Solution to Problem 428 | Relationship Between Load, Shear, and Moment

Problem 428
Beam loaded as shown in Fig. P-428.

 

Derivation of formula for volume of a frustum of pyramid/cone

Frustum of a pyramid and frustum of a cone
 

Frustum of a pyramid and frustum of a cone

 

The formula for frustum of a pyramid or frustum of a cone is given by
 

$V = \dfrac{h}{3} \left[ \, A_1 + A_2 + \sqrt{A_1A_2} \, \right]$

 

Where:
h = perpendicular distance between A1 and A2 (h is called the altitude of the frustum)
A1 = area of the lower base
A2 = area of the upper base
Note that A1 and A2 are parallel to each other.
 

Solution to Problem 427 | Relationship Between Load, Shear, and Moment

Problem 427
Beam loaded as shown in Fig. P-427.

 

Solution to Problem 426 | Relationship Between Load, Shear, and Moment

Problem 426
Cantilever beam acted upon by a uniformly distributed load and a couple as shown in Fig. P-426.

 

Solution to Problem 425 | Relationship Between Load, Shear, and Moment

Problem 425
Beam loaded as shown in Fig. P-425.

 

Relationship Between Load, Shear, and Moment

Shear and moment diagrams by shear and moment equationsThe vertical shear at C in the figure shown in previous section (also shown to the right) is taken as
$V_C = (\Sigma F_v)_L = R_1 - wx$
 

where R1 = R2 = wL/2
 

$V_c = \dfrac{wL}{2} - wx$
 

The moment at C is
$M_C = (\Sigma M_C) = \dfrac{wL}{2}x - wx \left( \dfrac{x}{2} \right)$

$M_C = \dfrac{wLx}{2} - \dfrac{wx^2}{2}$
 

If we differentiate M with respect to x:
$\dfrac{dM}{dx} = \dfrac{wL}{2} \cdot \dfrac{dx}{dx} - \dfrac{w}{2} \left( 2x \cdot \dfrac{dx}{dx} \right)$

$\dfrac{dM}{dx} = \dfrac{wL}{2} - wx = \text{shear}$
 

thus,

$\dfrac{dM}{dx} = V$

 

Solution to Problem 420 | Shear and Moment Diagrams

Problem 420
A total distributed load of 30 kips supported by a uniformly distributed reaction as shown in Fig. P-420.

 

Solution to Problem 419 | Shear and Moment Diagrams

Problem 419
Beam loaded as shown in Fig. P-419.

 

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