20 lbs d = 1 ft M = 20 lbs(1 ft) M = 20 ft*lbs Moment caused by B M=F*d F = 10 lbs d = 2 ft M = 10 lbs(2 ft) M = 20 ft*lbs

Static Equilibrium and the see-saw Static Equilibrium Forces are balanced and moments are balanced Fr Fe Dr De Point p - point of rotation Moment for the blue guy Force Fr * distance Dr = Fr*Dr De = Fe*De Moment for the red guy Force Fe * distance To solve for a mechanical system in static equilibrium we START with the equation that deals with moments being balanced: A moment is a Force * distance The distance is the displacement of the perpendicular component of force from the point of rotation (point p). For now all forces will be Perpendicular

Solving static equilibrium problems moments are positive & negative BY DEFINITION: Forces that cause a Counter Clockwise rotation about a point are considered positive moments Forces that cause a Clockwise rotation about a point are considered negative moments Fr +(Fr )(Dr) Dr The moment Fr*Dr is a positive moment about point p because it would cause a CCW rotation – (Fe)(De) Fe De The moment Fe*De is a negative moment about point p because it would cause a CW rotation

Solving equilibrium problems first class lever – see-saw Fr +(Fr )(Dr) Dr – (Fe)(De) Fe De Using the fact that in static equilibrium the sum of the moments about a point is equal to zero: use the formula and write an equation with proper signs +(Fr )(Dr) – (Fe)(De) = 0 Problem: If Fe = 10 lbs and De = 8 ft and Fr = 40 lbs. What is the distance Dr need to be for see-saw to be in static equilibrium?

Solving equation with one unknown Equation from static analysis and sum of moments Write down knowns and unknowns Fe = 10 lbs. De = 8 ft. Fr = 40 lbs. Dr. = ? +(Fr )(Dr) – (Fe)(De) = 0 (40 lbs. ) (Dr) – (10 lbs. )(8 ft. ) = 0 Get rid of negative sign by adding + (10 lbs. )(8 ft) same thing to both sides of equals sign (40 lbs. ) (Dr) = (10 lbs. )(8 ft) (40 lbs. ) Dr = (10)(8 ft) 40 Dr = 2 ft Parentheses not needed around Dr and simplify fraction Check that solution is correct

What about translational forces those in the X and Y directions? The forces in any system in static equilibrium need to balance. Another way of saying this is the forces need to sum up to equal zero. Y axis + pos Formulas - neg + pos Lets first consider the forces in the y – plane. Forces pointing down are negative and forces pointing up are positive X axis - neg The see-saw example below will need a REACTION force to be in static equilibrium. Therefore using this formula: Fr Fe Dr De F Reaction is up positive + F Reaction Fr is down negative - Fr Fe is down negative - Fe Summing the forces in the y direction yields

Solving equation with one unknown Equation from static analysis and sum of forces in y direction + F Reaction – Fr – Fe = 0 + F Reaction – 40 lbs. – 10 lbs. = 0 F Reaction – 40 lbs. - 10 lbs = 0 F reaction - 50 lbs. = 0 + 50 lbs Write down knowns and unknowns Fe = 10 lbs. F Reaction = ? Fr = 40 lbs. Substitute knowns into equation With units Collect like terms Add 50 lbs each side of equals F Reaction = 50 lbs. What about the forces in the X- Direction? NO FORCES in the X-direction for this simplification of a see-saw.

Types of reaction forces for simply supported beams Simply supported beam Pin connection Roller Need to understand the reaction forces for the different types of supports.

Free Body Diagram Simply supported beam What would you estimate the end reactions to be? P 1 Load = 100 lbs. Fax Y reaction force at A pin A B Fay Y reaction force at B roller Fby

Moment calculations for simply supported beam with point load P 1 Load = 100 lbs. 5’ Fax 10’ B A Fay Fby Which point is best use to calculate the sum of moments? We can choose any point to calculate the moments about. If we choose Point A we will have be able to eliminate 2 unknowns because the distance from the force Fax and Fay is zero to the point of rotation. Note: M=F*d if d=0 then product of F*0= 0. Fby(10 ft) = 100 lbs(5 ft) 10 ft Fax(0 ft) + Fay(0 ft) – P 1(5 ft) + Fby(10 ft) = 0 Fby = 10 lbs(5) Fax(0 ft) + Fay(0 ft) – 100 lbs(5 ft) + Fby(10 ft) = 0 substitute Fby = 50 lbs – 100 lbs(5 ft) + Fby(10 ft) = 0

Next equation to use in solution P 1 Load = 100 lbs. Fax A B Fby = 50 lbs Fay Write equation Substitute known values Collect like terms Solve for unknown + Fay + Fby -100 lbs = 0 + Fay + 50` lbs -100 lbs = 0 + Fay - 50 lbs = 0 Fay = 50 lbs Fax = 0 because there are no other forces in the x-direction therefore must be zero