A simply supported beam of length ? = 6 m carries two inclined loads ? of 140 kN, as shown in Figure Q2 (a). The beam is made from a T-section and the thickness for the two rectangular members is of 10 mm. All other dimensions are illustrated in Figure Q2 (b). Find the reactions.Draw the shear force diagram and provide the value of maximum shear force.Find the centroid.Note that the origin is set at point O.Find second moment of area.Derive a general formula for the computation of the vertical shear stress distributionthe cutting line A-A.Using formula derived in (e), determine the maximum vertical shear stress, andvertical shear stress in the web at the top web/flange intersection. 100 mmPA-100 mm30°30°A7 L/3L/3L/3_

Answered: Find the reactions. Draw the shear…

Elements Of Electromagnetics

7th Edition

ISBN:9780190698614

Author:Sadiku, Matthew N. O.

Publisher:Sadiku, Matthew N. O.

ChapterMA: Math Assessment

Section: Chapter Questions

Problem 1.1MA

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A simply supported beam of length ? = 6 m carries two inclined loads ? of 140 kN, as shown in Figure Q2 (a). The beam is made from a T-section and the thickness for the two rectangular members is of 10 mm. All other dimensions are illustrated in Figure Q2 (b).

Find the reactions.
Draw the shear force diagram and provide the value of maximum shear force.
Find the centroid.
Note that the origin is set at point O.
Find second moment of area.
Derive a general formula for the computation of the vertical shear stress distribution
the cutting line A-A.
Using formula derived in (e), determine the maximum vertical shear stress, and
vertical shear stress in the web at the top web/flange intersection.

100 mm
P
A
-
100 mm
30°
30°
A
7 L/3
L/3
L/3_

Expert Solution

Step 1

Given,

Length of the beam, $L = 6 m$

Inclined load, $P = 140 k N$

Thickness of rectangular member, $t = 10 m m$

Step 2

a). Reactions;

From equilibrium, taking a moment about A;

$R_{B} \times L - P \sin 30 \times \frac{2 L}{3} - P \sin 30 \times \frac{L}{3} = 0 R_{B} = P \sin 30 = 140 \times \sin 30 R_{B} = 70 k N$

and,

$\sum_{} F_{v} = 0 R_{A} + R_{B} = P \sin 30 + P \sin 30 R_{A} + 70 = 140 \times \sin 30 + 140 \times \sin 30 R_{A} = 70 k N$

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