FIXED BEAM AT BOTH ENDS - DISTRIBUTED LOAD CALCULATOR


Fixed - Fixed Beam with Distributed Load Calculator to find forces, moments, stresses, deflections and slopes of a fixed beam at both ends which is subjected to a uniformly, uniformly varying, trapezoidal, triangular and partially distributed load.

Fixed - Fixed Beam with Distributed Load Calculator:

Fixed - Fixed Beam with Distributed Load
INPUT PARAMETERS
Parameter Value
Distributed load magnitude at a [wa]*
Distributed load magnitude at L [wb]*
Beam Length [L]
Distance a
Distance b
Distance x
Modulus of Elasticity [E]
Distance from neutral axis to extreme fibers [c]
Second moment of area [I ]**

Note : Use dot "." as decimal separator.

Note * : wa and wb are positive in downward direction as shown in the figure and negative in upward direction.

Note ** : For second moment of area calculations of structural beams, visit " Sectional Properties Calculators".

 


RESULTS
Parameter Value
Reaction Force 1  [R1] ---
Reaction Force 2 [R2] ---
Transverse Shear Force @ distance x [Vx] ---
Maximum Transverse Shear Force [Vmax] ---
Reaction Moment 1 [M1] ---
Reaction Moment 2 [M2] ---
Moment @ distance x [Mx] ---
Maximum Moment [Mmax] ---
Slope 1 [θ1] ---
Slope 2 [θ2] ---
Slope @ distance x [θx] ---
Maximum Slope [θmax] ---
End Deflection 1 [y1] ---
End Deflection 2 [y2] ---
Deflection @ distance x [yx] ---
Maximum Deflection [ymax] ---
Bending Stress @ distance x [σx] ---
Maximum Bending Stress [σmax] ---

Note * : R1 and R2 are vertical end reactions at the left and right, respectively, and are positive upward. Shear forces and deflections are positive in upward direction and negative in downward direction. All moments are positive when producing compression on the upper portion of the beam cross section. All slopes are positive when up and to the right.

Note: Stresses are positive numbers, and these are stress magnitudes in the beam. It does not distinguish between tension or compression of the structural beam. This distinction depends on which side of the beam's neutral plane c input corresponds.



Slope


Deflection



Moment




Shear Force

Definitions:

Distributed load: A load which acts evenly over a structural member or over a surface that supports the load.

Fixed support: Fixed supports can resist vertical and horizontal forces as well as a moment. Since they restrain both rotation and translation, they are also known as rigid supports.

Roller support: Roller supports are free to rotate and translate along the surface upon which the roller rests. The resulting reaction force is always a single force that is perpendicular to the surface. Roller supports are commonly located at one end of long bridges to allow the expansion and contraction of the structure due to temperature changes.

Fixed beam: A beam which is fixed at both ends.

Structural beam: A structural element that withstands loads and moments. General shapes are rectangular sections, I beams, wide flange beams and C channels.

Supplements:

Link Usage
Sectional Properties Calculator of Profiles Sectional properties needed for the structural beam stress analysis can be calculated with sectional properties calculator.

List of Equations:

Following fixed beam distributed load formulas are used for the calculations. Superposition principle is used if needed.

 

Fixed Beam with Distributed Load
Parameter Equation
Reaction Force 1 [R1] $${ R }_{ 1 }=\frac { { w }_{ a } }{ 2{ L }^{ 3 } } { (L-a) }^{ 3 }(L+a)+\frac { { w }_{ L }-{ w }_{ a } }{ 20{ L }^{ 3 } } { (L-a) }^{ 3 }(3L+2a)$$
Reaction Force 2 [R2] $${ R }_{ 2 }=\frac { { w }_{ a }+{ w }_{ L } }{ 2 } \cdot (L-a)-{ R }_{ 1 }$$
Shear force at distance x [V] $$V={ R }_{ 1 }-{ w }_{ a }\left< x-a \right> -\frac { { w }_{ L }-{ w }_{ a } }{ 2\cdot (L-a) } { \left< x-a \right> }^{ 2 }$$
Reaction Moment 1 [M1] $${ M }_{ 1 }=-\frac { { w }_{ a } }{ 12{ L }^{ 2 } } { (L-a) }^{ 3 }(L+3a)-\frac { { w }_{ L }-{ w }_{ a } }{ 60{ L }^{ 2 } } { (L-a) }^{ 3 }(2L+3a)$$
Reaction Moment 2 [M2] $${ M }_{ 2 }={ R }_{ 1 }L+{ M }_{ 1 }-\frac { { w }_{ a } }{ 2 } { (L-a) }^{ 2 }-\frac { { w }_{ L }-{ w }_{ a } }{ 6 } { (L-a) }^{ 2 }$$
Moment at distance x [M] $$M={ M }_{ 1 }+{ R }_{ 1 }x-\frac { { w }_{ a } }{ 2 } { \left< x-a \right> }^{ 2 }-\frac { { w }_{ L }-{ w }_{ a } }{ 6(L-a) } { \left< x-a \right> }^{ 3 }$$
Bending stress at distance x [σ] $$\sigma =\frac { Mc }{ I } $$
End Deflection 1 [y1] $${ y }_{ 1 }=0$$
End Deflection 2 [y2] $${ y }_{ 2 }=0$$
Deflection at distance x [y] $$y={ y }_{ 1 }+{ \theta }_{ 1 }x+\frac { { M }_{ 1 }{ x }^{ 2 } }{ 2EI } +\frac { { R }_{ 1 }{ x }^{ 3 } }{ 6EI } -\frac { { w }_{ a } }{ 24EI } { \left< x-a \right> }^{ 4 }-\frac { { w }_{ L }-{ w }_{ a } }{ 120EI(L-a) } { \left< x-a \right> }^{ 5 }$$
Slope 1 [θ1] $${ \theta }_{ 1 }=0$$
Slope 2 [θ2] $${ \theta }_{ 2 }=0$$
Slope [θ] $$\theta ={ \theta }_{ 1 }+\frac { { M }_{ 1 }x }{ EI } +\frac { { R }_{ 1 }{ x }^{ 2 } }{ 2EI } -\frac { { w }_{ a } }{ 6EI } { \left< x-a \right> }^{ 3 }-\frac { { w }_{ L }-{ w }_{ a } }{ 24EI(L-a) } { \left< x-a \right> }^{ 4 }$$

Reference: