Question

$\frac{dM}{dx} = V(x)$ $\frac{dV}{dx} = -q(x)$ Problem 1: The beam below is loaded with a distributed load $q(x) = 3q_0x(L - x)$. The beam is attached to a pin support at $A$ and a roller at $B$. Find all the reactions at points $A$ and $B$ in terms of $q_0$ and $L$. $q(x) = 3q_0x(L - x)$ A 3L/4 B L/4 x

Name: fracdmdx vx fracdvdx qx problem 1 the beam below is loaded with a distributed load qx 3q0xl x the beam is attached to a pin support at a and a roller at b find all the reactions at points a 38066
Uploaded: 2024-07-24T18:29:43-08:00
Duration: 3 min 14 s
Channel: Adi S
Description: fracdmdx vx fracdvdx qx problem 1 the beam below is loaded with a distributed load qx 3q0xl x the beam is attached to a pin support at a and a roller at b find all the reactions at points a 38066

          $\frac{dM}{dx} = V(x)$  $\frac{dV}{dx} = -q(x)$
Problem 1:
The beam below is loaded with a distributed load $q(x) = 3q_0x(L - x)$. The beam is attached to
a pin support at $A$ and a roller at $B$. Find all the reactions at points $A$ and $B$ in terms of $q_0$ and $L$.
$q(x) = 3q_0x(L - x)$
A
3L/4
B
L/4
x

$fracdmdx vx fracdvdx qx problem 1 the beam below is loaded with a distributed load qx 3q0xl x the beam is attached to a pin support at a and a roller at b find all the reactions at points a 38066$

Added by John D.

University Physics with Modern Physics

Hugh D. Young 14th Edition

Instant Answer

Solved by Expert Adi S

07/24/2024

Step 1

This is done by integrating the distributed load function $q(x)$ over the length of the beam, which is $L$. $$Q = \int_0^L q(x) dx = \int_0^L 3q_0x(L - x) dx = 3q_0 \int_0^L (Lx - x^2) dx$$ $$Q = 3q_0 \left[ \frac{Lx^2}{2} - \frac{x^3}{3} \right]_0^L = 3q_0 \left( Show more…

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$\frac{dM}{dx} = V(x)$ $\frac{dV}{dx} = -q(x)$ Problem 1: The beam below is loaded with a distributed load $q(x) = 3q_0x(L - x)$. The beam is attached to a pin support at $A$ and a roller at $B$. Find all the reactions at points $A$ and $B$ in terms of $q_0$ and $L$. $q(x) = 3q_0x(L - x)$ A 3L/4 B L/4 x