ALL 15/ A square lead slab of side 50 cm and thickness 10 cm is subjected to a shearing force (on its narrow fa of 9 x 10* N. The lower edge is riveted to the floor. How much will the upper edge be displaced ? (Shear modulus of lead 5.6 x 109 N m2) %3D (4) 1.6 cm (1) 0.16 mm (2) 1.6 mm (3) 0.16 cm Sol. The lead slab is fixed and force is applied parallel to the narrow face as shown in the figure.
ALL 15/ A square lead slab of side 50 cm and thickness 10 cm is subjected to a shearing force (on its narrow fa of 9 x 10* N. The lower edge is riveted to the floor. How much will the upper edge be displaced ? (Shear modulus of lead 5.6 x 109 N m2) %3D (4) 1.6 cm (1) 0.16 mm (2) 1.6 mm (3) 0.16 cm Sol. The lead slab is fixed and force is applied parallel to the narrow face as shown in the figure.
Chapter2: Loads On Structures
Section: Chapter Questions
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![15
ALLA
A square lead slab of side 50 cm and thickness 10 cm is subjected to a shearing force (on its narrow fa
of 9 x 104 N. The lower edge is riveted to the floor. How much will the upper edge be displaced ?
(Shear modulus of lead
5.6 x 109 N m2)
%3D
(1) 0.16 mm
(2) 1.6 mm
(3) 0.16 cm
(4) 1.6 cm
Sol. The lead slab is fixed and force is applied parallel to the narrow face as shown in the figure.
F-
AL
50 cm
Area of the face parallel to which this force is applied is
0.5 m x 0.1 m = 0.05 m?
A =
50 cm x 10 cm =
If AL is the displacement of the upper edge of the slab due to tangential force F, then](/v2/_next/image?url=https%3A%2F%2Fcontent.bartleby.com%2Fqna-images%2Fquestion%2F85ed2e77-7f1f-4661-864c-36fc725a2bfb%2Fe449c484-1514-434d-9185-ebb694a31949%2F890dyxp_processed.jpeg&w=3840&q=75)
Transcribed Image Text:15
ALLA
A square lead slab of side 50 cm and thickness 10 cm is subjected to a shearing force (on its narrow fa
of 9 x 104 N. The lower edge is riveted to the floor. How much will the upper edge be displaced ?
(Shear modulus of lead
5.6 x 109 N m2)
%3D
(1) 0.16 mm
(2) 1.6 mm
(3) 0.16 cm
(4) 1.6 cm
Sol. The lead slab is fixed and force is applied parallel to the narrow face as shown in the figure.
F-
AL
50 cm
Area of the face parallel to which this force is applied is
0.5 m x 0.1 m = 0.05 m?
A =
50 cm x 10 cm =
If AL is the displacement of the upper edge of the slab due to tangential force F, then
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