Question

13. File Response: P3: The beam supports the distributed loa... Question The beam supports the distributed icad shown. Determine the resultant internal loadings acting on the cross-section at point D. Assume the reactions at supports A and B are vertical

Answer 1

Given a simply supported beam with triangular loading

Length of beam (L) = 1.5 + 3 + 1.5 = 6m

the supports are at point A and point B

the triangular loading varies with 4KN/m at support A to zero at support B

we need to find resultant internal loading at point D

There are basically three types of internal loading namely- shear force , axial force and bending moment

so, we have to find shear force , axial force and bending moment at point D

• Calculation of support reactions at A and B

support A and support B will have only vertical support reaction [ given in the question ]

Let vertical support reaction at support A = V_A

and vertical support reaction at support B = V_B

Taking moment equilibrium about point B (ΣM_B = 0) , we get:

VA XL - 2 2 x L x W x x L) = 0 3

$\Rightarrow V_{A} \times 6 - \frac{1}{2} \times 6 \times 4 \times (\frac{2}{3} \times 6) = 0$

From here, V_A = 8 KN

also, from vertical equilibrium of forces(ΣF_y = 0) , we get:

V_A + V_B = (1/2) * L * W

8 + V_B = (1/2) * 6 * 4

therefore, V_B = 4 KN

• Shear force at point D

First we have to find ordinate of triangular loading (m) at point D

we can calculate this by similarity of triangles as shown in the figure

$\Rightarrow \frac{m}{1.5} = \frac{4}{4.5+1.5}$

m = 1 KN/m

   cutting a section at point D from right, we get:

Let shear force at section D be V_D as shown in the figure, from the equation of shear force we get:

$\Rightarrow V_{D} + \frac{1}{2} \times 1.5 \times 1 - V_{B} = 0$

  $\Rightarrow V_{D} + 0.75 - 4 = 0$

therefore, V_D = 3.25 KN

So, resultant shear at point D is 3.25 KN (downward)

• Axial or normal force at point D

as there is no horizontal loading on the beam, so there is no resisting axial force acting on the beam

So, axial force is zero at point D

• Bending moment at D

Let bending moment at point D is M_D (assume clockwise)

bending moment equation for point D can be written as :

$\Rightarrow M_{D} + \frac{1}{2} \times 1 \times 1.5 \times \left ( \frac{1}{3} \times 1.5 \right ) - (4 \times 1.5) = 0$

$\Rightarrow M_{D} + 0.375 - 6 = 0$

From here, M_D = 5.625 KN-m

As the sign of M_D is positive, so our assumption of clockwise moment is correct

So, moment at point D is 5.625 KN-m