rectangular load

Problem 847 | Continuous Beams with Fixed Ends

Problem 847
Compute the moments over the supports and sketch the shear diagram for the continuous beam shown in Fig. P-847.
 

847-shear-diagram.gif

 

Problem 846 | Continuous Beams with Fixed Ends

Problem 846
Sketch the shear diagram for the continuous beam shown in Fig. P-846.
 

846-shear-diagram.gif

 

Problem 845 | Continuous Beams with Fixed Ends

Problem 845
Compute the moments over the supports for the beam shown in Fig. P-845 and then draw the shear diagram.
 

845-shear-diagram-imaginary-span.gif

 

Problem 822 | Continuous Beam by Three-Moment Equation

Problem 822
Solve Prob. 821 if the concentrated load is replaced by a uniformly distributed load of intensity wo over the middle span.
 

822-beta-alpha-span-uniform-load.gif

 

Answers:
$M_2 = -\dfrac{w_o L^2}{4} \cdot \dfrac{1 + 2\beta}{4(\alpha + 1)(1 + \beta) - 1}$

$M_3 = -\dfrac{w_o L^2}{4} \cdot \dfrac{1 + 2\alpha}{4(1 + \alpha)(1 + \beta) - 1}$
 

Problem 820 | Continuous Beam by Three-Moment Equation

Problem 820
Solve Prob. 819 if the concentrated load is replaced by a uniformly distributed load of intensity wo over the first span.
 

820-continuous-beam-uniform-load.gif

 

Problem 334 | Equilibrium of Parallel Force System

Problem 334
Determine the reactions for the beam loaded as shown in Fig. P-334.
 

334-point-rectangular-triangular-loads.gif

 

Solution to Problem 691 | Beam Deflection by Method of Superposition

Problem 691
Determine the midspan deflection for the beam shown in Fig. P-691. (Hint: Apply Case No. 7 and integrate.)
 

Solution to Problem 689 | Beam Deflection by Method of Superposition

Problem 689
The beam shown in Fig. P-689 has a rectangular cross section 4 inches wide by 8 inches deep. Compute the value of P that will limit the midspan deflection to 0.5 inch. Use E = 1.5 × 106 psi.

Solution to Problem 687 | Beam Deflection by Method of Superposition

Problem 687
Determine the midspan deflection of the beam shown in Fig. P-687 if E = 10 GPa and I = 20 × 106 mm4.
 

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